YES Termination w.r.t. Q proof of ../tpdb/TRS/secret06/aprove/division.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:

division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

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

division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof

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

division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(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:

DIVISION(x, y) → DIV(x, y, 0)
DIV(x, y, z) → LT(x, y)
LT(s(x), s(y)) → LT(x, y)
MINUS(s(x), s(y)) → MINUS(x, y)
INC(s(x)) → INC(x)
DIV(x, y, z) → INC(z)
IF(false, x, s(y), z) → DIV(minus(x, s(y)), s(y), z)
DIV(x, y, z) → IF(lt(x, y), x, y, inc(z))
IF(false, x, s(y), z) → MINUS(x, s(y))

The TRS R consists of the following rules:

division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

DIVISION(x, y) → DIV(x, y, 0)
DIV(x, y, z) → LT(x, y)
LT(s(x), s(y)) → LT(x, y)
MINUS(s(x), s(y)) → MINUS(x, y)
INC(s(x)) → INC(x)
DIV(x, y, z) → INC(z)
IF(false, x, s(y), z) → DIV(minus(x, s(y)), s(y), z)
DIV(x, y, z) → IF(lt(x, y), x, y, inc(z))
IF(false, x, s(y), z) → MINUS(x, s(y))

The TRS R consists of the following rules:

division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

INC(s(x)) → INC(x)

The TRS R consists of the following rules:

division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

INC(s(x)) → INC(x)

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

division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(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.

division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

INC(s(x)) → INC(x)

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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
              ↳ QDP
              ↳ QDP

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

LT(s(x), s(y)) → LT(x, y)

The TRS R consists of the following rules:

division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
              ↳ QDP
              ↳ QDP

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

LT(s(x), s(y)) → LT(x, y)

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

division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(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.

division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof
              ↳ QDP
              ↳ QDP

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

LT(s(x), s(y)) → LT(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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
              ↳ QDP

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

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
              ↳ QDP

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

MINUS(s(x), s(y)) → MINUS(x, y)

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

division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(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.

division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof
              ↳ QDP

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

MINUS(s(x), s(y)) → MINUS(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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof

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

DIV(x, y, z) → IF(lt(x, y), x, y, inc(z))
IF(false, x, s(y), z) → DIV(minus(x, s(y)), s(y), z)

The TRS R consists of the following rules:

division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof

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

DIV(x, y, z) → IF(lt(x, y), x, y, inc(z))
IF(false, x, s(y), z) → DIV(minus(x, s(y)), s(y), z)

The TRS R consists of the following rules:

minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(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.

division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing

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

IF(false, x, s(y), z) → DIV(minus(x, s(y)), s(y), z)
DIV(x, y, z) → IF(lt(x, y), x, y, inc(z))

The TRS R consists of the following rules:

minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule DIV(x, y, z) → IF(lt(x, y), x, y, inc(z)) at position [0] we obtained the following new rules:

DIV(s(x0), s(x1), y2) → IF(lt(x0, x1), s(x0), s(x1), inc(y2))
DIV(0, s(x0), y2) → IF(true, 0, s(x0), inc(y2))
DIV(x0, 0, y2) → IF(false, x0, 0, inc(y2))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof

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

DIV(x0, 0, y2) → IF(false, x0, 0, inc(y2))
DIV(s(x0), s(x1), y2) → IF(lt(x0, x1), s(x0), s(x1), inc(y2))
DIV(0, s(x0), y2) → IF(true, 0, s(x0), inc(y2))
IF(false, x, s(y), z) → DIV(minus(x, s(y)), s(y), z)

The TRS R consists of the following rules:

minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(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 2 less nodes.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Narrowing

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

DIV(s(x0), s(x1), y2) → IF(lt(x0, x1), s(x0), s(x1), inc(y2))
IF(false, x, s(y), z) → DIV(minus(x, s(y)), s(y), z)

The TRS R consists of the following rules:

minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule IF(false, x, s(y), z) → DIV(minus(x, s(y)), s(y), z) at position [0] we obtained the following new rules:

IF(false, s(x0), s(x1), y2) → DIV(minus(x0, x1), s(x1), y2)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Narrowing
                                  ↳ QDP
                                    ↳ QDPOrderProof
                                    ↳ QDPOrderProof

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

IF(false, s(x0), s(x1), y2) → DIV(minus(x0, x1), s(x1), y2)
DIV(s(x0), s(x1), y2) → IF(lt(x0, x1), s(x0), s(x1), inc(y2))

The TRS R consists of the following rules:

minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(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.


IF(false, s(x0), s(x1), y2) → DIV(minus(x0, x1), s(x1), y2)
DIV(s(x0), s(x1), y2) → IF(lt(x0, x1), s(x0), s(x1), inc(y2))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
IF(x1, x2, x3, x4)  =  IF(x1, x2, x3, x4)
false  =  false
s(x1)  =  s(x1)
DIV(x1, x2, x3)  =  DIV(x1, x2, x3)
minus(x1, x2)  =  x1
lt(x1, x2)  =  lt
inc(x1)  =  inc(x1)
0  =  0
true  =  true

Recursive path order with status [2].
Quasi-Precedence:
[IF4, DIV3] > [false, s1, lt, inc1] > [0, true]

Status:
true: multiset
DIV3: [1,2,3]
false: multiset
IF4: [2,1,4,3]
lt: []
0: multiset
inc1: [1]
s1: [1]


The following usable rules [17] were oriented:

minus(s(x), s(y)) → minus(x, y)
lt(s(x), s(y)) → lt(x, y)
inc(s(x)) → s(inc(x))
lt(0, s(y)) → true
minus(x, 0) → x
inc(0) → s(0)
lt(x, 0) → false



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Narrowing
                                  ↳ QDP
                                    ↳ QDPOrderProof
                                      ↳ QDP
                                        ↳ PisEmptyProof
                                    ↳ QDPOrderProof

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

minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(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.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


IF(false, s(x0), s(x1), y2) → DIV(minus(x0, x1), s(x1), y2)
The remaining pairs can at least be oriented weakly.

DIV(s(x0), s(x1), y2) → IF(lt(x0, x1), s(x0), s(x1), inc(y2))
Used ordering: Polynomial interpretation [25]:

POL(0) = 0   
POL(DIV(x1, x2, x3)) = x1   
POL(IF(x1, x2, x3, x4)) = x2   
POL(false) = 0   
POL(inc(x1)) = 0   
POL(lt(x1, x2)) = 0   
POL(minus(x1, x2)) = x1   
POL(s(x1)) = 1 + x1   
POL(true) = 0   

The following usable rules [17] were oriented:

minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Narrowing
                                  ↳ QDP
                                    ↳ QDPOrderProof
                                    ↳ QDPOrderProof
                                      ↳ QDP

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

DIV(s(x0), s(x1), y2) → IF(lt(x0, x1), s(x0), s(x1), inc(y2))

The TRS R consists of the following rules:

minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))

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