YES Termination w.r.t. Q proof of ../tpdb/TRS/Beerendonk/9.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:

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

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

Q is empty.

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

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

The TRS R 2 is

cond(true, x, y, z) → cond(and(gr(x, z), gr(y, z)), p(x), p(y), z)

The signature Sigma is {cond}

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof

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

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

The set Q consists of the following terms:

cond(true, x0, x1, x2)
and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
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:

COND(true, x, y, z) → AND(gr(x, z), gr(y, z))
COND(true, x, y, z) → COND(and(gr(x, z), gr(y, z)), p(x), p(y), z)
COND(true, x, y, z) → P(y)
GR(s(x), s(y)) → GR(x, y)
COND(true, x, y, z) → GR(y, z)
COND(true, x, y, z) → P(x)
COND(true, x, y, z) → GR(x, z)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

cond(true, x0, x1, x2)
and(true, true)
and(x0, false)
and(false, x0)
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.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

COND(true, x, y, z) → AND(gr(x, z), gr(y, z))
COND(true, x, y, z) → COND(and(gr(x, z), gr(y, z)), p(x), p(y), z)
COND(true, x, y, z) → P(y)
GR(s(x), s(y)) → GR(x, y)
COND(true, x, y, z) → GR(y, z)
COND(true, x, y, z) → P(x)
COND(true, x, y, z) → GR(x, z)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

cond(true, x0, x1, x2)
and(true, true)
and(x0, false)
and(false, x0)
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 2 SCCs with 5 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:

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

The set Q consists of the following terms:

cond(true, x0, x1, x2)
and(true, true)
and(x0, false)
and(false, x0)
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.
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:

cond(true, x0, x1, x2)
and(true, true)
and(x0, false)
and(false, x0)
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.

cond(true, x0, x1, x2)
and(true, true)
and(x0, false)
and(false, x0)
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
                ↳ 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: 



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof

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

COND(true, x, y, z) → COND(and(gr(x, z), gr(y, z)), p(x), p(y), z)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

cond(true, x0, x1, x2)
and(true, true)
and(x0, false)
and(false, x0)
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.
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
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof

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

COND(true, x, y, z) → COND(and(gr(x, z), gr(y, z)), p(x), p(y), z)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

cond(true, x0, x1, x2)
and(true, true)
and(x0, false)
and(false, x0)
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.

cond(true, x0, x1, x2)



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing

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

COND(true, x, y, z) → COND(and(gr(x, z), gr(y, z)), p(x), p(y), z)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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 COND(true, x, y, z) → COND(and(gr(x, z), gr(y, z)), p(x), p(y), z) at position [0] we obtained the following new rules:

COND(true, 0, y1, 0) → COND(and(false, gr(y1, 0)), p(0), p(y1), 0)
COND(true, y0, 0, 0) → COND(and(gr(y0, 0), false), p(y0), p(0), 0)
COND(true, s(x0), y1, s(x1)) → COND(and(gr(x0, x1), gr(y1, s(x1))), p(s(x0)), p(y1), s(x1))
COND(true, y0, s(x0), 0) → COND(and(gr(y0, 0), true), p(y0), p(s(x0)), 0)
COND(true, 0, y1, x0) → COND(and(false, gr(y1, x0)), p(0), p(y1), x0)
COND(true, y0, 0, x0) → COND(and(gr(y0, x0), false), p(y0), p(0), x0)
COND(true, y0, s(x0), s(x1)) → COND(and(gr(y0, s(x1)), gr(x0, x1)), p(y0), p(s(x0)), s(x1))
COND(true, s(x0), y1, 0) → COND(and(true, gr(y1, 0)), p(s(x0)), p(y1), 0)



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Rewriting

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

COND(true, y0, 0, 0) → COND(and(gr(y0, 0), false), p(y0), p(0), 0)
COND(true, 0, y1, 0) → COND(and(false, gr(y1, 0)), p(0), p(y1), 0)
COND(true, s(x0), y1, s(x1)) → COND(and(gr(x0, x1), gr(y1, s(x1))), p(s(x0)), p(y1), s(x1))
COND(true, y0, s(x0), 0) → COND(and(gr(y0, 0), true), p(y0), p(s(x0)), 0)
COND(true, y0, 0, x0) → COND(and(gr(y0, x0), false), p(y0), p(0), x0)
COND(true, 0, y1, x0) → COND(and(false, gr(y1, x0)), p(0), p(y1), x0)
COND(true, y0, s(x0), s(x1)) → COND(and(gr(y0, s(x1)), gr(x0, x1)), p(y0), p(s(x0)), s(x1))
COND(true, s(x0), y1, 0) → COND(and(true, gr(y1, 0)), p(s(x0)), p(y1), 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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 rewriting [15] the rule COND(true, 0, y1, 0) → COND(and(false, gr(y1, 0)), p(0), p(y1), 0) at position [0] we obtained the following new rules:

COND(true, 0, y1, 0) → COND(false, p(0), p(y1), 0)



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Rewriting
                              ↳ QDP
                                ↳ DependencyGraphProof

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

COND(true, 0, y1, 0) → COND(false, p(0), p(y1), 0)
COND(true, y0, 0, 0) → COND(and(gr(y0, 0), false), p(y0), p(0), 0)
COND(true, s(x0), y1, s(x1)) → COND(and(gr(x0, x1), gr(y1, s(x1))), p(s(x0)), p(y1), s(x1))
COND(true, y0, s(x0), 0) → COND(and(gr(y0, 0), true), p(y0), p(s(x0)), 0)
COND(true, 0, y1, x0) → COND(and(false, gr(y1, x0)), p(0), p(y1), x0)
COND(true, y0, 0, x0) → COND(and(gr(y0, x0), false), p(y0), p(0), x0)
COND(true, y0, s(x0), s(x1)) → COND(and(gr(y0, s(x1)), gr(x0, x1)), p(y0), p(s(x0)), s(x1))
COND(true, s(x0), y1, 0) → COND(and(true, gr(y1, 0)), p(s(x0)), p(y1), 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Rewriting
                              ↳ QDP
                                ↳ DependencyGraphProof
                                  ↳ QDP
                                    ↳ Rewriting

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

COND(true, y0, 0, 0) → COND(and(gr(y0, 0), false), p(y0), p(0), 0)
COND(true, s(x0), y1, s(x1)) → COND(and(gr(x0, x1), gr(y1, s(x1))), p(s(x0)), p(y1), s(x1))
COND(true, y0, s(x0), 0) → COND(and(gr(y0, 0), true), p(y0), p(s(x0)), 0)
COND(true, y0, 0, x0) → COND(and(gr(y0, x0), false), p(y0), p(0), x0)
COND(true, 0, y1, x0) → COND(and(false, gr(y1, x0)), p(0), p(y1), x0)
COND(true, y0, s(x0), s(x1)) → COND(and(gr(y0, s(x1)), gr(x0, x1)), p(y0), p(s(x0)), s(x1))
COND(true, s(x0), y1, 0) → COND(and(true, gr(y1, 0)), p(s(x0)), p(y1), 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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 rewriting [15] the rule COND(true, y0, 0, 0) → COND(and(gr(y0, 0), false), p(y0), p(0), 0) at position [0] we obtained the following new rules:

COND(true, y0, 0, 0) → COND(false, p(y0), p(0), 0)



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Rewriting
                              ↳ QDP
                                ↳ DependencyGraphProof
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ DependencyGraphProof

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

COND(true, y0, 0, 0) → COND(false, p(y0), p(0), 0)
COND(true, s(x0), y1, s(x1)) → COND(and(gr(x0, x1), gr(y1, s(x1))), p(s(x0)), p(y1), s(x1))
COND(true, y0, s(x0), 0) → COND(and(gr(y0, 0), true), p(y0), p(s(x0)), 0)
COND(true, 0, y1, x0) → COND(and(false, gr(y1, x0)), p(0), p(y1), x0)
COND(true, y0, 0, x0) → COND(and(gr(y0, x0), false), p(y0), p(0), x0)
COND(true, y0, s(x0), s(x1)) → COND(and(gr(y0, s(x1)), gr(x0, x1)), p(y0), p(s(x0)), s(x1))
COND(true, s(x0), y1, 0) → COND(and(true, gr(y1, 0)), p(s(x0)), p(y1), 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Rewriting
                              ↳ QDP
                                ↳ DependencyGraphProof
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ DependencyGraphProof
                                          ↳ QDP
                                            ↳ Rewriting

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

COND(true, s(x0), y1, s(x1)) → COND(and(gr(x0, x1), gr(y1, s(x1))), p(s(x0)), p(y1), s(x1))
COND(true, y0, s(x0), 0) → COND(and(gr(y0, 0), true), p(y0), p(s(x0)), 0)
COND(true, y0, 0, x0) → COND(and(gr(y0, x0), false), p(y0), p(0), x0)
COND(true, 0, y1, x0) → COND(and(false, gr(y1, x0)), p(0), p(y1), x0)
COND(true, y0, s(x0), s(x1)) → COND(and(gr(y0, s(x1)), gr(x0, x1)), p(y0), p(s(x0)), s(x1))
COND(true, s(x0), y1, 0) → COND(and(true, gr(y1, 0)), p(s(x0)), p(y1), 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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 rewriting [15] the rule COND(true, y0, s(x0), 0) → COND(and(gr(y0, 0), true), p(y0), p(s(x0)), 0) at position [2] we obtained the following new rules:

COND(true, y0, s(x0), 0) → COND(and(gr(y0, 0), true), p(y0), x0, 0)



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Rewriting
                              ↳ QDP
                                ↳ DependencyGraphProof
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ DependencyGraphProof
                                          ↳ QDP
                                            ↳ Rewriting
                                              ↳ QDP
                                                ↳ Rewriting

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

COND(true, y0, s(x0), 0) → COND(and(gr(y0, 0), true), p(y0), x0, 0)
COND(true, s(x0), y1, s(x1)) → COND(and(gr(x0, x1), gr(y1, s(x1))), p(s(x0)), p(y1), s(x1))
COND(true, 0, y1, x0) → COND(and(false, gr(y1, x0)), p(0), p(y1), x0)
COND(true, y0, 0, x0) → COND(and(gr(y0, x0), false), p(y0), p(0), x0)
COND(true, y0, s(x0), s(x1)) → COND(and(gr(y0, s(x1)), gr(x0, x1)), p(y0), p(s(x0)), s(x1))
COND(true, s(x0), y1, 0) → COND(and(true, gr(y1, 0)), p(s(x0)), p(y1), 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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 rewriting [15] the rule COND(true, 0, y1, x0) → COND(and(false, gr(y1, x0)), p(0), p(y1), x0) at position [0] we obtained the following new rules:

COND(true, 0, y1, x0) → COND(false, p(0), p(y1), x0)



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Rewriting
                              ↳ QDP
                                ↳ DependencyGraphProof
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ DependencyGraphProof
                                          ↳ QDP
                                            ↳ Rewriting
                                              ↳ QDP
                                                ↳ Rewriting
                                                  ↳ QDP
                                                    ↳ DependencyGraphProof

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

COND(true, y0, s(x0), 0) → COND(and(gr(y0, 0), true), p(y0), x0, 0)
COND(true, s(x0), y1, s(x1)) → COND(and(gr(x0, x1), gr(y1, s(x1))), p(s(x0)), p(y1), s(x1))
COND(true, 0, y1, x0) → COND(false, p(0), p(y1), x0)
COND(true, y0, 0, x0) → COND(and(gr(y0, x0), false), p(y0), p(0), x0)
COND(true, y0, s(x0), s(x1)) → COND(and(gr(y0, s(x1)), gr(x0, x1)), p(y0), p(s(x0)), s(x1))
COND(true, s(x0), y1, 0) → COND(and(true, gr(y1, 0)), p(s(x0)), p(y1), 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Rewriting
                              ↳ QDP
                                ↳ DependencyGraphProof
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ DependencyGraphProof
                                          ↳ QDP
                                            ↳ Rewriting
                                              ↳ QDP
                                                ↳ Rewriting
                                                  ↳ QDP
                                                    ↳ DependencyGraphProof
                                                      ↳ QDP
                                                        ↳ Rewriting

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

COND(true, y0, s(x0), 0) → COND(and(gr(y0, 0), true), p(y0), x0, 0)
COND(true, s(x0), y1, s(x1)) → COND(and(gr(x0, x1), gr(y1, s(x1))), p(s(x0)), p(y1), s(x1))
COND(true, y0, 0, x0) → COND(and(gr(y0, x0), false), p(y0), p(0), x0)
COND(true, y0, s(x0), s(x1)) → COND(and(gr(y0, s(x1)), gr(x0, x1)), p(y0), p(s(x0)), s(x1))
COND(true, s(x0), y1, 0) → COND(and(true, gr(y1, 0)), p(s(x0)), p(y1), 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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 rewriting [15] the rule COND(true, s(x0), y1, s(x1)) → COND(and(gr(x0, x1), gr(y1, s(x1))), p(s(x0)), p(y1), s(x1)) at position [1] we obtained the following new rules:

COND(true, s(x0), y1, s(x1)) → COND(and(gr(x0, x1), gr(y1, s(x1))), x0, p(y1), s(x1))



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Rewriting
                              ↳ QDP
                                ↳ DependencyGraphProof
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ DependencyGraphProof
                                          ↳ QDP
                                            ↳ Rewriting
                                              ↳ QDP
                                                ↳ Rewriting
                                                  ↳ QDP
                                                    ↳ DependencyGraphProof
                                                      ↳ QDP
                                                        ↳ Rewriting
                                                          ↳ QDP
                                                            ↳ Rewriting

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

COND(true, s(x0), y1, s(x1)) → COND(and(gr(x0, x1), gr(y1, s(x1))), x0, p(y1), s(x1))
COND(true, y0, s(x0), 0) → COND(and(gr(y0, 0), true), p(y0), x0, 0)
COND(true, y0, 0, x0) → COND(and(gr(y0, x0), false), p(y0), p(0), x0)
COND(true, y0, s(x0), s(x1)) → COND(and(gr(y0, s(x1)), gr(x0, x1)), p(y0), p(s(x0)), s(x1))
COND(true, s(x0), y1, 0) → COND(and(true, gr(y1, 0)), p(s(x0)), p(y1), 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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 rewriting [15] the rule COND(true, y0, 0, x0) → COND(and(gr(y0, x0), false), p(y0), p(0), x0) at position [0] we obtained the following new rules:

COND(true, y0, 0, x0) → COND(false, p(y0), p(0), x0)



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Rewriting
                              ↳ QDP
                                ↳ DependencyGraphProof
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ DependencyGraphProof
                                          ↳ QDP
                                            ↳ Rewriting
                                              ↳ QDP
                                                ↳ Rewriting
                                                  ↳ QDP
                                                    ↳ DependencyGraphProof
                                                      ↳ QDP
                                                        ↳ Rewriting
                                                          ↳ QDP
                                                            ↳ Rewriting
                                                              ↳ QDP
                                                                ↳ DependencyGraphProof

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

COND(true, y0, s(x0), 0) → COND(and(gr(y0, 0), true), p(y0), x0, 0)
COND(true, s(x0), y1, s(x1)) → COND(and(gr(x0, x1), gr(y1, s(x1))), x0, p(y1), s(x1))
COND(true, y0, 0, x0) → COND(false, p(y0), p(0), x0)
COND(true, y0, s(x0), s(x1)) → COND(and(gr(y0, s(x1)), gr(x0, x1)), p(y0), p(s(x0)), s(x1))
COND(true, s(x0), y1, 0) → COND(and(true, gr(y1, 0)), p(s(x0)), p(y1), 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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 2 SCCs with 1 less node.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Rewriting
                              ↳ QDP
                                ↳ DependencyGraphProof
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ DependencyGraphProof
                                          ↳ QDP
                                            ↳ Rewriting
                                              ↳ QDP
                                                ↳ Rewriting
                                                  ↳ QDP
                                                    ↳ DependencyGraphProof
                                                      ↳ QDP
                                                        ↳ Rewriting
                                                          ↳ QDP
                                                            ↳ Rewriting
                                                              ↳ QDP
                                                                ↳ DependencyGraphProof
                                                                  ↳ AND
                                                                    ↳ QDP
                                                                      ↳ UsableRulesProof
                                                                    ↳ QDP

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

COND(true, y0, s(x0), 0) → COND(and(gr(y0, 0), true), p(y0), x0, 0)
COND(true, s(x0), y1, 0) → COND(and(true, gr(y1, 0)), p(s(x0)), p(y1), 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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.
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
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Rewriting
                              ↳ QDP
                                ↳ DependencyGraphProof
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ DependencyGraphProof
                                          ↳ QDP
                                            ↳ Rewriting
                                              ↳ QDP
                                                ↳ Rewriting
                                                  ↳ QDP
                                                    ↳ DependencyGraphProof
                                                      ↳ QDP
                                                        ↳ Rewriting
                                                          ↳ QDP
                                                            ↳ Rewriting
                                                              ↳ QDP
                                                                ↳ DependencyGraphProof
                                                                  ↳ AND
                                                                    ↳ QDP
                                                                      ↳ UsableRulesProof
                                                                        ↳ QDP
                                                                          ↳ Rewriting
                                                                    ↳ QDP

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

COND(true, y0, s(x0), 0) → COND(and(gr(y0, 0), true), p(y0), x0, 0)
COND(true, s(x0), y1, 0) → COND(and(true, gr(y1, 0)), p(s(x0)), p(y1), 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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 rewriting [15] the rule COND(true, s(x0), y1, 0) → COND(and(true, gr(y1, 0)), p(s(x0)), p(y1), 0) at position [1] we obtained the following new rules:

COND(true, s(x0), y1, 0) → COND(and(true, gr(y1, 0)), x0, p(y1), 0)



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Rewriting
                              ↳ QDP
                                ↳ DependencyGraphProof
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ DependencyGraphProof
                                          ↳ QDP
                                            ↳ Rewriting
                                              ↳ QDP
                                                ↳ Rewriting
                                                  ↳ QDP
                                                    ↳ DependencyGraphProof
                                                      ↳ QDP
                                                        ↳ Rewriting
                                                          ↳ QDP
                                                            ↳ Rewriting
                                                              ↳ QDP
                                                                ↳ DependencyGraphProof
                                                                  ↳ AND
                                                                    ↳ QDP
                                                                      ↳ UsableRulesProof
                                                                        ↳ QDP
                                                                          ↳ Rewriting
                                                                            ↳ QDP
                                                                              ↳ QDPOrderProof
                                                                    ↳ QDP

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

COND(true, y0, s(x0), 0) → COND(and(gr(y0, 0), true), p(y0), x0, 0)
COND(true, s(x0), y1, 0) → COND(and(true, gr(y1, 0)), x0, p(y1), 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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 use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


COND(true, y0, s(x0), 0) → COND(and(gr(y0, 0), true), p(y0), x0, 0)
COND(true, s(x0), y1, 0) → COND(and(true, gr(y1, 0)), x0, p(y1), 0)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25]:

POL(0) = 0   
POL(COND(x1, x2, x3, x4)) = x2 + x3   
POL(and(x1, x2)) = 0   
POL(false) = 0   
POL(gr(x1, x2)) = 0   
POL(p(x1)) = x1   
POL(s(x1)) = 1 + x1   
POL(true) = 0   

The following usable rules [17] were oriented:

p(s(x)) → x
p(0) → 0



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Rewriting
                              ↳ QDP
                                ↳ DependencyGraphProof
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ DependencyGraphProof
                                          ↳ QDP
                                            ↳ Rewriting
                                              ↳ QDP
                                                ↳ Rewriting
                                                  ↳ QDP
                                                    ↳ DependencyGraphProof
                                                      ↳ QDP
                                                        ↳ Rewriting
                                                          ↳ QDP
                                                            ↳ Rewriting
                                                              ↳ QDP
                                                                ↳ DependencyGraphProof
                                                                  ↳ AND
                                                                    ↳ QDP
                                                                      ↳ UsableRulesProof
                                                                        ↳ QDP
                                                                          ↳ Rewriting
                                                                            ↳ QDP
                                                                              ↳ QDPOrderProof
                                                                                ↳ QDP
                                                                                  ↳ PisEmptyProof
                                                                    ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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 TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Rewriting
                              ↳ QDP
                                ↳ DependencyGraphProof
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ DependencyGraphProof
                                          ↳ QDP
                                            ↳ Rewriting
                                              ↳ QDP
                                                ↳ Rewriting
                                                  ↳ QDP
                                                    ↳ DependencyGraphProof
                                                      ↳ QDP
                                                        ↳ Rewriting
                                                          ↳ QDP
                                                            ↳ Rewriting
                                                              ↳ QDP
                                                                ↳ DependencyGraphProof
                                                                  ↳ AND
                                                                    ↳ QDP
                                                                    ↳ QDP
                                                                      ↳ Rewriting

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

COND(true, s(x0), y1, s(x1)) → COND(and(gr(x0, x1), gr(y1, s(x1))), x0, p(y1), s(x1))
COND(true, y0, s(x0), s(x1)) → COND(and(gr(y0, s(x1)), gr(x0, x1)), p(y0), p(s(x0)), s(x1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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 rewriting [15] the rule COND(true, y0, s(x0), s(x1)) → COND(and(gr(y0, s(x1)), gr(x0, x1)), p(y0), p(s(x0)), s(x1)) at position [2] we obtained the following new rules:

COND(true, y0, s(x0), s(x1)) → COND(and(gr(y0, s(x1)), gr(x0, x1)), p(y0), x0, s(x1))



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Rewriting
                              ↳ QDP
                                ↳ DependencyGraphProof
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ DependencyGraphProof
                                          ↳ QDP
                                            ↳ Rewriting
                                              ↳ QDP
                                                ↳ Rewriting
                                                  ↳ QDP
                                                    ↳ DependencyGraphProof
                                                      ↳ QDP
                                                        ↳ Rewriting
                                                          ↳ QDP
                                                            ↳ Rewriting
                                                              ↳ QDP
                                                                ↳ DependencyGraphProof
                                                                  ↳ AND
                                                                    ↳ QDP
                                                                    ↳ QDP
                                                                      ↳ Rewriting
                                                                        ↳ QDP
                                                                          ↳ QDPOrderProof

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

COND(true, s(x0), y1, s(x1)) → COND(and(gr(x0, x1), gr(y1, s(x1))), x0, p(y1), s(x1))
COND(true, y0, s(x0), s(x1)) → COND(and(gr(y0, s(x1)), gr(x0, x1)), p(y0), x0, s(x1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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 use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


COND(true, y0, s(x0), s(x1)) → COND(and(gr(y0, s(x1)), gr(x0, x1)), p(y0), x0, s(x1))
The remaining pairs can at least be oriented weakly.

COND(true, s(x0), y1, s(x1)) → COND(and(gr(x0, x1), gr(y1, s(x1))), x0, p(y1), s(x1))
Used ordering: Polynomial interpretation [25]:

POL(0) = 0   
POL(COND(x1, x2, x3, x4)) = x3   
POL(and(x1, x2)) = 1 + x1 + x2   
POL(false) = 1   
POL(gr(x1, x2)) = 0   
POL(p(x1)) = x1   
POL(s(x1)) = 1 + x1   
POL(true) = 0   

The following usable rules [17] were oriented:

p(s(x)) → x
p(0) → 0



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Rewriting
                              ↳ QDP
                                ↳ DependencyGraphProof
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ DependencyGraphProof
                                          ↳ QDP
                                            ↳ Rewriting
                                              ↳ QDP
                                                ↳ Rewriting
                                                  ↳ QDP
                                                    ↳ DependencyGraphProof
                                                      ↳ QDP
                                                        ↳ Rewriting
                                                          ↳ QDP
                                                            ↳ Rewriting
                                                              ↳ QDP
                                                                ↳ DependencyGraphProof
                                                                  ↳ AND
                                                                    ↳ QDP
                                                                    ↳ QDP
                                                                      ↳ Rewriting
                                                                        ↳ QDP
                                                                          ↳ QDPOrderProof
                                                                            ↳ QDP
                                                                              ↳ QDPSizeChangeProof

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

COND(true, s(x0), y1, s(x1)) → COND(and(gr(x0, x1), gr(y1, s(x1))), x0, p(y1), s(x1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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