Termination w.r.t. Q proof of ../tpdb/TRS/Beerendonk/24.trs

YES Termination w.r.t. Q proof of ../tpdb/TRS/Beerendonk/24.trs
Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond3(gr(x, 0), x, y)
cond2(false, x, y) → cond4(gr(y, 0), x, y)
cond3(true, x, y) → cond3(gr(x, 0), p(x), y)
cond3(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
cond4(true, x, y) → cond4(gr(y, 0), x, p(y))
cond4(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

Q is empty.

↳ QTRS

  ↳ AAECC Innermost

Q restricted rewrite system:
The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond3(gr(x, 0), x, y)
cond2(false, x, y) → cond4(gr(y, 0), x, y)
cond3(true, x, y) → cond3(gr(x, 0), p(x), y)
cond3(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
cond4(true, x, y) → cond4(gr(y, 0), x, p(y))
cond4(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

Q is empty.

We have applied [19,8] to switch to innermost. The TRS R 1 is

gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

The TRS R 2 is

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond3(gr(x, 0), x, y)
cond2(false, x, y) → cond4(gr(y, 0), x, y)
cond3(true, x, y) → cond3(gr(x, 0), p(x), y)
cond3(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
cond4(true, x, y) → cond4(gr(y, 0), x, p(y))
cond4(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)

The signature Sigma is {cond3, cond2, cond4, cond1}

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond3(gr(x, 0), x, y)
cond2(false, x, y) → cond4(gr(y, 0), x, y)
cond3(true, x, y) → cond3(gr(x, 0), p(x), y)
cond3(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
cond4(true, x, y) → cond4(gr(y, 0), x, p(y))
cond4(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
cond4(true, x0, x1)
cond4(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
p(0)
p(s(x0))

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

COND4(true, x, y) → P(y)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND3(false, x, y) → AND(gr(x, 0), gr(y, 0))
COND4(false, x, y) → GR(y, 0)
COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND3(true, x, y) → P(x)
COND4(false, x, y) → GR(x, 0)
COND2(false, x, y) → COND4(gr(y, 0), x, y)
COND4(false, x, y) → AND(gr(x, 0), gr(y, 0))
COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND2(true, x, y) → COND3(gr(x, 0), x, y)
COND3(false, x, y) → GR(x, 0)
COND4(true, x, y) → GR(y, 0)
GR(s(x), s(y)) → GR(x, y)
COND2(true, x, y) → GR(x, 0)
COND1(true, x, y) → GR(x, y)
COND2(false, x, y) → GR(y, 0)
COND3(true, x, y) → GR(x, 0)
COND3(false, x, y) → GR(y, 0)
COND4(true, x, y) → COND4(gr(y, 0), x, p(y))
COND3(true, x, y) → COND3(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond3(gr(x, 0), x, y)
cond2(false, x, y) → cond4(gr(y, 0), x, y)
cond3(true, x, y) → cond3(gr(x, 0), p(x), y)
cond3(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
cond4(true, x, y) → cond4(gr(y, 0), x, p(y))
cond4(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
cond4(true, x0, x1)
cond4(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

COND4(true, x, y) → P(y)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND3(false, x, y) → AND(gr(x, 0), gr(y, 0))
COND4(false, x, y) → GR(y, 0)
COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND3(true, x, y) → P(x)
COND4(false, x, y) → GR(x, 0)
COND2(false, x, y) → COND4(gr(y, 0), x, y)
COND4(false, x, y) → AND(gr(x, 0), gr(y, 0))
COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND2(true, x, y) → COND3(gr(x, 0), x, y)
COND3(false, x, y) → GR(x, 0)
COND4(true, x, y) → GR(y, 0)
GR(s(x), s(y)) → GR(x, y)
COND2(true, x, y) → GR(x, 0)
COND1(true, x, y) → GR(x, y)
COND2(false, x, y) → GR(y, 0)
COND3(true, x, y) → GR(x, 0)
COND3(false, x, y) → GR(y, 0)
COND4(true, x, y) → COND4(gr(y, 0), x, p(y))
COND3(true, x, y) → COND3(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond3(gr(x, 0), x, y)
cond2(false, x, y) → cond4(gr(y, 0), x, y)
cond3(true, x, y) → cond3(gr(x, 0), p(x), y)
cond3(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
cond4(true, x, y) → cond4(gr(y, 0), x, p(y))
cond4(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
cond4(true, x0, x1)
cond4(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 13 less nodes.

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

GR(s(x), s(y)) → GR(x, y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond3(gr(x, 0), x, y)
cond2(false, x, y) → cond4(gr(y, 0), x, y)
cond3(true, x, y) → cond3(gr(x, 0), p(x), y)
cond3(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
cond4(true, x, y) → cond4(gr(y, 0), x, p(y))
cond4(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
cond4(true, x0, x1)
cond4(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

GR(s(x), s(y)) → GR(x, y)

R is empty.
The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
cond4(true, x0, x1)
cond4(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
cond4(true, x0, x1)
cond4(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
p(0)
p(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

GR(s(x), s(y)) → GR(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

GR(s(x), s(y)) → GR(x, y)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

Q DP problem:
The TRS P consists of the following rules:

COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND2(true, x, y) → COND3(gr(x, 0), x, y)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND4(true, x, y) → COND4(gr(y, 0), x, p(y))
COND2(false, x, y) → COND4(gr(y, 0), x, y)
COND3(true, x, y) → COND3(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond3(gr(x, 0), x, y)
cond2(false, x, y) → cond4(gr(y, 0), x, y)
cond3(true, x, y) → cond3(gr(x, 0), p(x), y)
cond3(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
cond4(true, x, y) → cond4(gr(y, 0), x, p(y))
cond4(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
cond4(true, x0, x1)
cond4(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
p(0)
p(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

Q DP problem:
The TRS P consists of the following rules:

COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND2(true, x, y) → COND3(gr(x, 0), x, y)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND4(true, x, y) → COND4(gr(y, 0), x, p(y))
COND2(false, x, y) → COND4(gr(y, 0), x, y)
COND3(true, x, y) → COND3(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
cond4(true, x0, x1)
cond4(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
cond4(true, x0, x1)
cond4(false, x0, x1)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

Q DP problem:
The TRS P consists of the following rules:

COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND2(true, x, y) → COND3(gr(x, 0), x, y)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND3(true, x, y) → COND3(gr(x, 0), p(x), y)
COND2(false, x, y) → COND4(gr(y, 0), x, y)
COND4(true, x, y) → COND4(gr(y, 0), x, p(y))

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
The DP Problem is simplified using the Induction Calculus [18] with the following steps:
Note that final constraints are written in bold face.

For Pair COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y) the following chains were created:

We consider the chain COND1(true, x, y) → COND2(gr(x, y), x, y), COND2(true, x, y) → COND3(gr(x, 0), x, y), COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y) which results in the following constraint:
(1)    (COND2(gr(x22, x23), x22, x23)=COND2(true, x24, x25)∧COND3(gr(x24, 0), x24, x25)=COND3(false, x26, x27) ⇒ COND3(false, x26, x27)≥COND1(and(gr(x26, 0), gr(x27, 0)), x26, x27))

We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint:
(2)    (gr(x24, x23)=true∧0=x894∧gr(x24, x894)=false ⇒ COND3(false, x24, x23)≥COND1(and(gr(x24, 0), gr(x23, 0)), x24, x23))

We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gr(x24, x23)=true which results in the following new constraints:
(3)    (true=true∧0=x894∧gr(s(x896), x894)=false ⇒ COND3(false, s(x896), 0)≥COND1(and(gr(s(x896), 0), gr(0, 0)), s(x896), 0))

(4)    (gr(x897, x898)=true∧0=x894∧gr(s(x897), x894)=false∧(∀x899:gr(x897, x898)=true∧0=x899∧gr(x897, x899)=false ⇒ COND3(false, x897, x898)≥COND1(and(gr(x897, 0), gr(x898, 0)), x897, x898)) ⇒ COND3(false, s(x897), s(x898))≥COND1(and(gr(s(x897), 0), gr(s(x898), 0)), s(x897), s(x898)))

We simplified constraint (3) using rules (I), (II), (VII) which results in the following new constraint:
(5)    (0=x894∧s(x896)=x900∧gr(x900, x894)=false ⇒ COND3(false, s(x896), 0)≥COND1(and(gr(s(x896), 0), gr(0, 0)), s(x896), 0))

We simplified constraint (4) using rule (VII) which results in the following new constraint:
(6)    (gr(x897, x898)=true∧0=x894∧s(x897)=x906∧gr(x906, x894)=false∧(∀x899:gr(x897, x898)=true∧0=x899∧gr(x897, x899)=false ⇒ COND3(false, x897, x898)≥COND1(and(gr(x897, 0), gr(x898, 0)), x897, x898)) ⇒ COND3(false, s(x897), s(x898))≥COND1(and(gr(s(x897), 0), gr(s(x898), 0)), s(x897), s(x898)))

We simplified constraint (5) using rule (V) (with possible (I) afterwards) using induction on gr(x900, x894)=false which results in the following new constraints:
(7)    (false=false∧0=x901∧s(x896)=0 ⇒ COND3(false, s(x896), 0)≥COND1(and(gr(s(x896), 0), gr(0, 0)), s(x896), 0))

(8)    (gr(x903, x904)=false∧0=s(x904)∧s(x896)=s(x903)∧(∀x905:gr(x903, x904)=false∧0=x904∧s(x905)=x903 ⇒ COND3(false, s(x905), 0)≥COND1(and(gr(s(x905), 0), gr(0, 0)), s(x905), 0)) ⇒ COND3(false, s(x896), 0)≥COND1(and(gr(s(x896), 0), gr(0, 0)), s(x896), 0))

We solved constraint (7) using rules (I), (II).We solved constraint (8) using rules (I), (II).We simplified constraint (6) using rule (V) (with possible (I) afterwards) using induction on gr(x906, x894)=false which results in the following new constraints:
(9)    (false=false∧gr(x897, x898)=true∧0=x907∧s(x897)=0∧(∀x899:gr(x897, x898)=true∧0=x899∧gr(x897, x899)=false ⇒ COND3(false, x897, x898)≥COND1(and(gr(x897, 0), gr(x898, 0)), x897, x898)) ⇒ COND3(false, s(x897), s(x898))≥COND1(and(gr(s(x897), 0), gr(s(x898), 0)), s(x897), s(x898)))

(10)    (gr(x909, x910)=false∧gr(x897, x898)=true∧0=s(x910)∧s(x897)=s(x909)∧(∀x899:gr(x897, x898)=true∧0=x899∧gr(x897, x899)=false ⇒ COND3(false, x897, x898)≥COND1(and(gr(x897, 0), gr(x898, 0)), x897, x898))∧(∀x911,x912,x913:gr(x909, x910)=false∧gr(x911, x912)=true∧0=x910∧s(x911)=x909∧(∀x913:gr(x911, x912)=true∧0=x913∧gr(x911, x913)=false ⇒ COND3(false, x911, x912)≥COND1(and(gr(x911, 0), gr(x912, 0)), x911, x912)) ⇒ COND3(false, s(x911), s(x912))≥COND1(and(gr(s(x911), 0), gr(s(x912), 0)), s(x911), s(x912))) ⇒ COND3(false, s(x897), s(x898))≥COND1(and(gr(s(x897), 0), gr(s(x898), 0)), s(x897), s(x898)))

We solved constraint (9) using rules (I), (II).We solved constraint (10) using rules (I), (II).
We consider the chain COND2(true, x, y) → COND3(gr(x, 0), x, y), COND3(true, x, y) → COND3(gr(x, 0), p(x), y), COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y) which results in the following constraint:
(11)    (COND3(gr(x108, 0), x108, x109)=COND3(true, x110, x111)∧COND3(gr(x110, 0), p(x110), x111)=COND3(false, x112, x113) ⇒ COND3(false, x112, x113)≥COND1(and(gr(x112, 0), gr(x113, 0)), x112, x113))

We simplified constraint (11) using rules (I), (II), (III), (VII) which results in the following new constraint:
(12)    (0=x914∧gr(x110, x914)=true∧0=x915∧gr(x110, x915)=false∧p(x110)=x112 ⇒ COND3(false, x112, x109)≥COND1(and(gr(x112, 0), gr(x109, 0)), x112, x109))

We simplified constraint (12) using rule (V) (with possible (I) afterwards) using induction on gr(x110, x914)=true which results in the following new constraints:
(13)    (true=true∧0=0∧0=x915∧gr(s(x917), x915)=false∧p(s(x917))=x112 ⇒ COND3(false, x112, x109)≥COND1(and(gr(x112, 0), gr(x109, 0)), x112, x109))

(14)    (gr(x918, x919)=true∧0=s(x919)∧0=x915∧gr(s(x918), x915)=false∧p(s(x918))=x112∧(∀x920,x921,x922:gr(x918, x919)=true∧0=x919∧0=x920∧gr(x918, x920)=false∧p(x918)=x921 ⇒ COND3(false, x921, x922)≥COND1(and(gr(x921, 0), gr(x922, 0)), x921, x922)) ⇒ COND3(false, x112, x109)≥COND1(and(gr(x112, 0), gr(x109, 0)), x112, x109))

We simplified constraint (13) using rules (I), (II), (VII) which results in the following new constraint:
(15)    (0=x915∧s(x917)=x923∧gr(x923, x915)=false∧s(x917)=x924∧p(x924)=x112 ⇒ COND3(false, x112, x109)≥COND1(and(gr(x112, 0), gr(x109, 0)), x112, x109))

We solved constraint (14) using rules (I), (II).We simplified constraint (15) using rule (V) (with possible (I) afterwards) using induction on gr(x923, x915)=false which results in the following new constraints:
(16)    (false=false∧0=x925∧s(x917)=0∧s(x917)=x924∧p(x924)=x112 ⇒ COND3(false, x112, x109)≥COND1(and(gr(x112, 0), gr(x109, 0)), x112, x109))

(17)    (gr(x927, x928)=false∧0=s(x928)∧s(x917)=s(x927)∧s(x917)=x924∧p(x924)=x112∧(∀x929,x930,x931,x932:gr(x927, x928)=false∧0=x928∧s(x929)=x927∧s(x929)=x930∧p(x930)=x931 ⇒ COND3(false, x931, x932)≥COND1(and(gr(x931, 0), gr(x932, 0)), x931, x932)) ⇒ COND3(false, x112, x109)≥COND1(and(gr(x112, 0), gr(x109, 0)), x112, x109))

We solved constraint (16) using rules (I), (II).We solved constraint (17) using rules (I), (II).
We consider the chain COND3(true, x, y) → COND3(gr(x, 0), p(x), y), COND3(true, x, y) → COND3(gr(x, 0), p(x), y), COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y) which results in the following constraint:
(18)    (COND3(gr(x122, 0), p(x122), x123)=COND3(true, x124, x125)∧COND3(gr(x124, 0), p(x124), x125)=COND3(false, x126, x127) ⇒ COND3(false, x126, x127)≥COND1(and(gr(x126, 0), gr(x127, 0)), x126, x127))

We simplified constraint (18) using rules (I), (II), (III), (VII) which results in the following new constraint:
(19)    (0=x933∧gr(x122, x933)=true∧p(x122)=x124∧0=x934∧gr(x124, x934)=false∧p(x124)=x126 ⇒ COND3(false, x126, x123)≥COND1(and(gr(x126, 0), gr(x123, 0)), x126, x123))

We simplified constraint (19) using rule (V) (with possible (I) afterwards) using induction on gr(x122, x933)=true which results in the following new constraints:
(20)    (true=true∧0=0∧p(s(x936))=x124∧0=x934∧gr(x124, x934)=false∧p(x124)=x126 ⇒ COND3(false, x126, x123)≥COND1(and(gr(x126, 0), gr(x123, 0)), x126, x123))

(21)    (gr(x937, x938)=true∧0=s(x938)∧p(s(x937))=x124∧0=x934∧gr(x124, x934)=false∧p(x124)=x126∧(∀x939,x940,x941,x942:gr(x937, x938)=true∧0=x938∧p(x937)=x939∧0=x940∧gr(x939, x940)=false∧p(x939)=x941 ⇒ COND3(false, x941, x942)≥COND1(and(gr(x941, 0), gr(x942, 0)), x941, x942)) ⇒ COND3(false, x126, x123)≥COND1(and(gr(x126, 0), gr(x123, 0)), x126, x123))

We simplified constraint (20) using rules (I), (II), (VII) which results in the following new constraint:
(22)    (s(x936)=x943∧p(x943)=x124∧0=x934∧gr(x124, x934)=false∧p(x124)=x126 ⇒ COND3(false, x126, x123)≥COND1(and(gr(x126, 0), gr(x123, 0)), x126, x123))

We solved constraint (21) using rules (I), (II).We simplified constraint (22) using rule (V) (with possible (I) afterwards) using induction on gr(x124, x934)=false which results in the following new constraints:
(23)    (false=false∧s(x936)=x943∧p(x943)=0∧0=x944∧p(0)=x126 ⇒ COND3(false, x126, x123)≥COND1(and(gr(x126, 0), gr(x123, 0)), x126, x123))

(24)    (gr(x946, x947)=false∧s(x936)=x943∧p(x943)=s(x946)∧0=s(x947)∧p(s(x946))=x126∧(∀x948,x949,x950,x951:gr(x946, x947)=false∧s(x948)=x949∧p(x949)=x946∧0=x947∧p(x946)=x950 ⇒ COND3(false, x950, x951)≥COND1(and(gr(x950, 0), gr(x951, 0)), x950, x951)) ⇒ COND3(false, x126, x123)≥COND1(and(gr(x126, 0), gr(x123, 0)), x126, x123))

We simplified constraint (23) using rules (I), (II), (IV), (VII) which results in the following new constraint:
(25)    (s(x936)=x943∧p(x943)=0∧0=x952∧p(x952)=x126 ⇒ COND3(false, x126, x123)≥COND1(and(gr(x126, 0), gr(x123, 0)), x126, x123))

We solved constraint (24) using rules (I), (II).We simplified constraint (25) using rule (V) (with possible (I) afterwards) using induction on p(x943)=0 which results in the following new constraints:
(26)    (0=0∧s(x936)=0∧0=x952∧p(x952)=x126 ⇒ COND3(false, x126, x123)≥COND1(and(gr(x126, 0), gr(x123, 0)), x126, x123))

(27)    (x953=0∧s(x936)=s(x953)∧0=x952∧p(x952)=x126 ⇒ COND3(false, x126, x123)≥COND1(and(gr(x126, 0), gr(x123, 0)), x126, x123))

We solved constraint (26) using rules (I), (II).We simplified constraint (27) using rules (I), (II), (III), (IV) which results in the following new constraint:
(28)    (COND3(false, x126, x123)≥COND1(and(gr(x126, 0), gr(x123, 0)), x126, x123))

For Pair COND2(true, x, y) → COND3(gr(x, 0), x, y) the following chains were created:

We consider the chain COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y), COND1(true, x, y) → COND2(gr(x, y), x, y), COND2(true, x, y) → COND3(gr(x, 0), x, y) which results in the following constraint:
(29)    (COND1(and(gr(x162, 0), gr(x163, 0)), x162, x163)=COND1(true, x164, x165)∧COND2(gr(x164, x165), x164, x165)=COND2(true, x166, x167) ⇒ COND2(true, x166, x167)≥COND3(gr(x166, 0), x166, x167))

We simplified constraint (29) using rules (I), (II), (III), (VII) which results in the following new constraint:
(30)    (0=x956∧gr(x164, x956)=x954∧0=x957∧gr(x165, x957)=x955∧and(x954, x955)=true∧gr(x164, x165)=true ⇒ COND2(true, x164, x165)≥COND3(gr(x164, 0), x164, x165))

We simplified constraint (30) using rule (V) (with possible (I) afterwards) using induction on and(x954, x955)=true which results in the following new constraint:
(31)    (true=true∧0=x956∧gr(x164, x956)=true∧0=x957∧gr(x165, x957)=true∧gr(x164, x165)=true ⇒ COND2(true, x164, x165)≥COND3(gr(x164, 0), x164, x165))

We simplified constraint (31) using rules (I), (II) which results in the following new constraint:
(32)    (0=x956∧gr(x164, x956)=true∧0=x957∧gr(x165, x957)=true∧gr(x164, x165)=true ⇒ COND2(true, x164, x165)≥COND3(gr(x164, 0), x164, x165))

We simplified constraint (32) using rule (V) (with possible (I) afterwards) using induction on gr(x164, x165)=true which results in the following new constraints:
(33)    (true=true∧0=x956∧gr(s(x961), x956)=true∧0=x957∧gr(0, x957)=true ⇒ COND2(true, s(x961), 0)≥COND3(gr(s(x961), 0), s(x961), 0))

(34)    (gr(x962, x963)=true∧0=x956∧gr(s(x962), x956)=true∧0=x957∧gr(s(x963), x957)=true∧(∀x964,x965:gr(x962, x963)=true∧0=x964∧gr(x962, x964)=true∧0=x965∧gr(x963, x965)=true ⇒ COND2(true, x962, x963)≥COND3(gr(x962, 0), x962, x963)) ⇒ COND2(true, s(x962), s(x963))≥COND3(gr(s(x962), 0), s(x962), s(x963)))

We simplified constraint (33) using rules (I), (II), (VII) which results in the following new constraint:
(35)    (0=x956∧s(x961)=x966∧gr(x966, x956)=true∧0=x957∧0=x967∧gr(x967, x957)=true ⇒ COND2(true, s(x961), 0)≥COND3(gr(s(x961), 0), s(x961), 0))

We simplified constraint (34) using rule (VII) which results in the following new constraint:
(36)    (gr(x962, x963)=true∧0=x956∧s(x962)=x980∧gr(x980, x956)=true∧0=x957∧s(x963)=x981∧gr(x981, x957)=true∧(∀x964,x965:gr(x962, x963)=true∧0=x964∧gr(x962, x964)=true∧0=x965∧gr(x963, x965)=true ⇒ COND2(true, x962, x963)≥COND3(gr(x962, 0), x962, x963)) ⇒ COND2(true, s(x962), s(x963))≥COND3(gr(s(x962), 0), s(x962), s(x963)))

We simplified constraint (35) using rule (V) (with possible (I) afterwards) using induction on gr(x966, x956)=true which results in the following new constraints:
(37)    (true=true∧0=0∧s(x961)=s(x969)∧0=x957∧0=x967∧gr(x967, x957)=true ⇒ COND2(true, s(x961), 0)≥COND3(gr(s(x961), 0), s(x961), 0))

(38)    (gr(x970, x971)=true∧0=s(x971)∧s(x961)=s(x970)∧0=x957∧0=x967∧gr(x967, x957)=true∧(∀x972,x973,x974:gr(x970, x971)=true∧0=x971∧s(x972)=x970∧0=x973∧0=x974∧gr(x974, x973)=true ⇒ COND2(true, s(x972), 0)≥COND3(gr(s(x972), 0), s(x972), 0)) ⇒ COND2(true, s(x961), 0)≥COND3(gr(s(x961), 0), s(x961), 0))

We simplified constraint (37) using rules (I), (II), (IV) which results in the following new constraint:
(39)    (0=x957∧0=x967∧gr(x967, x957)=true ⇒ COND2(true, s(x961), 0)≥COND3(gr(s(x961), 0), s(x961), 0))

We solved constraint (38) using rules (I), (II).We simplified constraint (39) using rule (V) (with possible (I) afterwards) using induction on gr(x967, x957)=true which results in the following new constraints:
(40)    (true=true∧0=0∧0=s(x976) ⇒ COND2(true, s(x961), 0)≥COND3(gr(s(x961), 0), s(x961), 0))

(41)    (gr(x977, x978)=true∧0=s(x978)∧0=s(x977)∧(∀x979:gr(x977, x978)=true∧0=x978∧0=x977 ⇒ COND2(true, s(x979), 0)≥COND3(gr(s(x979), 0), s(x979), 0)) ⇒ COND2(true, s(x961), 0)≥COND3(gr(s(x961), 0), s(x961), 0))

We solved constraint (40) using rules (I), (II).We solved constraint (41) using rules (I), (II).We simplified constraint (36) using rule (V) (with possible (I) afterwards) using induction on gr(x980, x956)=true which results in the following new constraints:
(42)    (true=true∧gr(x962, x963)=true∧0=0∧s(x962)=s(x983)∧0=x957∧s(x963)=x981∧gr(x981, x957)=true∧(∀x964,x965:gr(x962, x963)=true∧0=x964∧gr(x962, x964)=true∧0=x965∧gr(x963, x965)=true ⇒ COND2(true, x962, x963)≥COND3(gr(x962, 0), x962, x963)) ⇒ COND2(true, s(x962), s(x963))≥COND3(gr(s(x962), 0), s(x962), s(x963)))

(43)    (gr(x984, x985)=true∧gr(x962, x963)=true∧0=s(x985)∧s(x962)=s(x984)∧0=x957∧s(x963)=x981∧gr(x981, x957)=true∧(∀x964,x965:gr(x962, x963)=true∧0=x964∧gr(x962, x964)=true∧0=x965∧gr(x963, x965)=true ⇒ COND2(true, x962, x963)≥COND3(gr(x962, 0), x962, x963))∧(∀x986,x987,x988,x989,x990,x991:gr(x984, x985)=true∧gr(x986, x987)=true∧0=x985∧s(x986)=x984∧0=x988∧s(x987)=x989∧gr(x989, x988)=true∧(∀x990,x991:gr(x986, x987)=true∧0=x990∧gr(x986, x990)=true∧0=x991∧gr(x987, x991)=true ⇒ COND2(true, x986, x987)≥COND3(gr(x986, 0), x986, x987)) ⇒ COND2(true, s(x986), s(x987))≥COND3(gr(s(x986), 0), s(x986), s(x987))) ⇒ COND2(true, s(x962), s(x963))≥COND3(gr(s(x962), 0), s(x962), s(x963)))

We simplified constraint (42) using rules (I), (II), (IV) which results in the following new constraint:
(44)    (gr(x962, x963)=true∧0=x957∧s(x963)=x981∧gr(x981, x957)=true∧(∀x964,x965:gr(x962, x963)=true∧0=x964∧gr(x962, x964)=true∧0=x965∧gr(x963, x965)=true ⇒ COND2(true, x962, x963)≥COND3(gr(x962, 0), x962, x963)) ⇒ COND2(true, s(x962), s(x963))≥COND3(gr(s(x962), 0), s(x962), s(x963)))

We solved constraint (43) using rules (I), (II).We simplified constraint (44) using rule (V) (with possible (I) afterwards) using induction on gr(x981, x957)=true which results in the following new constraints:
(45)    (true=true∧gr(x962, x963)=true∧0=0∧s(x963)=s(x993)∧(∀x964,x965:gr(x962, x963)=true∧0=x964∧gr(x962, x964)=true∧0=x965∧gr(x963, x965)=true ⇒ COND2(true, x962, x963)≥COND3(gr(x962, 0), x962, x963)) ⇒ COND2(true, s(x962), s(x963))≥COND3(gr(s(x962), 0), s(x962), s(x963)))

(46)    (gr(x994, x995)=true∧gr(x962, x963)=true∧0=s(x995)∧s(x963)=s(x994)∧(∀x964,x965:gr(x962, x963)=true∧0=x964∧gr(x962, x964)=true∧0=x965∧gr(x963, x965)=true ⇒ COND2(true, x962, x963)≥COND3(gr(x962, 0), x962, x963))∧(∀x996,x997,x998,x999:gr(x994, x995)=true∧gr(x996, x997)=true∧0=x995∧s(x997)=x994∧(∀x998,x999:gr(x996, x997)=true∧0=x998∧gr(x996, x998)=true∧0=x999∧gr(x997, x999)=true ⇒ COND2(true, x996, x997)≥COND3(gr(x996, 0), x996, x997)) ⇒ COND2(true, s(x996), s(x997))≥COND3(gr(s(x996), 0), s(x996), s(x997))) ⇒ COND2(true, s(x962), s(x963))≥COND3(gr(s(x962), 0), s(x962), s(x963)))

We simplified constraint (45) using rules (I), (II), (IV) which results in the following new constraint:
(47)    (gr(x962, x963)=true ⇒ COND2(true, s(x962), s(x963))≥COND3(gr(s(x962), 0), s(x962), s(x963)))

We solved constraint (46) using rules (I), (II).We simplified constraint (47) using rule (V) (with possible (I) afterwards) using induction on gr(x962, x963)=true which results in the following new constraints:
(48)    (true=true ⇒ COND2(true, s(s(x1001)), s(0))≥COND3(gr(s(s(x1001)), 0), s(s(x1001)), s(0)))

(49)    (gr(x1002, x1003)=true∧(gr(x1002, x1003)=true ⇒ COND2(true, s(x1002), s(x1003))≥COND3(gr(s(x1002), 0), s(x1002), s(x1003))) ⇒ COND2(true, s(s(x1002)), s(s(x1003)))≥COND3(gr(s(s(x1002)), 0), s(s(x1002)), s(s(x1003))))

We simplified constraint (48) using rules (I), (II) which results in the following new constraint:
(50)    (COND2(true, s(s(x1001)), s(0))≥COND3(gr(s(s(x1001)), 0), s(s(x1001)), s(0)))

We simplified constraint (49) using rule (VI) where we applied the induction hypothesis (gr(x1002, x1003)=true ⇒ COND2(true, s(x1002), s(x1003))≥COND3(gr(s(x1002), 0), s(x1002), s(x1003))) with σ = [ ] which results in the following new constraint:
(51)    (COND2(true, s(x1002), s(x1003))≥COND3(gr(s(x1002), 0), s(x1002), s(x1003)) ⇒ COND2(true, s(s(x1002)), s(s(x1003)))≥COND3(gr(s(s(x1002)), 0), s(s(x1002)), s(s(x1003))))
We consider the chain COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y), COND1(true, x, y) → COND2(gr(x, y), x, y), COND2(true, x, y) → COND3(gr(x, 0), x, y) which results in the following constraint:
(52)    (COND1(and(gr(x172, 0), gr(x173, 0)), x172, x173)=COND1(true, x174, x175)∧COND2(gr(x174, x175), x174, x175)=COND2(true, x176, x177) ⇒ COND2(true, x176, x177)≥COND3(gr(x176, 0), x176, x177))

We simplified constraint (52) using rules (I), (II), (III), (VII) which results in the following new constraint:
(53)    (0=x1006∧gr(x174, x1006)=x1004∧0=x1007∧gr(x175, x1007)=x1005∧and(x1004, x1005)=true∧gr(x174, x175)=true ⇒ COND2(true, x174, x175)≥COND3(gr(x174, 0), x174, x175))

We simplified constraint (53) using rule (V) (with possible (I) afterwards) using induction on and(x1004, x1005)=true which results in the following new constraint:
(54)    (true=true∧0=x1006∧gr(x174, x1006)=true∧0=x1007∧gr(x175, x1007)=true∧gr(x174, x175)=true ⇒ COND2(true, x174, x175)≥COND3(gr(x174, 0), x174, x175))

We simplified constraint (54) using rules (I), (II) which results in the following new constraint:
(55)    (0=x1006∧gr(x174, x1006)=true∧0=x1007∧gr(x175, x1007)=true∧gr(x174, x175)=true ⇒ COND2(true, x174, x175)≥COND3(gr(x174, 0), x174, x175))

We simplified constraint (55) using rule (V) (with possible (I) afterwards) using induction on gr(x174, x175)=true which results in the following new constraints:
(56)    (true=true∧0=x1006∧gr(s(x1011), x1006)=true∧0=x1007∧gr(0, x1007)=true ⇒ COND2(true, s(x1011), 0)≥COND3(gr(s(x1011), 0), s(x1011), 0))

(57)    (gr(x1012, x1013)=true∧0=x1006∧gr(s(x1012), x1006)=true∧0=x1007∧gr(s(x1013), x1007)=true∧(∀x1014,x1015:gr(x1012, x1013)=true∧0=x1014∧gr(x1012, x1014)=true∧0=x1015∧gr(x1013, x1015)=true ⇒ COND2(true, x1012, x1013)≥COND3(gr(x1012, 0), x1012, x1013)) ⇒ COND2(true, s(x1012), s(x1013))≥COND3(gr(s(x1012), 0), s(x1012), s(x1013)))

We simplified constraint (56) using rules (I), (II), (VII) which results in the following new constraint:
(58)    (0=x1006∧s(x1011)=x1016∧gr(x1016, x1006)=true∧0=x1007∧0=x1017∧gr(x1017, x1007)=true ⇒ COND2(true, s(x1011), 0)≥COND3(gr(s(x1011), 0), s(x1011), 0))

We simplified constraint (57) using rule (VII) which results in the following new constraint:
(59)    (gr(x1012, x1013)=true∧0=x1006∧s(x1012)=x1030∧gr(x1030, x1006)=true∧0=x1007∧s(x1013)=x1031∧gr(x1031, x1007)=true∧(∀x1014,x1015:gr(x1012, x1013)=true∧0=x1014∧gr(x1012, x1014)=true∧0=x1015∧gr(x1013, x1015)=true ⇒ COND2(true, x1012, x1013)≥COND3(gr(x1012, 0), x1012, x1013)) ⇒ COND2(true, s(x1012), s(x1013))≥COND3(gr(s(x1012), 0), s(x1012), s(x1013)))

We simplified constraint (58) using rule (V) (with possible (I) afterwards) using induction on gr(x1016, x1006)=true which results in the following new constraints:
(60)    (true=true∧0=0∧s(x1011)=s(x1019)∧0=x1007∧0=x1017∧gr(x1017, x1007)=true ⇒ COND2(true, s(x1011), 0)≥COND3(gr(s(x1011), 0), s(x1011), 0))

(61)    (gr(x1020, x1021)=true∧0=s(x1021)∧s(x1011)=s(x1020)∧0=x1007∧0=x1017∧gr(x1017, x1007)=true∧(∀x1022,x1023,x1024:gr(x1020, x1021)=true∧0=x1021∧s(x1022)=x1020∧0=x1023∧0=x1024∧gr(x1024, x1023)=true ⇒ COND2(true, s(x1022), 0)≥COND3(gr(s(x1022), 0), s(x1022), 0)) ⇒ COND2(true, s(x1011), 0)≥COND3(gr(s(x1011), 0), s(x1011), 0))

We simplified constraint (60) using rules (I), (II), (IV) which results in the following new constraint:
(62)    (0=x1007∧0=x1017∧gr(x1017, x1007)=true ⇒ COND2(true, s(x1011), 0)≥COND3(gr(s(x1011), 0), s(x1011), 0))

We solved constraint (61) using rules (I), (II).We simplified constraint (62) using rule (V) (with possible (I) afterwards) using induction on gr(x1017, x1007)=true which results in the following new constraints:
(63)    (true=true∧0=0∧0=s(x1026) ⇒ COND2(true, s(x1011), 0)≥COND3(gr(s(x1011), 0), s(x1011), 0))

(64)    (gr(x1027, x1028)=true∧0=s(x1028)∧0=s(x1027)∧(∀x1029:gr(x1027, x1028)=true∧0=x1028∧0=x1027 ⇒ COND2(true, s(x1029), 0)≥COND3(gr(s(x1029), 0), s(x1029), 0)) ⇒ COND2(true, s(x1011), 0)≥COND3(gr(s(x1011), 0), s(x1011), 0))

We solved constraint (63) using rules (I), (II).We solved constraint (64) using rules (I), (II).We simplified constraint (59) using rule (V) (with possible (I) afterwards) using induction on gr(x1030, x1006)=true which results in the following new constraints:
(65)    (true=true∧gr(x1012, x1013)=true∧0=0∧s(x1012)=s(x1033)∧0=x1007∧s(x1013)=x1031∧gr(x1031, x1007)=true∧(∀x1014,x1015:gr(x1012, x1013)=true∧0=x1014∧gr(x1012, x1014)=true∧0=x1015∧gr(x1013, x1015)=true ⇒ COND2(true, x1012, x1013)≥COND3(gr(x1012, 0), x1012, x1013)) ⇒ COND2(true, s(x1012), s(x1013))≥COND3(gr(s(x1012), 0), s(x1012), s(x1013)))

(66)    (gr(x1034, x1035)=true∧gr(x1012, x1013)=true∧0=s(x1035)∧s(x1012)=s(x1034)∧0=x1007∧s(x1013)=x1031∧gr(x1031, x1007)=true∧(∀x1014,x1015:gr(x1012, x1013)=true∧0=x1014∧gr(x1012, x1014)=true∧0=x1015∧gr(x1013, x1015)=true ⇒ COND2(true, x1012, x1013)≥COND3(gr(x1012, 0), x1012, x1013))∧(∀x1036,x1037,x1038,x1039,x1040,x1041:gr(x1034, x1035)=true∧gr(x1036, x1037)=true∧0=x1035∧s(x1036)=x1034∧0=x1038∧s(x1037)=x1039∧gr(x1039, x1038)=true∧(∀x1040,x1041:gr(x1036, x1037)=true∧0=x1040∧gr(x1036, x1040)=true∧0=x1041∧gr(x1037, x1041)=true ⇒ COND2(true, x1036, x1037)≥COND3(gr(x1036, 0), x1036, x1037)) ⇒ COND2(true, s(x1036), s(x1037))≥COND3(gr(s(x1036), 0), s(x1036), s(x1037))) ⇒ COND2(true, s(x1012), s(x1013))≥COND3(gr(s(x1012), 0), s(x1012), s(x1013)))

We simplified constraint (65) using rules (I), (II), (IV) which results in the following new constraint:
(67)    (gr(x1012, x1013)=true∧0=x1007∧s(x1013)=x1031∧gr(x1031, x1007)=true∧(∀x1014,x1015:gr(x1012, x1013)=true∧0=x1014∧gr(x1012, x1014)=true∧0=x1015∧gr(x1013, x1015)=true ⇒ COND2(true, x1012, x1013)≥COND3(gr(x1012, 0), x1012, x1013)) ⇒ COND2(true, s(x1012), s(x1013))≥COND3(gr(s(x1012), 0), s(x1012), s(x1013)))

We solved constraint (66) using rules (I), (II).We simplified constraint (67) using rule (V) (with possible (I) afterwards) using induction on gr(x1031, x1007)=true which results in the following new constraints:
(68)    (true=true∧gr(x1012, x1013)=true∧0=0∧s(x1013)=s(x1043)∧(∀x1014,x1015:gr(x1012, x1013)=true∧0=x1014∧gr(x1012, x1014)=true∧0=x1015∧gr(x1013, x1015)=true ⇒ COND2(true, x1012, x1013)≥COND3(gr(x1012, 0), x1012, x1013)) ⇒ COND2(true, s(x1012), s(x1013))≥COND3(gr(s(x1012), 0), s(x1012), s(x1013)))

(69)    (gr(x1044, x1045)=true∧gr(x1012, x1013)=true∧0=s(x1045)∧s(x1013)=s(x1044)∧(∀x1014,x1015:gr(x1012, x1013)=true∧0=x1014∧gr(x1012, x1014)=true∧0=x1015∧gr(x1013, x1015)=true ⇒ COND2(true, x1012, x1013)≥COND3(gr(x1012, 0), x1012, x1013))∧(∀x1046,x1047,x1048,x1049:gr(x1044, x1045)=true∧gr(x1046, x1047)=true∧0=x1045∧s(x1047)=x1044∧(∀x1048,x1049:gr(x1046, x1047)=true∧0=x1048∧gr(x1046, x1048)=true∧0=x1049∧gr(x1047, x1049)=true ⇒ COND2(true, x1046, x1047)≥COND3(gr(x1046, 0), x1046, x1047)) ⇒ COND2(true, s(x1046), s(x1047))≥COND3(gr(s(x1046), 0), s(x1046), s(x1047))) ⇒ COND2(true, s(x1012), s(x1013))≥COND3(gr(s(x1012), 0), s(x1012), s(x1013)))

We simplified constraint (68) using rules (I), (II), (IV) which results in the following new constraint:
(70)    (gr(x1012, x1013)=true ⇒ COND2(true, s(x1012), s(x1013))≥COND3(gr(s(x1012), 0), s(x1012), s(x1013)))

We solved constraint (69) using rules (I), (II).We simplified constraint (70) using rule (V) (with possible (I) afterwards) using induction on gr(x1012, x1013)=true which results in the following new constraints:
(71)    (true=true ⇒ COND2(true, s(s(x1051)), s(0))≥COND3(gr(s(s(x1051)), 0), s(s(x1051)), s(0)))

(72)    (gr(x1052, x1053)=true∧(gr(x1052, x1053)=true ⇒ COND2(true, s(x1052), s(x1053))≥COND3(gr(s(x1052), 0), s(x1052), s(x1053))) ⇒ COND2(true, s(s(x1052)), s(s(x1053)))≥COND3(gr(s(s(x1052)), 0), s(s(x1052)), s(s(x1053))))

We simplified constraint (71) using rules (I), (II) which results in the following new constraint:
(73)    (COND2(true, s(s(x1051)), s(0))≥COND3(gr(s(s(x1051)), 0), s(s(x1051)), s(0)))

We simplified constraint (72) using rule (VI) where we applied the induction hypothesis (gr(x1052, x1053)=true ⇒ COND2(true, s(x1052), s(x1053))≥COND3(gr(s(x1052), 0), s(x1052), s(x1053))) with σ = [ ] which results in the following new constraint:
(74)    (COND2(true, s(x1052), s(x1053))≥COND3(gr(s(x1052), 0), s(x1052), s(x1053)) ⇒ COND2(true, s(s(x1052)), s(s(x1053)))≥COND3(gr(s(s(x1052)), 0), s(s(x1052)), s(s(x1053))))

For Pair COND1(true, x, y) → COND2(gr(x, y), x, y) the following chains were created:

We consider the chain COND2(true, x, y) → COND3(gr(x, 0), x, y), COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y), COND1(true, x, y) → COND2(gr(x, y), x, y) which results in the following constraint:
(75)    (COND3(gr(x256, 0), x256, x257)=COND3(false, x258, x259)∧COND1(and(gr(x258, 0), gr(x259, 0)), x258, x259)=COND1(true, x260, x261) ⇒ COND1(true, x260, x261)≥COND2(gr(x260, x261), x260, x261))

We simplified constraint (75) using rules (I), (II), (III), (VII) which results in the following new constraint:
(76)    (0=x1054∧gr(x258, x1054)=false∧0=x1057∧gr(x258, x1057)=x1055∧0=x1058∧gr(x259, x1058)=x1056∧and(x1055, x1056)=true ⇒ COND1(true, x258, x259)≥COND2(gr(x258, x259), x258, x259))

We simplified constraint (76) using rule (V) (with possible (I) afterwards) using induction on and(x1055, x1056)=true which results in the following new constraint:
(77)    (true=true∧0=x1054∧gr(x258, x1054)=false∧0=x1057∧gr(x258, x1057)=true∧0=x1058∧gr(x259, x1058)=true ⇒ COND1(true, x258, x259)≥COND2(gr(x258, x259), x258, x259))

We simplified constraint (77) using rules (I), (II) which results in the following new constraint:
(78)    (0=x1054∧gr(x258, x1054)=false∧0=x1057∧gr(x258, x1057)=true∧0=x1058∧gr(x259, x1058)=true ⇒ COND1(true, x258, x259)≥COND2(gr(x258, x259), x258, x259))

We simplified constraint (78) using rule (V) (with possible (I) afterwards) using induction on gr(x258, x1054)=false which results in the following new constraints:
(79)    (false=false∧0=x1061∧0=x1057∧gr(0, x1057)=true∧0=x1058∧gr(x259, x1058)=true ⇒ COND1(true, 0, x259)≥COND2(gr(0, x259), 0, x259))

(80)    (gr(x1063, x1064)=false∧0=s(x1064)∧0=x1057∧gr(s(x1063), x1057)=true∧0=x1058∧gr(x259, x1058)=true∧(∀x1065,x1066,x1067:gr(x1063, x1064)=false∧0=x1064∧0=x1065∧gr(x1063, x1065)=true∧0=x1066∧gr(x1067, x1066)=true ⇒ COND1(true, x1063, x1067)≥COND2(gr(x1063, x1067), x1063, x1067)) ⇒ COND1(true, s(x1063), x259)≥COND2(gr(s(x1063), x259), s(x1063), x259))

We simplified constraint (79) using rules (I), (II), (IV), (VII) which results in the following new constraint:
(81)    (0=x1057∧0=x1068∧gr(x1068, x1057)=true∧0=x1058∧gr(x259, x1058)=true ⇒ COND1(true, 0, x259)≥COND2(gr(0, x259), 0, x259))

We solved constraint (80) using rules (I), (II).We simplified constraint (81) using rule (V) (with possible (I) afterwards) using induction on gr(x1068, x1057)=true which results in the following new constraints:
(82)    (true=true∧0=0∧0=s(x1070)∧0=x1058∧gr(x259, x1058)=true ⇒ COND1(true, 0, x259)≥COND2(gr(0, x259), 0, x259))

(83)    (gr(x1071, x1072)=true∧0=s(x1072)∧0=s(x1071)∧0=x1058∧gr(x259, x1058)=true∧(∀x1073,x1074:gr(x1071, x1072)=true∧0=x1072∧0=x1071∧0=x1073∧gr(x1074, x1073)=true ⇒ COND1(true, 0, x1074)≥COND2(gr(0, x1074), 0, x1074)) ⇒ COND1(true, 0, x259)≥COND2(gr(0, x259), 0, x259))

We solved constraint (82) using rules (I), (II).We solved constraint (83) using rules (I), (II).
We consider the chain COND3(true, x, y) → COND3(gr(x, 0), p(x), y), COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y), COND1(true, x, y) → COND2(gr(x, y), x, y) which results in the following constraint:
(84)    (COND3(gr(x270, 0), p(x270), x271)=COND3(false, x272, x273)∧COND1(and(gr(x272, 0), gr(x273, 0)), x272, x273)=COND1(true, x274, x275) ⇒ COND1(true, x274, x275)≥COND2(gr(x274, x275), x274, x275))

We simplified constraint (84) using rules (I), (II), (III), (VII) which results in the following new constraint:
(85)    (0=x1075∧gr(x270, x1075)=false∧p(x270)=x272∧0=x1078∧gr(x272, x1078)=x1076∧0=x1079∧gr(x273, x1079)=x1077∧and(x1076, x1077)=true ⇒ COND1(true, x272, x273)≥COND2(gr(x272, x273), x272, x273))

We simplified constraint (85) using rule (V) (with possible (I) afterwards) using induction on and(x1076, x1077)=true which results in the following new constraint:
(86)    (true=true∧0=x1075∧gr(x270, x1075)=false∧p(x270)=x272∧0=x1078∧gr(x272, x1078)=true∧0=x1079∧gr(x273, x1079)=true ⇒ COND1(true, x272, x273)≥COND2(gr(x272, x273), x272, x273))

We simplified constraint (86) using rules (I), (II) which results in the following new constraint:
(87)    (0=x1075∧gr(x270, x1075)=false∧p(x270)=x272∧0=x1078∧gr(x272, x1078)=true∧0=x1079∧gr(x273, x1079)=true ⇒ COND1(true, x272, x273)≥COND2(gr(x272, x273), x272, x273))

We simplified constraint (87) using rule (V) (with possible (I) afterwards) using induction on gr(x270, x1075)=false which results in the following new constraints:
(88)    (false=false∧0=x1082∧p(0)=x272∧0=x1078∧gr(x272, x1078)=true∧0=x1079∧gr(x273, x1079)=true ⇒ COND1(true, x272, x273)≥COND2(gr(x272, x273), x272, x273))

(89)    (gr(x1084, x1085)=false∧0=s(x1085)∧p(s(x1084))=x272∧0=x1078∧gr(x272, x1078)=true∧0=x1079∧gr(x273, x1079)=true∧(∀x1086,x1087,x1088,x1089:gr(x1084, x1085)=false∧0=x1085∧p(x1084)=x1086∧0=x1087∧gr(x1086, x1087)=true∧0=x1088∧gr(x1089, x1088)=true ⇒ COND1(true, x1086, x1089)≥COND2(gr(x1086, x1089), x1086, x1089)) ⇒ COND1(true, x272, x273)≥COND2(gr(x272, x273), x272, x273))

We simplified constraint (88) using rules (I), (II), (IV), (VII) which results in the following new constraint:
(90)    (0=x1090∧p(x1090)=x272∧0=x1078∧gr(x272, x1078)=true∧0=x1079∧gr(x273, x1079)=true ⇒ COND1(true, x272, x273)≥COND2(gr(x272, x273), x272, x273))

We solved constraint (89) using rules (I), (II).We simplified constraint (90) using rule (V) (with possible (I) afterwards) using induction on gr(x272, x1078)=true which results in the following new constraints:
(91)    (true=true∧0=x1090∧p(x1090)=s(x1092)∧0=0∧0=x1079∧gr(x273, x1079)=true ⇒ COND1(true, s(x1092), x273)≥COND2(gr(s(x1092), x273), s(x1092), x273))

(92)    (gr(x1093, x1094)=true∧0=x1090∧p(x1090)=s(x1093)∧0=s(x1094)∧0=x1079∧gr(x273, x1079)=true∧(∀x1095,x1096,x1097:gr(x1093, x1094)=true∧0=x1095∧p(x1095)=x1093∧0=x1094∧0=x1096∧gr(x1097, x1096)=true ⇒ COND1(true, x1093, x1097)≥COND2(gr(x1093, x1097), x1093, x1097)) ⇒ COND1(true, s(x1093), x273)≥COND2(gr(s(x1093), x273), s(x1093), x273))

We simplified constraint (91) using rules (I), (II) which results in the following new constraint:
(93)    (0=x1090∧p(x1090)=s(x1092)∧0=x1079∧gr(x273, x1079)=true ⇒ COND1(true, s(x1092), x273)≥COND2(gr(s(x1092), x273), s(x1092), x273))

We solved constraint (92) using rules (I), (II).We simplified constraint (93) using rule (V) (with possible (I) afterwards) using induction on p(x1090)=s(x1092) which results in the following new constraint:
(94)    (x1098=s(x1092)∧0=s(x1098)∧0=x1079∧gr(x273, x1079)=true ⇒ COND1(true, s(x1092), x273)≥COND2(gr(s(x1092), x273), s(x1092), x273))

We solved constraint (94) using rules (I), (II).
We consider the chain COND4(true, x, y) → COND4(gr(y, 0), x, p(y)), COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y), COND1(true, x, y) → COND2(gr(x, y), x, y) which results in the following constraint:
(95)    (COND4(gr(x319, 0), x318, p(x319))=COND4(false, x320, x321)∧COND1(and(gr(x320, 0), gr(x321, 0)), x320, x321)=COND1(true, x322, x323) ⇒ COND1(true, x322, x323)≥COND2(gr(x322, x323), x322, x323))

We simplified constraint (95) using rules (I), (II), (III), (VII) which results in the following new constraint:
(96)    (0=x1099∧gr(x319, x1099)=false∧p(x319)=x321∧0=x1102∧gr(x320, x1102)=x1100∧0=x1103∧gr(x321, x1103)=x1101∧and(x1100, x1101)=true ⇒ COND1(true, x320, x321)≥COND2(gr(x320, x321), x320, x321))

We simplified constraint (96) using rule (V) (with possible (I) afterwards) using induction on and(x1100, x1101)=true which results in the following new constraint:
(97)    (true=true∧0=x1099∧gr(x319, x1099)=false∧p(x319)=x321∧0=x1102∧gr(x320, x1102)=true∧0=x1103∧gr(x321, x1103)=true ⇒ COND1(true, x320, x321)≥COND2(gr(x320, x321), x320, x321))

We simplified constraint (97) using rules (I), (II) which results in the following new constraint:
(98)    (0=x1099∧gr(x319, x1099)=false∧p(x319)=x321∧0=x1102∧gr(x320, x1102)=true∧0=x1103∧gr(x321, x1103)=true ⇒ COND1(true, x320, x321)≥COND2(gr(x320, x321), x320, x321))

We simplified constraint (98) using rule (V) (with possible (I) afterwards) using induction on gr(x319, x1099)=false which results in the following new constraints:
(99)    (false=false∧0=x1106∧p(0)=x321∧0=x1102∧gr(x320, x1102)=true∧0=x1103∧gr(x321, x1103)=true ⇒ COND1(true, x320, x321)≥COND2(gr(x320, x321), x320, x321))

(100)    (gr(x1108, x1109)=false∧0=s(x1109)∧p(s(x1108))=x321∧0=x1102∧gr(x320, x1102)=true∧0=x1103∧gr(x321, x1103)=true∧(∀x1110,x1111,x1112,x1113:gr(x1108, x1109)=false∧0=x1109∧p(x1108)=x1110∧0=x1111∧gr(x1112, x1111)=true∧0=x1113∧gr(x1110, x1113)=true ⇒ COND1(true, x1112, x1110)≥COND2(gr(x1112, x1110), x1112, x1110)) ⇒ COND1(true, x320, x321)≥COND2(gr(x320, x321), x320, x321))

We simplified constraint (99) using rules (I), (II), (IV), (VII) which results in the following new constraint:
(101)    (0=x1114∧p(x1114)=x321∧0=x1102∧gr(x320, x1102)=true∧0=x1103∧gr(x321, x1103)=true ⇒ COND1(true, x320, x321)≥COND2(gr(x320, x321), x320, x321))

We solved constraint (100) using rules (I), (II).We simplified constraint (101) using rule (V) (with possible (I) afterwards) using induction on gr(x320, x1102)=true which results in the following new constraints:
(102)    (true=true∧0=x1114∧p(x1114)=x321∧0=0∧0=x1103∧gr(x321, x1103)=true ⇒ COND1(true, s(x1116), x321)≥COND2(gr(s(x1116), x321), s(x1116), x321))

(103)    (gr(x1117, x1118)=true∧0=x1114∧p(x1114)=x321∧0=s(x1118)∧0=x1103∧gr(x321, x1103)=true∧(∀x1119,x1120,x1121:gr(x1117, x1118)=true∧0=x1119∧p(x1119)=x1120∧0=x1118∧0=x1121∧gr(x1120, x1121)=true ⇒ COND1(true, x1117, x1120)≥COND2(gr(x1117, x1120), x1117, x1120)) ⇒ COND1(true, s(x1117), x321)≥COND2(gr(s(x1117), x321), s(x1117), x321))

We simplified constraint (102) using rules (I), (II) which results in the following new constraint:
(104)    (0=x1114∧p(x1114)=x321∧0=x1103∧gr(x321, x1103)=true ⇒ COND1(true, s(x1116), x321)≥COND2(gr(s(x1116), x321), s(x1116), x321))

We solved constraint (103) using rules (I), (II).We simplified constraint (104) using rule (V) (with possible (I) afterwards) using induction on gr(x321, x1103)=true which results in the following new constraints:
(105)    (true=true∧0=x1114∧p(x1114)=s(x1123)∧0=0 ⇒ COND1(true, s(x1116), s(x1123))≥COND2(gr(s(x1116), s(x1123)), s(x1116), s(x1123)))

(106)    (gr(x1124, x1125)=true∧0=x1114∧p(x1114)=s(x1124)∧0=s(x1125)∧(∀x1126,x1127:gr(x1124, x1125)=true∧0=x1126∧p(x1126)=x1124∧0=x1125 ⇒ COND1(true, s(x1127), x1124)≥COND2(gr(s(x1127), x1124), s(x1127), x1124)) ⇒ COND1(true, s(x1116), s(x1124))≥COND2(gr(s(x1116), s(x1124)), s(x1116), s(x1124)))

We simplified constraint (105) using rules (I), (II) which results in the following new constraint:
(107)    (0=x1114∧p(x1114)=s(x1123) ⇒ COND1(true, s(x1116), s(x1123))≥COND2(gr(s(x1116), s(x1123)), s(x1116), s(x1123)))

We solved constraint (106) using rules (I), (II).We simplified constraint (107) using rule (V) (with possible (I) afterwards) using induction on p(x1114)=s(x1123) which results in the following new constraint:
(108)    (x1128=s(x1123)∧0=s(x1128) ⇒ COND1(true, s(x1116), s(x1123))≥COND2(gr(s(x1116), s(x1123)), s(x1116), s(x1123)))

We solved constraint (108) using rules (I), (II).
We consider the chain COND2(false, x, y) → COND4(gr(y, 0), x, y), COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y), COND1(true, x, y) → COND2(gr(x, y), x, y) which results in the following constraint:
(109)    (COND4(gr(x325, 0), x324, x325)=COND4(false, x326, x327)∧COND1(and(gr(x326, 0), gr(x327, 0)), x326, x327)=COND1(true, x328, x329) ⇒ COND1(true, x328, x329)≥COND2(gr(x328, x329), x328, x329))

We simplified constraint (109) using rules (I), (II), (III), (VII) which results in the following new constraint:
(110)    (0=x1129∧gr(x327, x1129)=false∧0=x1132∧gr(x326, x1132)=x1130∧0=x1133∧gr(x327, x1133)=x1131∧and(x1130, x1131)=true ⇒ COND1(true, x326, x327)≥COND2(gr(x326, x327), x326, x327))

We simplified constraint (110) using rule (V) (with possible (I) afterwards) using induction on and(x1130, x1131)=true which results in the following new constraint:
(111)    (true=true∧0=x1129∧gr(x327, x1129)=false∧0=x1132∧gr(x326, x1132)=true∧0=x1133∧gr(x327, x1133)=true ⇒ COND1(true, x326, x327)≥COND2(gr(x326, x327), x326, x327))

We simplified constraint (111) using rules (I), (II) which results in the following new constraint:
(112)    (0=x1129∧gr(x327, x1129)=false∧0=x1132∧gr(x326, x1132)=true∧0=x1133∧gr(x327, x1133)=true ⇒ COND1(true, x326, x327)≥COND2(gr(x326, x327), x326, x327))

We simplified constraint (112) using rule (V) (with possible (I) afterwards) using induction on gr(x327, x1129)=false which results in the following new constraints:
(113)    (false=false∧0=x1136∧0=x1132∧gr(x326, x1132)=true∧0=x1133∧gr(0, x1133)=true ⇒ COND1(true, x326, 0)≥COND2(gr(x326, 0), x326, 0))

(114)    (gr(x1138, x1139)=false∧0=s(x1139)∧0=x1132∧gr(x326, x1132)=true∧0=x1133∧gr(s(x1138), x1133)=true∧(∀x1140,x1141,x1142:gr(x1138, x1139)=false∧0=x1139∧0=x1140∧gr(x1141, x1140)=true∧0=x1142∧gr(x1138, x1142)=true ⇒ COND1(true, x1141, x1138)≥COND2(gr(x1141, x1138), x1141, x1138)) ⇒ COND1(true, x326, s(x1138))≥COND2(gr(x326, s(x1138)), x326, s(x1138)))

We simplified constraint (113) using rules (I), (II), (IV), (VII) which results in the following new constraint:
(115)    (0=x1132∧gr(x326, x1132)=true∧0=x1133∧0=x1143∧gr(x1143, x1133)=true ⇒ COND1(true, x326, 0)≥COND2(gr(x326, 0), x326, 0))

We solved constraint (114) using rules (I), (II).We simplified constraint (115) using rule (V) (with possible (I) afterwards) using induction on gr(x326, x1132)=true which results in the following new constraints:
(116)    (true=true∧0=0∧0=x1133∧0=x1143∧gr(x1143, x1133)=true ⇒ COND1(true, s(x1145), 0)≥COND2(gr(s(x1145), 0), s(x1145), 0))

(117)    (gr(x1146, x1147)=true∧0=s(x1147)∧0=x1133∧0=x1143∧gr(x1143, x1133)=true∧(∀x1148,x1149:gr(x1146, x1147)=true∧0=x1147∧0=x1148∧0=x1149∧gr(x1149, x1148)=true ⇒ COND1(true, x1146, 0)≥COND2(gr(x1146, 0), x1146, 0)) ⇒ COND1(true, s(x1146), 0)≥COND2(gr(s(x1146), 0), s(x1146), 0))

We simplified constraint (116) using rules (I), (II) which results in the following new constraint:
(118)    (0=x1133∧0=x1143∧gr(x1143, x1133)=true ⇒ COND1(true, s(x1145), 0)≥COND2(gr(s(x1145), 0), s(x1145), 0))

We solved constraint (117) using rules (I), (II).We simplified constraint (118) using rule (V) (with possible (I) afterwards) using induction on gr(x1143, x1133)=true which results in the following new constraints:
(119)    (true=true∧0=0∧0=s(x1151) ⇒ COND1(true, s(x1145), 0)≥COND2(gr(s(x1145), 0), s(x1145), 0))

(120)    (gr(x1152, x1153)=true∧0=s(x1153)∧0=s(x1152)∧(∀x1154:gr(x1152, x1153)=true∧0=x1153∧0=x1152 ⇒ COND1(true, s(x1154), 0)≥COND2(gr(s(x1154), 0), s(x1154), 0)) ⇒ COND1(true, s(x1145), 0)≥COND2(gr(s(x1145), 0), s(x1145), 0))

We solved constraint (119) using rules (I), (II).We solved constraint (120) using rules (I), (II).

For Pair COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y) the following chains were created:

We consider the chain COND4(true, x, y) → COND4(gr(y, 0), x, p(y)), COND4(true, x, y) → COND4(gr(y, 0), x, p(y)), COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y) which results in the following constraint:
(121)    (COND4(gr(x463, 0), x462, p(x463))=COND4(true, x464, x465)∧COND4(gr(x465, 0), x464, p(x465))=COND4(false, x466, x467) ⇒ COND4(false, x466, x467)≥COND1(and(gr(x466, 0), gr(x467, 0)), x466, x467))

We simplified constraint (121) using rules (I), (II), (III), (VII) which results in the following new constraint:
(122)    (0=x1155∧gr(x463, x1155)=true∧p(x463)=x465∧0=x1156∧gr(x465, x1156)=false∧p(x465)=x467 ⇒ COND4(false, x462, x467)≥COND1(and(gr(x462, 0), gr(x467, 0)), x462, x467))

We simplified constraint (122) using rule (V) (with possible (I) afterwards) using induction on gr(x463, x1155)=true which results in the following new constraints:
(123)    (true=true∧0=0∧p(s(x1158))=x465∧0=x1156∧gr(x465, x1156)=false∧p(x465)=x467 ⇒ COND4(false, x462, x467)≥COND1(and(gr(x462, 0), gr(x467, 0)), x462, x467))

(124)    (gr(x1159, x1160)=true∧0=s(x1160)∧p(s(x1159))=x465∧0=x1156∧gr(x465, x1156)=false∧p(x465)=x467∧(∀x1161,x1162,x1163,x1164:gr(x1159, x1160)=true∧0=x1160∧p(x1159)=x1161∧0=x1162∧gr(x1161, x1162)=false∧p(x1161)=x1163 ⇒ COND4(false, x1164, x1163)≥COND1(and(gr(x1164, 0), gr(x1163, 0)), x1164, x1163)) ⇒ COND4(false, x462, x467)≥COND1(and(gr(x462, 0), gr(x467, 0)), x462, x467))

We simplified constraint (123) using rules (I), (II), (VII) which results in the following new constraint:
(125)    (s(x1158)=x1165∧p(x1165)=x465∧0=x1156∧gr(x465, x1156)=false∧p(x465)=x467 ⇒ COND4(false, x462, x467)≥COND1(and(gr(x462, 0), gr(x467, 0)), x462, x467))

We solved constraint (124) using rules (I), (II).We simplified constraint (125) using rule (V) (with possible (I) afterwards) using induction on gr(x465, x1156)=false which results in the following new constraints:
(126)    (false=false∧s(x1158)=x1165∧p(x1165)=0∧0=x1166∧p(0)=x467 ⇒ COND4(false, x462, x467)≥COND1(and(gr(x462, 0), gr(x467, 0)), x462, x467))

(127)    (gr(x1168, x1169)=false∧s(x1158)=x1165∧p(x1165)=s(x1168)∧0=s(x1169)∧p(s(x1168))=x467∧(∀x1170,x1171,x1172,x1173:gr(x1168, x1169)=false∧s(x1170)=x1171∧p(x1171)=x1168∧0=x1169∧p(x1168)=x1172 ⇒ COND4(false, x1173, x1172)≥COND1(and(gr(x1173, 0), gr(x1172, 0)), x1173, x1172)) ⇒ COND4(false, x462, x467)≥COND1(and(gr(x462, 0), gr(x467, 0)), x462, x467))

We simplified constraint (126) using rules (I), (II), (IV), (VII) which results in the following new constraint:
(128)    (s(x1158)=x1165∧p(x1165)=0∧0=x1174∧p(x1174)=x467 ⇒ COND4(false, x462, x467)≥COND1(and(gr(x462, 0), gr(x467, 0)), x462, x467))

We solved constraint (127) using rules (I), (II).We simplified constraint (128) using rule (V) (with possible (I) afterwards) using induction on p(x1165)=0 which results in the following new constraints:
(129)    (0=0∧s(x1158)=0∧0=x1174∧p(x1174)=x467 ⇒ COND4(false, x462, x467)≥COND1(and(gr(x462, 0), gr(x467, 0)), x462, x467))

(130)    (x1175=0∧s(x1158)=s(x1175)∧0=x1174∧p(x1174)=x467 ⇒ COND4(false, x462, x467)≥COND1(and(gr(x462, 0), gr(x467, 0)), x462, x467))

We solved constraint (129) using rules (I), (II).We simplified constraint (130) using rules (I), (II), (III), (IV) which results in the following new constraint:
(131)    (COND4(false, x462, x467)≥COND1(and(gr(x462, 0), gr(x467, 0)), x462, x467))
We consider the chain COND2(false, x, y) → COND4(gr(y, 0), x, y), COND4(true, x, y) → COND4(gr(y, 0), x, p(y)), COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y) which results in the following constraint:
(132)    (COND4(gr(x469, 0), x468, x469)=COND4(true, x470, x471)∧COND4(gr(x471, 0), x470, p(x471))=COND4(false, x472, x473) ⇒ COND4(false, x472, x473)≥COND1(and(gr(x472, 0), gr(x473, 0)), x472, x473))

We simplified constraint (132) using rules (I), (II), (III), (VII) which results in the following new constraint:
(133)    (0=x1176∧gr(x471, x1176)=true∧0=x1177∧gr(x471, x1177)=false∧p(x471)=x473 ⇒ COND4(false, x468, x473)≥COND1(and(gr(x468, 0), gr(x473, 0)), x468, x473))

We simplified constraint (133) using rule (V) (with possible (I) afterwards) using induction on gr(x471, x1176)=true which results in the following new constraints:
(134)    (true=true∧0=0∧0=x1177∧gr(s(x1179), x1177)=false∧p(s(x1179))=x473 ⇒ COND4(false, x468, x473)≥COND1(and(gr(x468, 0), gr(x473, 0)), x468, x473))

(135)    (gr(x1180, x1181)=true∧0=s(x1181)∧0=x1177∧gr(s(x1180), x1177)=false∧p(s(x1180))=x473∧(∀x1182,x1183,x1184:gr(x1180, x1181)=true∧0=x1181∧0=x1182∧gr(x1180, x1182)=false∧p(x1180)=x1183 ⇒ COND4(false, x1184, x1183)≥COND1(and(gr(x1184, 0), gr(x1183, 0)), x1184, x1183)) ⇒ COND4(false, x468, x473)≥COND1(and(gr(x468, 0), gr(x473, 0)), x468, x473))

We simplified constraint (134) using rules (I), (II), (VII) which results in the following new constraint:
(136)    (0=x1177∧s(x1179)=x1185∧gr(x1185, x1177)=false∧s(x1179)=x1186∧p(x1186)=x473 ⇒ COND4(false, x468, x473)≥COND1(and(gr(x468, 0), gr(x473, 0)), x468, x473))

We solved constraint (135) using rules (I), (II).We simplified constraint (136) using rule (V) (with possible (I) afterwards) using induction on gr(x1185, x1177)=false which results in the following new constraints:
(137)    (false=false∧0=x1187∧s(x1179)=0∧s(x1179)=x1186∧p(x1186)=x473 ⇒ COND4(false, x468, x473)≥COND1(and(gr(x468, 0), gr(x473, 0)), x468, x473))

(138)    (gr(x1189, x1190)=false∧0=s(x1190)∧s(x1179)=s(x1189)∧s(x1179)=x1186∧p(x1186)=x473∧(∀x1191,x1192,x1193,x1194:gr(x1189, x1190)=false∧0=x1190∧s(x1191)=x1189∧s(x1191)=x1192∧p(x1192)=x1193 ⇒ COND4(false, x1194, x1193)≥COND1(and(gr(x1194, 0), gr(x1193, 0)), x1194, x1193)) ⇒ COND4(false, x468, x473)≥COND1(and(gr(x468, 0), gr(x473, 0)), x468, x473))

We solved constraint (137) using rules (I), (II).We solved constraint (138) using rules (I), (II).
We consider the chain COND1(true, x, y) → COND2(gr(x, y), x, y), COND2(false, x, y) → COND4(gr(y, 0), x, y), COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y) which results in the following constraint:
(139)    (COND2(gr(x480, x481), x480, x481)=COND2(false, x482, x483)∧COND4(gr(x483, 0), x482, x483)=COND4(false, x484, x485) ⇒ COND4(false, x484, x485)≥COND1(and(gr(x484, 0), gr(x485, 0)), x484, x485))

We simplified constraint (139) using rules (I), (II), (III), (VII) which results in the following new constraint:
(140)    (gr(x480, x483)=false∧0=x1195∧gr(x483, x1195)=false ⇒ COND4(false, x480, x483)≥COND1(and(gr(x480, 0), gr(x483, 0)), x480, x483))

We simplified constraint (140) using rule (V) (with possible (I) afterwards) using induction on gr(x480, x483)=false which results in the following new constraints:
(141)    (false=false∧0=x1195∧gr(x1196, x1195)=false ⇒ COND4(false, 0, x1196)≥COND1(and(gr(0, 0), gr(x1196, 0)), 0, x1196))

(142)    (gr(x1198, x1199)=false∧0=x1195∧gr(s(x1199), x1195)=false∧(∀x1200:gr(x1198, x1199)=false∧0=x1200∧gr(x1199, x1200)=false ⇒ COND4(false, x1198, x1199)≥COND1(and(gr(x1198, 0), gr(x1199, 0)), x1198, x1199)) ⇒ COND4(false, s(x1198), s(x1199))≥COND1(and(gr(s(x1198), 0), gr(s(x1199), 0)), s(x1198), s(x1199)))

We simplified constraint (141) using rules (I), (II) which results in the following new constraint:
(143)    (0=x1195∧gr(x1196, x1195)=false ⇒ COND4(false, 0, x1196)≥COND1(and(gr(0, 0), gr(x1196, 0)), 0, x1196))

We simplified constraint (142) using rule (VII) which results in the following new constraint:
(144)    (gr(x1198, x1199)=false∧0=x1195∧s(x1199)=x1205∧gr(x1205, x1195)=false∧(∀x1200:gr(x1198, x1199)=false∧0=x1200∧gr(x1199, x1200)=false ⇒ COND4(false, x1198, x1199)≥COND1(and(gr(x1198, 0), gr(x1199, 0)), x1198, x1199)) ⇒ COND4(false, s(x1198), s(x1199))≥COND1(and(gr(s(x1198), 0), gr(s(x1199), 0)), s(x1198), s(x1199)))

We simplified constraint (143) using rule (V) (with possible (I) afterwards) using induction on gr(x1196, x1195)=false which results in the following new constraints:
(145)    (false=false∧0=x1201 ⇒ COND4(false, 0, 0)≥COND1(and(gr(0, 0), gr(0, 0)), 0, 0))

(146)    (gr(x1203, x1204)=false∧0=s(x1204)∧(gr(x1203, x1204)=false∧0=x1204 ⇒ COND4(false, 0, x1203)≥COND1(and(gr(0, 0), gr(x1203, 0)), 0, x1203)) ⇒ COND4(false, 0, s(x1203))≥COND1(and(gr(0, 0), gr(s(x1203), 0)), 0, s(x1203)))

We simplified constraint (145) using rules (I), (II), (IV) which results in the following new constraint:
(147)    (COND4(false, 0, 0)≥COND1(and(gr(0, 0), gr(0, 0)), 0, 0))

We solved constraint (146) using rules (I), (II).We simplified constraint (144) using rule (V) (with possible (I) afterwards) using induction on gr(x1205, x1195)=false which results in the following new constraints:
(148)    (false=false∧gr(x1198, x1199)=false∧0=x1206∧s(x1199)=0∧(∀x1200:gr(x1198, x1199)=false∧0=x1200∧gr(x1199, x1200)=false ⇒ COND4(false, x1198, x1199)≥COND1(and(gr(x1198, 0), gr(x1199, 0)), x1198, x1199)) ⇒ COND4(false, s(x1198), s(x1199))≥COND1(and(gr(s(x1198), 0), gr(s(x1199), 0)), s(x1198), s(x1199)))

(149)    (gr(x1208, x1209)=false∧gr(x1198, x1199)=false∧0=s(x1209)∧s(x1199)=s(x1208)∧(∀x1200:gr(x1198, x1199)=false∧0=x1200∧gr(x1199, x1200)=false ⇒ COND4(false, x1198, x1199)≥COND1(and(gr(x1198, 0), gr(x1199, 0)), x1198, x1199))∧(∀x1210,x1211,x1212:gr(x1208, x1209)=false∧gr(x1210, x1211)=false∧0=x1209∧s(x1211)=x1208∧(∀x1212:gr(x1210, x1211)=false∧0=x1212∧gr(x1211, x1212)=false ⇒ COND4(false, x1210, x1211)≥COND1(and(gr(x1210, 0), gr(x1211, 0)), x1210, x1211)) ⇒ COND4(false, s(x1210), s(x1211))≥COND1(and(gr(s(x1210), 0), gr(s(x1211), 0)), s(x1210), s(x1211))) ⇒ COND4(false, s(x1198), s(x1199))≥COND1(and(gr(s(x1198), 0), gr(s(x1199), 0)), s(x1198), s(x1199)))

We solved constraint (148) using rules (I), (II).We solved constraint (149) using rules (I), (II).

For Pair COND4(true, x, y) → COND4(gr(y, 0), x, p(y)) the following chains were created:

We consider the chain COND4(true, x, y) → COND4(gr(y, 0), x, p(y)), COND4(true, x, y) → COND4(gr(y, 0), x, p(y)), COND4(true, x, y) → COND4(gr(y, 0), x, p(y)) which results in the following constraint:
(150)    (COND4(gr(x591, 0), x590, p(x591))=COND4(true, x592, x593)∧COND4(gr(x593, 0), x592, p(x593))=COND4(true, x594, x595) ⇒ COND4(true, x594, x595)≥COND4(gr(x595, 0), x594, p(x595)))

We simplified constraint (150) using rules (I), (II), (III), (VII) which results in the following new constraint:
(151)    (0=x1213∧gr(x591, x1213)=true∧p(x591)=x593∧0=x1214∧gr(x593, x1214)=true∧p(x593)=x595 ⇒ COND4(true, x590, x595)≥COND4(gr(x595, 0), x590, p(x595)))

We simplified constraint (151) using rule (V) (with possible (I) afterwards) using induction on gr(x591, x1213)=true which results in the following new constraints:
(152)    (true=true∧0=0∧p(s(x1216))=x593∧0=x1214∧gr(x593, x1214)=true∧p(x593)=x595 ⇒ COND4(true, x590, x595)≥COND4(gr(x595, 0), x590, p(x595)))

(153)    (gr(x1217, x1218)=true∧0=s(x1218)∧p(s(x1217))=x593∧0=x1214∧gr(x593, x1214)=true∧p(x593)=x595∧(∀x1219,x1220,x1221,x1222:gr(x1217, x1218)=true∧0=x1218∧p(x1217)=x1219∧0=x1220∧gr(x1219, x1220)=true∧p(x1219)=x1221 ⇒ COND4(true, x1222, x1221)≥COND4(gr(x1221, 0), x1222, p(x1221))) ⇒ COND4(true, x590, x595)≥COND4(gr(x595, 0), x590, p(x595)))

We simplified constraint (152) using rules (I), (II), (VII) which results in the following new constraint:
(154)    (s(x1216)=x1223∧p(x1223)=x593∧0=x1214∧gr(x593, x1214)=true∧p(x593)=x595 ⇒ COND4(true, x590, x595)≥COND4(gr(x595, 0), x590, p(x595)))

We solved constraint (153) using rules (I), (II).We simplified constraint (154) using rule (V) (with possible (I) afterwards) using induction on gr(x593, x1214)=true which results in the following new constraints:
(155)    (true=true∧s(x1216)=x1223∧p(x1223)=s(x1225)∧0=0∧p(s(x1225))=x595 ⇒ COND4(true, x590, x595)≥COND4(gr(x595, 0), x590, p(x595)))

(156)    (gr(x1226, x1227)=true∧s(x1216)=x1223∧p(x1223)=s(x1226)∧0=s(x1227)∧p(s(x1226))=x595∧(∀x1228,x1229,x1230,x1231:gr(x1226, x1227)=true∧s(x1228)=x1229∧p(x1229)=x1226∧0=x1227∧p(x1226)=x1230 ⇒ COND4(true, x1231, x1230)≥COND4(gr(x1230, 0), x1231, p(x1230))) ⇒ COND4(true, x590, x595)≥COND4(gr(x595, 0), x590, p(x595)))

We simplified constraint (155) using rules (I), (II), (VII) which results in the following new constraint:
(157)    (s(x1216)=x1223∧p(x1223)=s(x1225)∧s(x1225)=x1232∧p(x1232)=x595 ⇒ COND4(true, x590, x595)≥COND4(gr(x595, 0), x590, p(x595)))

We solved constraint (156) using rules (I), (II).We simplified constraint (157) using rule (V) (with possible (I) afterwards) using induction on p(x1223)=s(x1225) which results in the following new constraint:
(158)    (x1233=s(x1225)∧s(x1216)=s(x1233)∧s(x1225)=x1232∧p(x1232)=x595 ⇒ COND4(true, x590, x595)≥COND4(gr(x595, 0), x590, p(x595)))

We simplified constraint (158) using rules (I), (II), (III), (IV) which results in the following new constraint:
(159)    (COND4(true, x590, x595)≥COND4(gr(x595, 0), x590, p(x595)))
We consider the chain COND2(false, x, y) → COND4(gr(y, 0), x, y), COND4(true, x, y) → COND4(gr(y, 0), x, p(y)), COND4(true, x, y) → COND4(gr(y, 0), x, p(y)) which results in the following constraint:
(160)    (COND4(gr(x597, 0), x596, x597)=COND4(true, x598, x599)∧COND4(gr(x599, 0), x598, p(x599))=COND4(true, x600, x601) ⇒ COND4(true, x600, x601)≥COND4(gr(x601, 0), x600, p(x601)))

We simplified constraint (160) using rules (I), (II), (III), (VII) which results in the following new constraint:
(161)    (0=x1234∧gr(x599, x1234)=true∧0=x1235∧gr(x599, x1235)=true∧p(x599)=x601 ⇒ COND4(true, x596, x601)≥COND4(gr(x601, 0), x596, p(x601)))

We simplified constraint (161) using rule (V) (with possible (I) afterwards) using induction on gr(x599, x1234)=true which results in the following new constraints:
(162)    (true=true∧0=0∧0=x1235∧gr(s(x1237), x1235)=true∧p(s(x1237))=x601 ⇒ COND4(true, x596, x601)≥COND4(gr(x601, 0), x596, p(x601)))

(163)    (gr(x1238, x1239)=true∧0=s(x1239)∧0=x1235∧gr(s(x1238), x1235)=true∧p(s(x1238))=x601∧(∀x1240,x1241,x1242:gr(x1238, x1239)=true∧0=x1239∧0=x1240∧gr(x1238, x1240)=true∧p(x1238)=x1241 ⇒ COND4(true, x1242, x1241)≥COND4(gr(x1241, 0), x1242, p(x1241))) ⇒ COND4(true, x596, x601)≥COND4(gr(x601, 0), x596, p(x601)))

We simplified constraint (162) using rules (I), (II), (VII) which results in the following new constraint:
(164)    (0=x1235∧s(x1237)=x1243∧gr(x1243, x1235)=true∧s(x1237)=x1244∧p(x1244)=x601 ⇒ COND4(true, x596, x601)≥COND4(gr(x601, 0), x596, p(x601)))

We solved constraint (163) using rules (I), (II).We simplified constraint (164) using rule (V) (with possible (I) afterwards) using induction on gr(x1243, x1235)=true which results in the following new constraints:
(165)    (true=true∧0=0∧s(x1237)=s(x1246)∧s(x1237)=x1244∧p(x1244)=x601 ⇒ COND4(true, x596, x601)≥COND4(gr(x601, 0), x596, p(x601)))

(166)    (gr(x1247, x1248)=true∧0=s(x1248)∧s(x1237)=s(x1247)∧s(x1237)=x1244∧p(x1244)=x601∧(∀x1249,x1250,x1251,x1252:gr(x1247, x1248)=true∧0=x1248∧s(x1249)=x1247∧s(x1249)=x1250∧p(x1250)=x1251 ⇒ COND4(true, x1252, x1251)≥COND4(gr(x1251, 0), x1252, p(x1251))) ⇒ COND4(true, x596, x601)≥COND4(gr(x601, 0), x596, p(x601)))

We simplified constraint (165) using rules (I), (II), (IV) which results in the following new constraint:
(167)    (COND4(true, x596, x601)≥COND4(gr(x601, 0), x596, p(x601)))

We solved constraint (166) using rules (I), (II).
We consider the chain COND1(true, x, y) → COND2(gr(x, y), x, y), COND2(false, x, y) → COND4(gr(y, 0), x, y), COND4(true, x, y) → COND4(gr(y, 0), x, p(y)) which results in the following constraint:
(168)    (COND2(gr(x608, x609), x608, x609)=COND2(false, x610, x611)∧COND4(gr(x611, 0), x610, x611)=COND4(true, x612, x613) ⇒ COND4(true, x612, x613)≥COND4(gr(x613, 0), x612, p(x613)))

We simplified constraint (168) using rules (I), (II), (III), (VII) which results in the following new constraint:
(169)    (gr(x608, x611)=false∧0=x1253∧gr(x611, x1253)=true ⇒ COND4(true, x608, x611)≥COND4(gr(x611, 0), x608, p(x611)))

We simplified constraint (169) using rule (V) (with possible (I) afterwards) using induction on gr(x608, x611)=false which results in the following new constraints:
(170)    (false=false∧0=x1253∧gr(x1254, x1253)=true ⇒ COND4(true, 0, x1254)≥COND4(gr(x1254, 0), 0, p(x1254)))

(171)    (gr(x1256, x1257)=false∧0=x1253∧gr(s(x1257), x1253)=true∧(∀x1258:gr(x1256, x1257)=false∧0=x1258∧gr(x1257, x1258)=true ⇒ COND4(true, x1256, x1257)≥COND4(gr(x1257, 0), x1256, p(x1257))) ⇒ COND4(true, s(x1256), s(x1257))≥COND4(gr(s(x1257), 0), s(x1256), p(s(x1257))))

We simplified constraint (170) using rules (I), (II) which results in the following new constraint:
(172)    (0=x1253∧gr(x1254, x1253)=true ⇒ COND4(true, 0, x1254)≥COND4(gr(x1254, 0), 0, p(x1254)))

We simplified constraint (171) using rule (VII) which results in the following new constraint:
(173)    (gr(x1256, x1257)=false∧0=x1253∧s(x1257)=x1263∧gr(x1263, x1253)=true∧(∀x1258:gr(x1256, x1257)=false∧0=x1258∧gr(x1257, x1258)=true ⇒ COND4(true, x1256, x1257)≥COND4(gr(x1257, 0), x1256, p(x1257))) ⇒ COND4(true, s(x1256), s(x1257))≥COND4(gr(s(x1257), 0), s(x1256), p(s(x1257))))

We simplified constraint (172) using rule (V) (with possible (I) afterwards) using induction on gr(x1254, x1253)=true which results in the following new constraints:
(174)    (true=true∧0=0 ⇒ COND4(true, 0, s(x1260))≥COND4(gr(s(x1260), 0), 0, p(s(x1260))))

(175)    (gr(x1261, x1262)=true∧0=s(x1262)∧(gr(x1261, x1262)=true∧0=x1262 ⇒ COND4(true, 0, x1261)≥COND4(gr(x1261, 0), 0, p(x1261))) ⇒ COND4(true, 0, s(x1261))≥COND4(gr(s(x1261), 0), 0, p(s(x1261))))

We simplified constraint (174) using rules (I), (II) which results in the following new constraint:
(176)    (COND4(true, 0, s(x1260))≥COND4(gr(s(x1260), 0), 0, p(s(x1260))))

We solved constraint (175) using rules (I), (II).We simplified constraint (173) using rule (V) (with possible (I) afterwards) using induction on gr(x1263, x1253)=true which results in the following new constraints:
(177)    (true=true∧gr(x1256, x1257)=false∧0=0∧s(x1257)=s(x1265)∧(∀x1258:gr(x1256, x1257)=false∧0=x1258∧gr(x1257, x1258)=true ⇒ COND4(true, x1256, x1257)≥COND4(gr(x1257, 0), x1256, p(x1257))) ⇒ COND4(true, s(x1256), s(x1257))≥COND4(gr(s(x1257), 0), s(x1256), p(s(x1257))))

(178)    (gr(x1266, x1267)=true∧gr(x1256, x1257)=false∧0=s(x1267)∧s(x1257)=s(x1266)∧(∀x1258:gr(x1256, x1257)=false∧0=x1258∧gr(x1257, x1258)=true ⇒ COND4(true, x1256, x1257)≥COND4(gr(x1257, 0), x1256, p(x1257)))∧(∀x1268,x1269,x1270:gr(x1266, x1267)=true∧gr(x1268, x1269)=false∧0=x1267∧s(x1269)=x1266∧(∀x1270:gr(x1268, x1269)=false∧0=x1270∧gr(x1269, x1270)=true ⇒ COND4(true, x1268, x1269)≥COND4(gr(x1269, 0), x1268, p(x1269))) ⇒ COND4(true, s(x1268), s(x1269))≥COND4(gr(s(x1269), 0), s(x1268), p(s(x1269)))) ⇒ COND4(true, s(x1256), s(x1257))≥COND4(gr(s(x1257), 0), s(x1256), p(s(x1257))))

We simplified constraint (177) using rules (I), (II), (IV) which results in the following new constraint:
(179)    (gr(x1256, x1257)=false ⇒ COND4(true, s(x1256), s(x1257))≥COND4(gr(s(x1257), 0), s(x1256), p(s(x1257))))

We solved constraint (178) using rules (I), (II).We simplified constraint (179) using rule (V) (with possible (I) afterwards) using induction on gr(x1256, x1257)=false which results in the following new constraints:
(180)    (false=false ⇒ COND4(true, s(0), s(x1271))≥COND4(gr(s(x1271), 0), s(0), p(s(x1271))))

(181)    (gr(x1273, x1274)=false∧(gr(x1273, x1274)=false ⇒ COND4(true, s(x1273), s(x1274))≥COND4(gr(s(x1274), 0), s(x1273), p(s(x1274)))) ⇒ COND4(true, s(s(x1273)), s(s(x1274)))≥COND4(gr(s(s(x1274)), 0), s(s(x1273)), p(s(s(x1274)))))

We simplified constraint (180) using rules (I), (II) which results in the following new constraint:
(182)    (COND4(true, s(0), s(x1271))≥COND4(gr(s(x1271), 0), s(0), p(s(x1271))))

We simplified constraint (181) using rule (VI) where we applied the induction hypothesis (gr(x1273, x1274)=false ⇒ COND4(true, s(x1273), s(x1274))≥COND4(gr(s(x1274), 0), s(x1273), p(s(x1274)))) with σ = [ ] which results in the following new constraint:
(183)    (COND4(true, s(x1273), s(x1274))≥COND4(gr(s(x1274), 0), s(x1273), p(s(x1274))) ⇒ COND4(true, s(s(x1273)), s(s(x1274)))≥COND4(gr(s(s(x1274)), 0), s(s(x1273)), p(s(s(x1274)))))

For Pair COND2(false, x, y) → COND4(gr(y, 0), x, y) the following chains were created:

We consider the chain COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y), COND1(true, x, y) → COND2(gr(x, y), x, y), COND2(false, x, y) → COND4(gr(y, 0), x, y) which results in the following constraint:
(184)    (COND1(and(gr(x674, 0), gr(x675, 0)), x674, x675)=COND1(true, x676, x677)∧COND2(gr(x676, x677), x676, x677)=COND2(false, x678, x679) ⇒ COND2(false, x678, x679)≥COND4(gr(x679, 0), x678, x679))

We simplified constraint (184) using rules (I), (II), (III), (VII) which results in the following new constraint:
(185)    (0=x1277∧gr(x676, x1277)=x1275∧0=x1278∧gr(x677, x1278)=x1276∧and(x1275, x1276)=true∧gr(x676, x677)=false ⇒ COND2(false, x676, x677)≥COND4(gr(x677, 0), x676, x677))

We simplified constraint (185) using rule (V) (with possible (I) afterwards) using induction on and(x1275, x1276)=true which results in the following new constraint:
(186)    (true=true∧0=x1277∧gr(x676, x1277)=true∧0=x1278∧gr(x677, x1278)=true∧gr(x676, x677)=false ⇒ COND2(false, x676, x677)≥COND4(gr(x677, 0), x676, x677))

We simplified constraint (186) using rules (I), (II) which results in the following new constraint:
(187)    (0=x1277∧gr(x676, x1277)=true∧0=x1278∧gr(x677, x1278)=true∧gr(x676, x677)=false ⇒ COND2(false, x676, x677)≥COND4(gr(x677, 0), x676, x677))

We simplified constraint (187) using rule (V) (with possible (I) afterwards) using induction on gr(x676, x677)=false which results in the following new constraints:
(188)    (false=false∧0=x1277∧gr(0, x1277)=true∧0=x1278∧gr(x1281, x1278)=true ⇒ COND2(false, 0, x1281)≥COND4(gr(x1281, 0), 0, x1281))

(189)    (gr(x1283, x1284)=false∧0=x1277∧gr(s(x1283), x1277)=true∧0=x1278∧gr(s(x1284), x1278)=true∧(∀x1285,x1286:gr(x1283, x1284)=false∧0=x1285∧gr(x1283, x1285)=true∧0=x1286∧gr(x1284, x1286)=true ⇒ COND2(false, x1283, x1284)≥COND4(gr(x1284, 0), x1283, x1284)) ⇒ COND2(false, s(x1283), s(x1284))≥COND4(gr(s(x1284), 0), s(x1283), s(x1284)))

We simplified constraint (188) using rules (I), (II), (VII) which results in the following new constraint:
(190)    (0=x1277∧0=x1287∧gr(x1287, x1277)=true∧0=x1278∧gr(x1281, x1278)=true ⇒ COND2(false, 0, x1281)≥COND4(gr(x1281, 0), 0, x1281))

We simplified constraint (189) using rule (VII) which results in the following new constraint:
(191)    (gr(x1283, x1284)=false∧0=x1277∧s(x1283)=x1294∧gr(x1294, x1277)=true∧0=x1278∧s(x1284)=x1295∧gr(x1295, x1278)=true∧(∀x1285,x1286:gr(x1283, x1284)=false∧0=x1285∧gr(x1283, x1285)=true∧0=x1286∧gr(x1284, x1286)=true ⇒ COND2(false, x1283, x1284)≥COND4(gr(x1284, 0), x1283, x1284)) ⇒ COND2(false, s(x1283), s(x1284))≥COND4(gr(s(x1284), 0), s(x1283), s(x1284)))

We simplified constraint (190) using rule (V) (with possible (I) afterwards) using induction on gr(x1287, x1277)=true which results in the following new constraints:
(192)    (true=true∧0=0∧0=s(x1289)∧0=x1278∧gr(x1281, x1278)=true ⇒ COND2(false, 0, x1281)≥COND4(gr(x1281, 0), 0, x1281))

(193)    (gr(x1290, x1291)=true∧0=s(x1291)∧0=s(x1290)∧0=x1278∧gr(x1281, x1278)=true∧(∀x1292,x1293:gr(x1290, x1291)=true∧0=x1291∧0=x1290∧0=x1292∧gr(x1293, x1292)=true ⇒ COND2(false, 0, x1293)≥COND4(gr(x1293, 0), 0, x1293)) ⇒ COND2(false, 0, x1281)≥COND4(gr(x1281, 0), 0, x1281))

We solved constraint (192) using rules (I), (II).We solved constraint (193) using rules (I), (II).We simplified constraint (191) using rule (V) (with possible (I) afterwards) using induction on gr(x1294, x1277)=true which results in the following new constraints:
(194)    (true=true∧gr(x1283, x1284)=false∧0=0∧s(x1283)=s(x1297)∧0=x1278∧s(x1284)=x1295∧gr(x1295, x1278)=true∧(∀x1285,x1286:gr(x1283, x1284)=false∧0=x1285∧gr(x1283, x1285)=true∧0=x1286∧gr(x1284, x1286)=true ⇒ COND2(false, x1283, x1284)≥COND4(gr(x1284, 0), x1283, x1284)) ⇒ COND2(false, s(x1283), s(x1284))≥COND4(gr(s(x1284), 0), s(x1283), s(x1284)))

(195)    (gr(x1298, x1299)=true∧gr(x1283, x1284)=false∧0=s(x1299)∧s(x1283)=s(x1298)∧0=x1278∧s(x1284)=x1295∧gr(x1295, x1278)=true∧(∀x1285,x1286:gr(x1283, x1284)=false∧0=x1285∧gr(x1283, x1285)=true∧0=x1286∧gr(x1284, x1286)=true ⇒ COND2(false, x1283, x1284)≥COND4(gr(x1284, 0), x1283, x1284))∧(∀x1300,x1301,x1302,x1303,x1304,x1305:gr(x1298, x1299)=true∧gr(x1300, x1301)=false∧0=x1299∧s(x1300)=x1298∧0=x1302∧s(x1301)=x1303∧gr(x1303, x1302)=true∧(∀x1304,x1305:gr(x1300, x1301)=false∧0=x1304∧gr(x1300, x1304)=true∧0=x1305∧gr(x1301, x1305)=true ⇒ COND2(false, x1300, x1301)≥COND4(gr(x1301, 0), x1300, x1301)) ⇒ COND2(false, s(x1300), s(x1301))≥COND4(gr(s(x1301), 0), s(x1300), s(x1301))) ⇒ COND2(false, s(x1283), s(x1284))≥COND4(gr(s(x1284), 0), s(x1283), s(x1284)))

We simplified constraint (194) using rules (I), (II), (IV) which results in the following new constraint:
(196)    (gr(x1283, x1284)=false∧0=x1278∧s(x1284)=x1295∧gr(x1295, x1278)=true∧(∀x1285,x1286:gr(x1283, x1284)=false∧0=x1285∧gr(x1283, x1285)=true∧0=x1286∧gr(x1284, x1286)=true ⇒ COND2(false, x1283, x1284)≥COND4(gr(x1284, 0), x1283, x1284)) ⇒ COND2(false, s(x1283), s(x1284))≥COND4(gr(s(x1284), 0), s(x1283), s(x1284)))

We solved constraint (195) using rules (I), (II).We simplified constraint (196) using rule (V) (with possible (I) afterwards) using induction on gr(x1295, x1278)=true which results in the following new constraints:
(197)    (true=true∧gr(x1283, x1284)=false∧0=0∧s(x1284)=s(x1307)∧(∀x1285,x1286:gr(x1283, x1284)=false∧0=x1285∧gr(x1283, x1285)=true∧0=x1286∧gr(x1284, x1286)=true ⇒ COND2(false, x1283, x1284)≥COND4(gr(x1284, 0), x1283, x1284)) ⇒ COND2(false, s(x1283), s(x1284))≥COND4(gr(s(x1284), 0), s(x1283), s(x1284)))

(198)    (gr(x1308, x1309)=true∧gr(x1283, x1284)=false∧0=s(x1309)∧s(x1284)=s(x1308)∧(∀x1285,x1286:gr(x1283, x1284)=false∧0=x1285∧gr(x1283, x1285)=true∧0=x1286∧gr(x1284, x1286)=true ⇒ COND2(false, x1283, x1284)≥COND4(gr(x1284, 0), x1283, x1284))∧(∀x1310,x1311,x1312,x1313:gr(x1308, x1309)=true∧gr(x1310, x1311)=false∧0=x1309∧s(x1311)=x1308∧(∀x1312,x1313:gr(x1310, x1311)=false∧0=x1312∧gr(x1310, x1312)=true∧0=x1313∧gr(x1311, x1313)=true ⇒ COND2(false, x1310, x1311)≥COND4(gr(x1311, 0), x1310, x1311)) ⇒ COND2(false, s(x1310), s(x1311))≥COND4(gr(s(x1311), 0), s(x1310), s(x1311))) ⇒ COND2(false, s(x1283), s(x1284))≥COND4(gr(s(x1284), 0), s(x1283), s(x1284)))

We simplified constraint (197) using rules (I), (II), (IV) which results in the following new constraint:
(199)    (gr(x1283, x1284)=false ⇒ COND2(false, s(x1283), s(x1284))≥COND4(gr(s(x1284), 0), s(x1283), s(x1284)))

We solved constraint (198) using rules (I), (II).We simplified constraint (199) using rule (V) (with possible (I) afterwards) using induction on gr(x1283, x1284)=false which results in the following new constraints:
(200)    (false=false ⇒ COND2(false, s(0), s(x1314))≥COND4(gr(s(x1314), 0), s(0), s(x1314)))

(201)    (gr(x1316, x1317)=false∧(gr(x1316, x1317)=false ⇒ COND2(false, s(x1316), s(x1317))≥COND4(gr(s(x1317), 0), s(x1316), s(x1317))) ⇒ COND2(false, s(s(x1316)), s(s(x1317)))≥COND4(gr(s(s(x1317)), 0), s(s(x1316)), s(s(x1317))))

We simplified constraint (200) using rules (I), (II) which results in the following new constraint:
(202)    (COND2(false, s(0), s(x1314))≥COND4(gr(s(x1314), 0), s(0), s(x1314)))

We simplified constraint (201) using rule (VI) where we applied the induction hypothesis (gr(x1316, x1317)=false ⇒ COND2(false, s(x1316), s(x1317))≥COND4(gr(s(x1317), 0), s(x1316), s(x1317))) with σ = [ ] which results in the following new constraint:
(203)    (COND2(false, s(x1316), s(x1317))≥COND4(gr(s(x1317), 0), s(x1316), s(x1317)) ⇒ COND2(false, s(s(x1316)), s(s(x1317)))≥COND4(gr(s(s(x1317)), 0), s(s(x1316)), s(s(x1317))))
We consider the chain COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y), COND1(true, x, y) → COND2(gr(x, y), x, y), COND2(false, x, y) → COND4(gr(y, 0), x, y) which results in the following constraint:
(204)    (COND1(and(gr(x684, 0), gr(x685, 0)), x684, x685)=COND1(true, x686, x687)∧COND2(gr(x686, x687), x686, x687)=COND2(false, x688, x689) ⇒ COND2(false, x688, x689)≥COND4(gr(x689, 0), x688, x689))

We simplified constraint (204) using rules (I), (II), (III), (VII) which results in the following new constraint:
(205)    (0=x1320∧gr(x686, x1320)=x1318∧0=x1321∧gr(x687, x1321)=x1319∧and(x1318, x1319)=true∧gr(x686, x687)=false ⇒ COND2(false, x686, x687)≥COND4(gr(x687, 0), x686, x687))

We simplified constraint (205) using rule (V) (with possible (I) afterwards) using induction on and(x1318, x1319)=true which results in the following new constraint:
(206)    (true=true∧0=x1320∧gr(x686, x1320)=true∧0=x1321∧gr(x687, x1321)=true∧gr(x686, x687)=false ⇒ COND2(false, x686, x687)≥COND4(gr(x687, 0), x686, x687))

We simplified constraint (206) using rules (I), (II) which results in the following new constraint:
(207)    (0=x1320∧gr(x686, x1320)=true∧0=x1321∧gr(x687, x1321)=true∧gr(x686, x687)=false ⇒ COND2(false, x686, x687)≥COND4(gr(x687, 0), x686, x687))

We simplified constraint (207) using rule (V) (with possible (I) afterwards) using induction on gr(x686, x687)=false which results in the following new constraints:
(208)    (false=false∧0=x1320∧gr(0, x1320)=true∧0=x1321∧gr(x1324, x1321)=true ⇒ COND2(false, 0, x1324)≥COND4(gr(x1324, 0), 0, x1324))

(209)    (gr(x1326, x1327)=false∧0=x1320∧gr(s(x1326), x1320)=true∧0=x1321∧gr(s(x1327), x1321)=true∧(∀x1328,x1329:gr(x1326, x1327)=false∧0=x1328∧gr(x1326, x1328)=true∧0=x1329∧gr(x1327, x1329)=true ⇒ COND2(false, x1326, x1327)≥COND4(gr(x1327, 0), x1326, x1327)) ⇒ COND2(false, s(x1326), s(x1327))≥COND4(gr(s(x1327), 0), s(x1326), s(x1327)))

We simplified constraint (208) using rules (I), (II), (VII) which results in the following new constraint:
(210)    (0=x1320∧0=x1330∧gr(x1330, x1320)=true∧0=x1321∧gr(x1324, x1321)=true ⇒ COND2(false, 0, x1324)≥COND4(gr(x1324, 0), 0, x1324))

We simplified constraint (209) using rule (VII) which results in the following new constraint:
(211)    (gr(x1326, x1327)=false∧0=x1320∧s(x1326)=x1337∧gr(x1337, x1320)=true∧0=x1321∧s(x1327)=x1338∧gr(x1338, x1321)=true∧(∀x1328,x1329:gr(x1326, x1327)=false∧0=x1328∧gr(x1326, x1328)=true∧0=x1329∧gr(x1327, x1329)=true ⇒ COND2(false, x1326, x1327)≥COND4(gr(x1327, 0), x1326, x1327)) ⇒ COND2(false, s(x1326), s(x1327))≥COND4(gr(s(x1327), 0), s(x1326), s(x1327)))

We simplified constraint (210) using rule (V) (with possible (I) afterwards) using induction on gr(x1330, x1320)=true which results in the following new constraints:
(212)    (true=true∧0=0∧0=s(x1332)∧0=x1321∧gr(x1324, x1321)=true ⇒ COND2(false, 0, x1324)≥COND4(gr(x1324, 0), 0, x1324))

(213)    (gr(x1333, x1334)=true∧0=s(x1334)∧0=s(x1333)∧0=x1321∧gr(x1324, x1321)=true∧(∀x1335,x1336:gr(x1333, x1334)=true∧0=x1334∧0=x1333∧0=x1335∧gr(x1336, x1335)=true ⇒ COND2(false, 0, x1336)≥COND4(gr(x1336, 0), 0, x1336)) ⇒ COND2(false, 0, x1324)≥COND4(gr(x1324, 0), 0, x1324))

We solved constraint (212) using rules (I), (II).We solved constraint (213) using rules (I), (II).We simplified constraint (211) using rule (V) (with possible (I) afterwards) using induction on gr(x1337, x1320)=true which results in the following new constraints:
(214)    (true=true∧gr(x1326, x1327)=false∧0=0∧s(x1326)=s(x1340)∧0=x1321∧s(x1327)=x1338∧gr(x1338, x1321)=true∧(∀x1328,x1329:gr(x1326, x1327)=false∧0=x1328∧gr(x1326, x1328)=true∧0=x1329∧gr(x1327, x1329)=true ⇒ COND2(false, x1326, x1327)≥COND4(gr(x1327, 0), x1326, x1327)) ⇒ COND2(false, s(x1326), s(x1327))≥COND4(gr(s(x1327), 0), s(x1326), s(x1327)))

(215)    (gr(x1341, x1342)=true∧gr(x1326, x1327)=false∧0=s(x1342)∧s(x1326)=s(x1341)∧0=x1321∧s(x1327)=x1338∧gr(x1338, x1321)=true∧(∀x1328,x1329:gr(x1326, x1327)=false∧0=x1328∧gr(x1326, x1328)=true∧0=x1329∧gr(x1327, x1329)=true ⇒ COND2(false, x1326, x1327)≥COND4(gr(x1327, 0), x1326, x1327))∧(∀x1343,x1344,x1345,x1346,x1347,x1348:gr(x1341, x1342)=true∧gr(x1343, x1344)=false∧0=x1342∧s(x1343)=x1341∧0=x1345∧s(x1344)=x1346∧gr(x1346, x1345)=true∧(∀x1347,x1348:gr(x1343, x1344)=false∧0=x1347∧gr(x1343, x1347)=true∧0=x1348∧gr(x1344, x1348)=true ⇒ COND2(false, x1343, x1344)≥COND4(gr(x1344, 0), x1343, x1344)) ⇒ COND2(false, s(x1343), s(x1344))≥COND4(gr(s(x1344), 0), s(x1343), s(x1344))) ⇒ COND2(false, s(x1326), s(x1327))≥COND4(gr(s(x1327), 0), s(x1326), s(x1327)))

We simplified constraint (214) using rules (I), (II), (IV) which results in the following new constraint:
(216)    (gr(x1326, x1327)=false∧0=x1321∧s(x1327)=x1338∧gr(x1338, x1321)=true∧(∀x1328,x1329:gr(x1326, x1327)=false∧0=x1328∧gr(x1326, x1328)=true∧0=x1329∧gr(x1327, x1329)=true ⇒ COND2(false, x1326, x1327)≥COND4(gr(x1327, 0), x1326, x1327)) ⇒ COND2(false, s(x1326), s(x1327))≥COND4(gr(s(x1327), 0), s(x1326), s(x1327)))

We solved constraint (215) using rules (I), (II).We simplified constraint (216) using rule (V) (with possible (I) afterwards) using induction on gr(x1338, x1321)=true which results in the following new constraints:
(217)    (true=true∧gr(x1326, x1327)=false∧0=0∧s(x1327)=s(x1350)∧(∀x1328,x1329:gr(x1326, x1327)=false∧0=x1328∧gr(x1326, x1328)=true∧0=x1329∧gr(x1327, x1329)=true ⇒ COND2(false, x1326, x1327)≥COND4(gr(x1327, 0), x1326, x1327)) ⇒ COND2(false, s(x1326), s(x1327))≥COND4(gr(s(x1327), 0), s(x1326), s(x1327)))

(218)    (gr(x1351, x1352)=true∧gr(x1326, x1327)=false∧0=s(x1352)∧s(x1327)=s(x1351)∧(∀x1328,x1329:gr(x1326, x1327)=false∧0=x1328∧gr(x1326, x1328)=true∧0=x1329∧gr(x1327, x1329)=true ⇒ COND2(false, x1326, x1327)≥COND4(gr(x1327, 0), x1326, x1327))∧(∀x1353,x1354,x1355,x1356:gr(x1351, x1352)=true∧gr(x1353, x1354)=false∧0=x1352∧s(x1354)=x1351∧(∀x1355,x1356:gr(x1353, x1354)=false∧0=x1355∧gr(x1353, x1355)=true∧0=x1356∧gr(x1354, x1356)=true ⇒ COND2(false, x1353, x1354)≥COND4(gr(x1354, 0), x1353, x1354)) ⇒ COND2(false, s(x1353), s(x1354))≥COND4(gr(s(x1354), 0), s(x1353), s(x1354))) ⇒ COND2(false, s(x1326), s(x1327))≥COND4(gr(s(x1327), 0), s(x1326), s(x1327)))

We simplified constraint (217) using rules (I), (II), (IV) which results in the following new constraint:
(219)    (gr(x1326, x1327)=false ⇒ COND2(false, s(x1326), s(x1327))≥COND4(gr(s(x1327), 0), s(x1326), s(x1327)))

We solved constraint (218) using rules (I), (II).We simplified constraint (219) using rule (V) (with possible (I) afterwards) using induction on gr(x1326, x1327)=false which results in the following new constraints:
(220)    (false=false ⇒ COND2(false, s(0), s(x1357))≥COND4(gr(s(x1357), 0), s(0), s(x1357)))

(221)    (gr(x1359, x1360)=false∧(gr(x1359, x1360)=false ⇒ COND2(false, s(x1359), s(x1360))≥COND4(gr(s(x1360), 0), s(x1359), s(x1360))) ⇒ COND2(false, s(s(x1359)), s(s(x1360)))≥COND4(gr(s(s(x1360)), 0), s(s(x1359)), s(s(x1360))))

We simplified constraint (220) using rules (I), (II) which results in the following new constraint:
(222)    (COND2(false, s(0), s(x1357))≥COND4(gr(s(x1357), 0), s(0), s(x1357)))

We simplified constraint (221) using rule (VI) where we applied the induction hypothesis (gr(x1359, x1360)=false ⇒ COND2(false, s(x1359), s(x1360))≥COND4(gr(s(x1360), 0), s(x1359), s(x1360))) with σ = [ ] which results in the following new constraint:
(223)    (COND2(false, s(x1359), s(x1360))≥COND4(gr(s(x1360), 0), s(x1359), s(x1360)) ⇒ COND2(false, s(s(x1359)), s(s(x1360)))≥COND4(gr(s(s(x1360)), 0), s(s(x1359)), s(s(x1360))))

For Pair COND3(true, x, y) → COND3(gr(x, 0), p(x), y) the following chains were created:

We consider the chain COND1(true, x, y) → COND2(gr(x, y), x, y), COND2(true, x, y) → COND3(gr(x, 0), x, y), COND3(true, x, y) → COND3(gr(x, 0), p(x), y) which results in the following constraint:
(224)    (COND2(gr(x788, x789), x788, x789)=COND2(true, x790, x791)∧COND3(gr(x790, 0), x790, x791)=COND3(true, x792, x793) ⇒ COND3(true, x792, x793)≥COND3(gr(x792, 0), p(x792), x793))

We simplified constraint (224) using rules (I), (II), (III), (VII) which results in the following new constraint:
(225)    (gr(x790, x789)=true∧0=x1361∧gr(x790, x1361)=true ⇒ COND3(true, x790, x789)≥COND3(gr(x790, 0), p(x790), x789))

We simplified constraint (225) using rule (V) (with possible (I) afterwards) using induction on gr(x790, x789)=true which results in the following new constraints:
(226)    (true=true∧0=x1361∧gr(s(x1363), x1361)=true ⇒ COND3(true, s(x1363), 0)≥COND3(gr(s(x1363), 0), p(s(x1363)), 0))

(227)    (gr(x1364, x1365)=true∧0=x1361∧gr(s(x1364), x1361)=true∧(∀x1366:gr(x1364, x1365)=true∧0=x1366∧gr(x1364, x1366)=true ⇒ COND3(true, x1364, x1365)≥COND3(gr(x1364, 0), p(x1364), x1365)) ⇒ COND3(true, s(x1364), s(x1365))≥COND3(gr(s(x1364), 0), p(s(x1364)), s(x1365)))

We simplified constraint (226) using rules (I), (II), (VII) which results in the following new constraint:
(228)    (0=x1361∧s(x1363)=x1367∧gr(x1367, x1361)=true ⇒ COND3(true, s(x1363), 0)≥COND3(gr(s(x1363), 0), p(s(x1363)), 0))

We simplified constraint (227) using rule (VII) which results in the following new constraint:
(229)    (gr(x1364, x1365)=true∧0=x1361∧s(x1364)=x1373∧gr(x1373, x1361)=true∧(∀x1366:gr(x1364, x1365)=true∧0=x1366∧gr(x1364, x1366)=true ⇒ COND3(true, x1364, x1365)≥COND3(gr(x1364, 0), p(x1364), x1365)) ⇒ COND3(true, s(x1364), s(x1365))≥COND3(gr(s(x1364), 0), p(s(x1364)), s(x1365)))

We simplified constraint (228) using rule (V) (with possible (I) afterwards) using induction on gr(x1367, x1361)=true which results in the following new constraints:
(230)    (true=true∧0=0∧s(x1363)=s(x1369) ⇒ COND3(true, s(x1363), 0)≥COND3(gr(s(x1363), 0), p(s(x1363)), 0))

(231)    (gr(x1370, x1371)=true∧0=s(x1371)∧s(x1363)=s(x1370)∧(∀x1372:gr(x1370, x1371)=true∧0=x1371∧s(x1372)=x1370 ⇒ COND3(true, s(x1372), 0)≥COND3(gr(s(x1372), 0), p(s(x1372)), 0)) ⇒ COND3(true, s(x1363), 0)≥COND3(gr(s(x1363), 0), p(s(x1363)), 0))

We simplified constraint (230) using rules (I), (II), (IV) which results in the following new constraint:
(232)    (COND3(true, s(x1363), 0)≥COND3(gr(s(x1363), 0), p(s(x1363)), 0))

We solved constraint (231) using rules (I), (II).We simplified constraint (229) using rule (V) (with possible (I) afterwards) using induction on gr(x1373, x1361)=true which results in the following new constraints:
(233)    (true=true∧gr(x1364, x1365)=true∧0=0∧s(x1364)=s(x1375)∧(∀x1366:gr(x1364, x1365)=true∧0=x1366∧gr(x1364, x1366)=true ⇒ COND3(true, x1364, x1365)≥COND3(gr(x1364, 0), p(x1364), x1365)) ⇒ COND3(true, s(x1364), s(x1365))≥COND3(gr(s(x1364), 0), p(s(x1364)), s(x1365)))

(234)    (gr(x1376, x1377)=true∧gr(x1364, x1365)=true∧0=s(x1377)∧s(x1364)=s(x1376)∧(∀x1366:gr(x1364, x1365)=true∧0=x1366∧gr(x1364, x1366)=true ⇒ COND3(true, x1364, x1365)≥COND3(gr(x1364, 0), p(x1364), x1365))∧(∀x1378,x1379,x1380:gr(x1376, x1377)=true∧gr(x1378, x1379)=true∧0=x1377∧s(x1378)=x1376∧(∀x1380:gr(x1378, x1379)=true∧0=x1380∧gr(x1378, x1380)=true ⇒ COND3(true, x1378, x1379)≥COND3(gr(x1378, 0), p(x1378), x1379)) ⇒ COND3(true, s(x1378), s(x1379))≥COND3(gr(s(x1378), 0), p(s(x1378)), s(x1379))) ⇒ COND3(true, s(x1364), s(x1365))≥COND3(gr(s(x1364), 0), p(s(x1364)), s(x1365)))

We simplified constraint (233) using rules (I), (II), (IV) which results in the following new constraint:
(235)    (gr(x1364, x1365)=true ⇒ COND3(true, s(x1364), s(x1365))≥COND3(gr(s(x1364), 0), p(s(x1364)), s(x1365)))

We solved constraint (234) using rules (I), (II).We simplified constraint (235) using rule (V) (with possible (I) afterwards) using induction on gr(x1364, x1365)=true which results in the following new constraints:
(236)    (true=true ⇒ COND3(true, s(s(x1382)), s(0))≥COND3(gr(s(s(x1382)), 0), p(s(s(x1382))), s(0)))

(237)    (gr(x1383, x1384)=true∧(gr(x1383, x1384)=true ⇒ COND3(true, s(x1383), s(x1384))≥COND3(gr(s(x1383), 0), p(s(x1383)), s(x1384))) ⇒ COND3(true, s(s(x1383)), s(s(x1384)))≥COND3(gr(s(s(x1383)), 0), p(s(s(x1383))), s(s(x1384))))

We simplified constraint (236) using rules (I), (II) which results in the following new constraint:
(238)    (COND3(true, s(s(x1382)), s(0))≥COND3(gr(s(s(x1382)), 0), p(s(s(x1382))), s(0)))

We simplified constraint (237) using rule (VI) where we applied the induction hypothesis (gr(x1383, x1384)=true ⇒ COND3(true, s(x1383), s(x1384))≥COND3(gr(s(x1383), 0), p(s(x1383)), s(x1384))) with σ = [ ] which results in the following new constraint:
(239)    (COND3(true, s(x1383), s(x1384))≥COND3(gr(s(x1383), 0), p(s(x1383)), s(x1384)) ⇒ COND3(true, s(s(x1383)), s(s(x1384)))≥COND3(gr(s(s(x1383)), 0), p(s(s(x1383))), s(s(x1384))))
We consider the chain COND2(true, x, y) → COND3(gr(x, 0), x, y), COND3(true, x, y) → COND3(gr(x, 0), p(x), y), COND3(true, x, y) → COND3(gr(x, 0), p(x), y) which results in the following constraint:
(240)    (COND3(gr(x874, 0), x874, x875)=COND3(true, x876, x877)∧COND3(gr(x876, 0), p(x876), x877)=COND3(true, x878, x879) ⇒ COND3(true, x878, x879)≥COND3(gr(x878, 0), p(x878), x879))

We simplified constraint (240) using rules (I), (II), (III), (VII) which results in the following new constraint:
(241)    (0=x1385∧gr(x876, x1385)=true∧0=x1386∧gr(x876, x1386)=true∧p(x876)=x878 ⇒ COND3(true, x878, x875)≥COND3(gr(x878, 0), p(x878), x875))

We simplified constraint (241) using rule (V) (with possible (I) afterwards) using induction on gr(x876, x1385)=true which results in the following new constraints:
(242)    (true=true∧0=0∧0=x1386∧gr(s(x1388), x1386)=true∧p(s(x1388))=x878 ⇒ COND3(true, x878, x875)≥COND3(gr(x878, 0), p(x878), x875))

(243)    (gr(x1389, x1390)=true∧0=s(x1390)∧0=x1386∧gr(s(x1389), x1386)=true∧p(s(x1389))=x878∧(∀x1391,x1392,x1393:gr(x1389, x1390)=true∧0=x1390∧0=x1391∧gr(x1389, x1391)=true∧p(x1389)=x1392 ⇒ COND3(true, x1392, x1393)≥COND3(gr(x1392, 0), p(x1392), x1393)) ⇒ COND3(true, x878, x875)≥COND3(gr(x878, 0), p(x878), x875))

We simplified constraint (242) using rules (I), (II), (VII) which results in the following new constraint:
(244)    (0=x1386∧s(x1388)=x1394∧gr(x1394, x1386)=true∧s(x1388)=x1395∧p(x1395)=x878 ⇒ COND3(true, x878, x875)≥COND3(gr(x878, 0), p(x878), x875))

We solved constraint (243) using rules (I), (II).We simplified constraint (244) using rule (V) (with possible (I) afterwards) using induction on gr(x1394, x1386)=true which results in the following new constraints:
(245)    (true=true∧0=0∧s(x1388)=s(x1397)∧s(x1388)=x1395∧p(x1395)=x878 ⇒ COND3(true, x878, x875)≥COND3(gr(x878, 0), p(x878), x875))

(246)    (gr(x1398, x1399)=true∧0=s(x1399)∧s(x1388)=s(x1398)∧s(x1388)=x1395∧p(x1395)=x878∧(∀x1400,x1401,x1402,x1403:gr(x1398, x1399)=true∧0=x1399∧s(x1400)=x1398∧s(x1400)=x1401∧p(x1401)=x1402 ⇒ COND3(true, x1402, x1403)≥COND3(gr(x1402, 0), p(x1402), x1403)) ⇒ COND3(true, x878, x875)≥COND3(gr(x878, 0), p(x878), x875))

We simplified constraint (245) using rules (I), (II), (IV) which results in the following new constraint:
(247)    (COND3(true, x878, x875)≥COND3(gr(x878, 0), p(x878), x875))

We solved constraint (246) using rules (I), (II).
We consider the chain COND3(true, x, y) → COND3(gr(x, 0), p(x), y), COND3(true, x, y) → COND3(gr(x, 0), p(x), y), COND3(true, x, y) → COND3(gr(x, 0), p(x), y) which results in the following constraint:
(248)    (COND3(gr(x888, 0), p(x888), x889)=COND3(true, x890, x891)∧COND3(gr(x890, 0), p(x890), x891)=COND3(true, x892, x893) ⇒ COND3(true, x892, x893)≥COND3(gr(x892, 0), p(x892), x893))

We simplified constraint (248) using rules (I), (II), (III), (VII) which results in the following new constraint:
(249)    (0=x1404∧gr(x888, x1404)=true∧p(x888)=x890∧0=x1405∧gr(x890, x1405)=true∧p(x890)=x892 ⇒ COND3(true, x892, x889)≥COND3(gr(x892, 0), p(x892), x889))

We simplified constraint (249) using rule (V) (with possible (I) afterwards) using induction on gr(x888, x1404)=true which results in the following new constraints:
(250)    (true=true∧0=0∧p(s(x1407))=x890∧0=x1405∧gr(x890, x1405)=true∧p(x890)=x892 ⇒ COND3(true, x892, x889)≥COND3(gr(x892, 0), p(x892), x889))

(251)    (gr(x1408, x1409)=true∧0=s(x1409)∧p(s(x1408))=x890∧0=x1405∧gr(x890, x1405)=true∧p(x890)=x892∧(∀x1410,x1411,x1412,x1413:gr(x1408, x1409)=true∧0=x1409∧p(x1408)=x1410∧0=x1411∧gr(x1410, x1411)=true∧p(x1410)=x1412 ⇒ COND3(true, x1412, x1413)≥COND3(gr(x1412, 0), p(x1412), x1413)) ⇒ COND3(true, x892, x889)≥COND3(gr(x892, 0), p(x892), x889))

We simplified constraint (250) using rules (I), (II), (VII) which results in the following new constraint:
(252)    (s(x1407)=x1414∧p(x1414)=x890∧0=x1405∧gr(x890, x1405)=true∧p(x890)=x892 ⇒ COND3(true, x892, x889)≥COND3(gr(x892, 0), p(x892), x889))

We solved constraint (251) using rules (I), (II).We simplified constraint (252) using rule (V) (with possible (I) afterwards) using induction on gr(x890, x1405)=true which results in the following new constraints:
(253)    (true=true∧s(x1407)=x1414∧p(x1414)=s(x1416)∧0=0∧p(s(x1416))=x892 ⇒ COND3(true, x892, x889)≥COND3(gr(x892, 0), p(x892), x889))

(254)    (gr(x1417, x1418)=true∧s(x1407)=x1414∧p(x1414)=s(x1417)∧0=s(x1418)∧p(s(x1417))=x892∧(∀x1419,x1420,x1421,x1422:gr(x1417, x1418)=true∧s(x1419)=x1420∧p(x1420)=x1417∧0=x1418∧p(x1417)=x1421 ⇒ COND3(true, x1421, x1422)≥COND3(gr(x1421, 0), p(x1421), x1422)) ⇒ COND3(true, x892, x889)≥COND3(gr(x892, 0), p(x892), x889))

We simplified constraint (253) using rules (I), (II), (VII) which results in the following new constraint:
(255)    (s(x1407)=x1414∧p(x1414)=s(x1416)∧s(x1416)=x1423∧p(x1423)=x892 ⇒ COND3(true, x892, x889)≥COND3(gr(x892, 0), p(x892), x889))

We solved constraint (254) using rules (I), (II).We simplified constraint (255) using rule (V) (with possible (I) afterwards) using induction on p(x1414)=s(x1416) which results in the following new constraint:
(256)    (x1424=s(x1416)∧s(x1407)=s(x1424)∧s(x1416)=x1423∧p(x1423)=x892 ⇒ COND3(true, x892, x889)≥COND3(gr(x892, 0), p(x892), x889))

We simplified constraint (256) using rules (I), (II), (III), (IV) which results in the following new constraint:
(257)    (COND3(true, x892, x889)≥COND3(gr(x892, 0), p(x892), x889))

To summarize, we get the following constraints P_≥ for the following pairs.

COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
- (COND3(false, x126, x123)≥COND1(and(gr(x126, 0), gr(x123, 0)), x126, x123))
COND2(true, x, y) → COND3(gr(x, 0), x, y)
- (COND2(true, s(s(x1001)), s(0))≥COND3(gr(s(s(x1001)), 0), s(s(x1001)), s(0)))
- (COND2(true, s(x1002), s(x1003))≥COND3(gr(s(x1002), 0), s(x1002), s(x1003)) ⇒ COND2(true, s(s(x1002)), s(s(x1003)))≥COND3(gr(s(s(x1002)), 0), s(s(x1002)), s(s(x1003))))
- (COND2(true, s(s(x1051)), s(0))≥COND3(gr(s(s(x1051)), 0), s(s(x1051)), s(0)))
- (COND2(true, s(x1052), s(x1053))≥COND3(gr(s(x1052), 0), s(x1052), s(x1053)) ⇒ COND2(true, s(s(x1052)), s(s(x1053)))≥COND3(gr(s(s(x1052)), 0), s(s(x1052)), s(s(x1053))))
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
- (COND4(false, x462, x467)≥COND1(and(gr(x462, 0), gr(x467, 0)), x462, x467))
- (COND4(false, 0, 0)≥COND1(and(gr(0, 0), gr(0, 0)), 0, 0))
COND4(true, x, y) → COND4(gr(y, 0), x, p(y))
- (COND4(true, x590, x595)≥COND4(gr(x595, 0), x590, p(x595)))
- (COND4(true, x596, x601)≥COND4(gr(x601, 0), x596, p(x601)))
- (COND4(true, 0, s(x1260))≥COND4(gr(s(x1260), 0), 0, p(s(x1260))))
- (COND4(true, s(0), s(x1271))≥COND4(gr(s(x1271), 0), s(0), p(s(x1271))))
- (COND4(true, s(x1273), s(x1274))≥COND4(gr(s(x1274), 0), s(x1273), p(s(x1274))) ⇒ COND4(true, s(s(x1273)), s(s(x1274)))≥COND4(gr(s(s(x1274)), 0), s(s(x1273)), p(s(s(x1274)))))
COND2(false, x, y) → COND4(gr(y, 0), x, y)
- (COND2(false, s(0), s(x1314))≥COND4(gr(s(x1314), 0), s(0), s(x1314)))
- (COND2(false, s(x1316), s(x1317))≥COND4(gr(s(x1317), 0), s(x1316), s(x1317)) ⇒ COND2(false, s(s(x1316)), s(s(x1317)))≥COND4(gr(s(s(x1317)), 0), s(s(x1316)), s(s(x1317))))
- (COND2(false, s(0), s(x1357))≥COND4(gr(s(x1357), 0), s(0), s(x1357)))
- (COND2(false, s(x1359), s(x1360))≥COND4(gr(s(x1360), 0), s(x1359), s(x1360)) ⇒ COND2(false, s(s(x1359)), s(s(x1360)))≥COND4(gr(s(s(x1360)), 0), s(s(x1359)), s(s(x1360))))
COND3(true, x, y) → COND3(gr(x, 0), p(x), y)
- (COND3(true, s(x1363), 0)≥COND3(gr(s(x1363), 0), p(s(x1363)), 0))
- (COND3(true, s(s(x1382)), s(0))≥COND3(gr(s(s(x1382)), 0), p(s(s(x1382))), s(0)))
- (COND3(true, s(x1383), s(x1384))≥COND3(gr(s(x1383), 0), p(s(x1383)), s(x1384)) ⇒ COND3(true, s(s(x1383)), s(s(x1384)))≥COND3(gr(s(s(x1383)), 0), p(s(s(x1383))), s(s(x1384))))
- (COND3(true, x878, x875)≥COND3(gr(x878, 0), p(x878), x875))
- (COND3(true, x892, x889)≥COND3(gr(x892, 0), p(x892), x889))

The constraints for P_> respective P_bound are constructed from P_≥ where we just replace every occurence of "t ≥ s" in P_≥ by "t > s" respective "t ≥ c". Here c stands for the fresh constant used for P_bound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation [18]:

POL(0) = 0
POL(COND1(x₁, x₂, x₃)) = -1 - 2·x₁·x₃ + x₁² + x₃²
POL(COND2(x₁, x₂, x₃)) = -1 + 2·x₂·x₃ + x₂² + x₃²
POL(COND3(x₁, x₂, x₃)) = -1 + x₃²
POL(COND4(x₁, x₂, x₃)) = -1 + x₃²
POL(and(x₁, x₂)) = 0
POL(c) = -1
POL(false) = 0
POL(gr(x₁, x₂)) = 0
POL(p(x₁)) = x₁
POL(s(x₁)) = x₁
POL(true) = 0

The following pairs are in P_>:

COND1(true, x, y) → COND2(gr(x, y), x, y)

The following pairs are in P_bound:

COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND2(true, x, y) → COND3(gr(x, 0), x, y)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND4(true, x, y) → COND4(gr(y, 0), x, p(y))
COND2(false, x, y) → COND4(gr(y, 0), x, y)
COND3(true, x, y) → COND3(gr(x, 0), p(x), y)

The following rules are usable:

and(true, true) ↔ true
and(false, x) ↔ false
p(s(x)) → x
and(x, false) ↔ false
p(0) → 0

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ QDP

                            ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND2(true, x, y) → COND3(gr(x, 0), x, y)
COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND4(true, x, y) → COND4(gr(y, 0), x, p(y))
COND2(false, x, y) → COND4(gr(y, 0), x, y)
COND3(true, x, y) → COND3(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 4 less nodes.

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                  ↳ UsableRulesProof

                                ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

COND4(true, x, y) → COND4(gr(y, 0), x, p(y))

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
p(0)
p(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ QReductionProof

                                ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

COND4(true, x, y) → COND4(gr(y, 0), x, p(y))

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

and(true, true)
and(false, x0)
and(x0, false)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ QReductionProof

                                        ↳ QDP

                                          ↳ Narrowing

                                ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

COND4(true, x, y) → COND4(gr(y, 0), x, p(y))

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule COND4(true, x, y) → COND4(gr(y, 0), x, p(y)) at position [0] we obtained the following new rules:

COND4(true, y0, 0) → COND4(false, y0, p(0))
COND4(true, y0, s(x0)) → COND4(true, y0, p(s(x0)))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ QReductionProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

COND4(true, y0, 0) → COND4(false, y0, p(0))
COND4(true, y0, s(x0)) → COND4(true, y0, p(s(x0)))

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ QReductionProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                                ↳ QDP

                                                  ↳ UsableRulesProof

                                ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

COND4(true, y0, s(x0)) → COND4(true, y0, p(s(x0)))

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ QReductionProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                                ↳ QDP

                                                  ↳ UsableRulesProof

                                                    ↳ QDP

                                                      ↳ QReductionProof

                                ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

COND4(true, y0, s(x0)) → COND4(true, y0, p(s(x0)))

The TRS R consists of the following rules:

p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ QReductionProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                                ↳ QDP

                                                  ↳ UsableRulesProof

                                                    ↳ QDP

                                                      ↳ QReductionProof

                                                        ↳ QDP

                                                          ↳ Rewriting

                                ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

COND4(true, y0, s(x0)) → COND4(true, y0, p(s(x0)))

The TRS R consists of the following rules:

p(s(x)) → x

The set Q consists of the following terms:

p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule COND4(true, y0, s(x0)) → COND4(true, y0, p(s(x0))) at position [2] we obtained the following new rules:

COND4(true, y0, s(x0)) → COND4(true, y0, x0)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ QReductionProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                                ↳ QDP

                                                  ↳ UsableRulesProof

                                                    ↳ QDP

                                                      ↳ QReductionProof

                                                        ↳ QDP

                                                          ↳ Rewriting

                                                            ↳ QDP

                                                              ↳ UsableRulesProof

                                ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

COND4(true, y0, s(x0)) → COND4(true, y0, x0)

The TRS R consists of the following rules:

p(s(x)) → x

The set Q consists of the following terms:

p(0)
p(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ QReductionProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                                ↳ QDP

                                                  ↳ UsableRulesProof

                                                    ↳ QDP

                                                      ↳ QReductionProof

                                                        ↳ QDP

                                                          ↳ Rewriting

                                                            ↳ QDP

                                                              ↳ UsableRulesProof

                                                                ↳ QDP

                                                                  ↳ QReductionProof

                                ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

COND4(true, y0, s(x0)) → COND4(true, y0, x0)

R is empty.
The set Q consists of the following terms:

p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

p(0)
p(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ QReductionProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                                ↳ QDP

                                                  ↳ UsableRulesProof

                                                    ↳ QDP

                                                      ↳ QReductionProof

                                                        ↳ QDP

                                                          ↳ Rewriting

                                                            ↳ QDP

                                                              ↳ UsableRulesProof

                                                                ↳ QDP

                                                                  ↳ QReductionProof

                                                                    ↳ QDP

                                                                      ↳ ForwardInstantiation

                                ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

COND4(true, y0, s(x0)) → COND4(true, y0, x0)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule COND4(true, y0, s(x0)) → COND4(true, y0, x0) we obtained the following new rules:

COND4(true, x0, s(s(y_1))) → COND4(true, x0, s(y_1))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ QReductionProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                                ↳ QDP

                                                  ↳ UsableRulesProof

                                                    ↳ QDP

                                                      ↳ QReductionProof

                                                        ↳ QDP

                                                          ↳ Rewriting

                                                            ↳ QDP

                                                              ↳ UsableRulesProof

                                                                ↳ QDP

                                                                  ↳ QReductionProof

                                                                    ↳ QDP

                                                                      ↳ ForwardInstantiation

                                                                        ↳ QDP

                                                                          ↳ QDPSizeChangeProof

                                ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

COND4(true, x0, s(s(y_1))) → COND4(true, x0, s(y_1))

From the DPs we obtained the following set of size-change graphs:

COND4(true, x0, s(s(y_1))) → COND4(true, x0, s(y_1))
The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                ↳ QDP

                                  ↳ UsableRulesProof

Q DP problem:
The TRS P consists of the following rules:

COND3(true, x, y) → COND3(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
p(0)
p(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ QReductionProof

Q DP problem:
The TRS P consists of the following rules:

COND3(true, x, y) → COND3(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

and(true, true)
and(false, x0)
and(x0, false)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ QReductionProof

                                        ↳ QDP

                                          ↳ Narrowing

Q DP problem:
The TRS P consists of the following rules:

COND3(true, x, y) → COND3(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule COND3(true, x, y) → COND3(gr(x, 0), p(x), y) at position [0] we obtained the following new rules:

COND3(true, s(x0), y1) → COND3(true, p(s(x0)), y1)
COND3(true, 0, y1) → COND3(false, p(0), y1)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ QReductionProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

COND3(true, s(x0), y1) → COND3(true, p(s(x0)), y1)
COND3(true, 0, y1) → COND3(false, p(0), y1)

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ QReductionProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                                ↳ QDP

                                                  ↳ UsableRulesProof

Q DP problem:
The TRS P consists of the following rules:

COND3(true, s(x0), y1) → COND3(true, p(s(x0)), y1)

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ QReductionProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                                ↳ QDP

                                                  ↳ UsableRulesProof

                                                    ↳ QDP

                                                      ↳ QReductionProof

Q DP problem:
The TRS P consists of the following rules:

COND3(true, s(x0), y1) → COND3(true, p(s(x0)), y1)

The TRS R consists of the following rules:

p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ QReductionProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                                ↳ QDP

                                                  ↳ UsableRulesProof

                                                    ↳ QDP

                                                      ↳ QReductionProof

                                                        ↳ QDP

                                                          ↳ Rewriting

Q DP problem:
The TRS P consists of the following rules:

COND3(true, s(x0), y1) → COND3(true, p(s(x0)), y1)

The TRS R consists of the following rules:

p(s(x)) → x

The set Q consists of the following terms:

p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule COND3(true, s(x0), y1) → COND3(true, p(s(x0)), y1) at position [1] we obtained the following new rules:

COND3(true, s(x0), y1) → COND3(true, x0, y1)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ QReductionProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                                ↳ QDP

                                                  ↳ UsableRulesProof

                                                    ↳ QDP

                                                      ↳ QReductionProof

                                                        ↳ QDP

                                                          ↳ Rewriting

                                                            ↳ QDP

                                                              ↳ UsableRulesProof

Q DP problem:
The TRS P consists of the following rules:

COND3(true, s(x0), y1) → COND3(true, x0, y1)

The TRS R consists of the following rules:

p(s(x)) → x

The set Q consists of the following terms:

p(0)
p(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ QReductionProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                                ↳ QDP

                                                  ↳ UsableRulesProof

                                                    ↳ QDP

                                                      ↳ QReductionProof

                                                        ↳ QDP

                                                          ↳ Rewriting

                                                            ↳ QDP

                                                              ↳ UsableRulesProof

                                                                ↳ QDP

                                                                  ↳ QReductionProof

Q DP problem:
The TRS P consists of the following rules:

COND3(true, s(x0), y1) → COND3(true, x0, y1)

R is empty.
The set Q consists of the following terms:

p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

p(0)
p(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ QReductionProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                                ↳ QDP

                                                  ↳ UsableRulesProof

                                                    ↳ QDP

                                                      ↳ QReductionProof

                                                        ↳ QDP

                                                          ↳ Rewriting

                                                            ↳ QDP

                                                              ↳ UsableRulesProof

                                                                ↳ QDP

                                                                  ↳ QReductionProof

                                                                    ↳ QDP

                                                                      ↳ ForwardInstantiation

Q DP problem:
The TRS P consists of the following rules:

COND3(true, s(x0), y1) → COND3(true, x0, y1)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule COND3(true, s(x0), y1) → COND3(true, x0, y1) we obtained the following new rules:

COND3(true, s(s(y_0)), x1) → COND3(true, s(y_0), x1)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ QReductionProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                                ↳ QDP

                                                  ↳ UsableRulesProof

                                                    ↳ QDP

                                                      ↳ QReductionProof

                                                        ↳ QDP

                                                          ↳ Rewriting

                                                            ↳ QDP

                                                              ↳ UsableRulesProof

                                                                ↳ QDP

                                                                  ↳ QReductionProof

                                                                    ↳ QDP

                                                                      ↳ ForwardInstantiation

                                                                        ↳ QDP

                                                                          ↳ QDPSizeChangeProof

Q DP problem:
The TRS P consists of the following rules:

COND3(true, s(s(y_0)), x1) → COND3(true, s(y_0), x1)

From the DPs we obtained the following set of size-change graphs:

COND3(true, s(s(y_0)), x1) → COND3(true, s(y_0), x1)
The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3