Termination w.r.t. Q proof of /tmp/tmpTJ-2VI/mod.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:

minus(0, y) → 0
minus(s(x), y) → if(gt(s(x), y), x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
mod(x, 0) → 0
mod(x, s(y)) → if1(lt(x, s(y)), x, s(y))
if1(true, x, y) → x
if1(false, x, y) → mod(minus(x, y), y)
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
lt(x, 0) → false
lt(0, s(x)) → true
lt(s(x), s(y)) → lt(x, y)

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

minus(0, y) → 0
minus(s(x), y) → if(gt(s(x), y), x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
mod(x, 0) → 0
mod(x, s(y)) → if1(lt(x, s(y)), x, s(y))
if1(true, x, y) → x
if1(false, x, y) → mod(minus(x, y), y)
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
lt(x, 0) → false
lt(0, s(x)) → true
lt(s(x), s(y)) → lt(x, y)

Q is empty.

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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

minus(0, y) → 0
minus(s(x), y) → if(gt(s(x), y), x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
mod(x, 0) → 0
mod(x, s(y)) → if1(lt(x, s(y)), x, s(y))
if1(true, x, y) → x
if1(false, x, y) → mod(minus(x, y), y)
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
lt(x, 0) → false
lt(0, s(x)) → true
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
mod(x0, 0)
mod(x0, s(x1))
if1(true, x0, x1)
if1(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

MINUS(s(x), y) → IF(gt(s(x), y), x, y)
MINUS(s(x), y) → GT(s(x), y)
IF(true, x, y) → MINUS(x, y)
MOD(x, s(y)) → IF1(lt(x, s(y)), x, s(y))
MOD(x, s(y)) → LT(x, s(y))
IF1(false, x, y) → MOD(minus(x, y), y)
IF1(false, x, y) → MINUS(x, y)
GT(s(x), s(y)) → GT(x, y)
LT(s(x), s(y)) → LT(x, y)

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), y) → if(gt(s(x), y), x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
mod(x, 0) → 0
mod(x, s(y)) → if1(lt(x, s(y)), x, s(y))
if1(true, x, y) → x
if1(false, x, y) → mod(minus(x, y), y)
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
lt(x, 0) → false
lt(0, s(x)) → true
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
mod(x0, 0)
mod(x0, s(x1))
if1(true, x0, x1)
if1(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

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:

MINUS(s(x), y) → IF(gt(s(x), y), x, y)
MINUS(s(x), y) → GT(s(x), y)
IF(true, x, y) → MINUS(x, y)
MOD(x, s(y)) → IF1(lt(x, s(y)), x, s(y))
MOD(x, s(y)) → LT(x, s(y))
IF1(false, x, y) → MOD(minus(x, y), y)
IF1(false, x, y) → MINUS(x, y)
GT(s(x), s(y)) → GT(x, y)
LT(s(x), s(y)) → LT(x, y)

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), y) → if(gt(s(x), y), x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
mod(x, 0) → 0
mod(x, s(y)) → if1(lt(x, s(y)), x, s(y))
if1(true, x, y) → x
if1(false, x, y) → mod(minus(x, y), y)
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
lt(x, 0) → false
lt(0, s(x)) → true
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
mod(x0, 0)
mod(x0, s(x1))
if1(true, x0, x1)
if1(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 3 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:

LT(s(x), s(y)) → LT(x, y)

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), y) → if(gt(s(x), y), x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
mod(x, 0) → 0
mod(x, s(y)) → if1(lt(x, s(y)), x, s(y))
if1(true, x, y) → x
if1(false, x, y) → mod(minus(x, y), y)
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
lt(x, 0) → false
lt(0, s(x)) → true
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
mod(x0, 0)
mod(x0, s(x1))
if1(true, x0, x1)
if1(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

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 [LPAR04] we can delete all non-usable rules [FROCOS05] 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:

LT(s(x), s(y)) → LT(x, y)

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

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
mod(x0, 0)
mod(x0, s(x1))
if1(true, x0, x1)
if1(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

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.[THIEMANN].

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
mod(x0, 0)
mod(x0, s(x1))
if1(true, x0, x1)
if1(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

↳ 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:

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 [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

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

↳ 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:

GT(s(x), s(y)) → GT(x, y)

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), y) → if(gt(s(x), y), x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
mod(x, 0) → 0
mod(x, s(y)) → if1(lt(x, s(y)), x, s(y))
if1(true, x, y) → x
if1(false, x, y) → mod(minus(x, y), y)
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
lt(x, 0) → false
lt(0, s(x)) → true
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
mod(x0, 0)
mod(x0, s(x1))
if1(true, x0, x1)
if1(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

↳ 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:

GT(s(x), s(y)) → GT(x, y)

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

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
mod(x0, 0)
mod(x0, s(x1))
if1(true, x0, x1)
if1(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

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.[THIEMANN].

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
mod(x0, 0)
mod(x0, s(x1))
if1(true, x0, x1)
if1(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

↳ 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:

GT(s(x), s(y)) → GT(x, y)

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

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

↳ 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:

IF(true, x, y) → MINUS(x, y)
MINUS(s(x), y) → IF(gt(s(x), y), x, y)

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), y) → if(gt(s(x), y), x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
mod(x, 0) → 0
mod(x, s(y)) → if1(lt(x, s(y)), x, s(y))
if1(true, x, y) → x
if1(false, x, y) → mod(minus(x, y), y)
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
lt(x, 0) → false
lt(0, s(x)) → true
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
mod(x0, 0)
mod(x0, s(x1))
if1(true, x0, x1)
if1(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

↳ 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:

IF(true, x, y) → MINUS(x, y)
MINUS(s(x), y) → IF(gt(s(x), y), x, y)

The TRS R consists of the following rules:

gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
gt(0, y) → false

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
mod(x0, 0)
mod(x0, s(x1))
if1(true, x0, x1)
if1(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

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.[THIEMANN].

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
mod(x0, 0)
mod(x0, s(x1))
if1(true, x0, x1)
if1(false, x0, x1)
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

↳ 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:

IF(true, x, y) → MINUS(x, y)
MINUS(s(x), y) → IF(gt(s(x), y), x, y)

The TRS R consists of the following rules:

gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
gt(0, y) → false

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

MINUS(s(x), y) → IF(gt(s(x), y), x, y)
The graph contains the following edges 1 > 2, 2 >= 3
IF(true, x, y) → MINUS(x, y)
The graph contains the following edges 2 >= 1, 3 >= 2

↳ 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:

IF1(false, x, y) → MOD(minus(x, y), y)
MOD(x, s(y)) → IF1(lt(x, s(y)), x, s(y))

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), y) → if(gt(s(x), y), x, y)
if(true, x, y) → s(minus(x, y))
if(false, x, y) → 0
mod(x, 0) → 0
mod(x, s(y)) → if1(lt(x, s(y)), x, s(y))
if1(true, x, y) → x
if1(false, x, y) → mod(minus(x, y), y)
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
lt(x, 0) → false
lt(0, s(x)) → true
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
mod(x0, 0)
mod(x0, s(x1))
if1(true, x0, x1)
if1(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

↳ 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:

IF1(false, x, y) → MOD(minus(x, y), y)
MOD(x, s(y)) → IF1(lt(x, s(y)), x, s(y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
mod(x0, 0)
mod(x0, s(x1))
if1(true, x0, x1)
if1(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

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.[THIEMANN].

mod(x0, 0)
mod(x0, s(x1))
if1(true, x0, x1)
if1(false, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Instantiation

                        ↳ Narrowing

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

IF1(false, x, y) → MOD(minus(x, y), y)
MOD(x, s(y)) → IF1(lt(x, s(y)), x, s(y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [LPAR04] the rule IF1(false, x, y) → MOD(minus(x, y), y) we obtained the following new rules [LPAR04]:

IF1(false, z0, s(z1)) → MOD(minus(z0, s(z1)), s(z1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Instantiation

                          ↳ QDP

                        ↳ Narrowing

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

MOD(x, s(y)) → IF1(lt(x, s(y)), x, s(y))
IF1(false, z0, s(z1)) → MOD(minus(z0, s(z1)), s(z1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [LPAR04] the rule MOD(x, s(y)) → IF1(lt(x, s(y)), x, s(y)) at position [0] we obtained the following new rules [LPAR04]:

MOD(0, s(x0)) → IF1(true, 0, s(x0))
MOD(s(x0), s(x1)) → IF1(lt(x0, x1), s(x0), s(x1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Instantiation

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

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

IF1(false, x, y) → MOD(minus(x, y), y)
MOD(0, s(x0)) → IF1(true, 0, s(x0))
MOD(s(x0), s(x1)) → IF1(lt(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Instantiation

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

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

MOD(s(x0), s(x1)) → IF1(lt(x0, x1), s(x0), s(x1))
IF1(false, x, y) → MOD(minus(x, y), y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [LPAR04] the rule IF1(false, x, y) → MOD(minus(x, y), y) at position [0] we obtained the following new rules [LPAR04]:

IF1(false, 0, x0) → MOD(0, x0)
IF1(false, s(x0), x1) → MOD(if(gt(s(x0), x1), x0, x1), x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Instantiation

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

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

MOD(s(x0), s(x1)) → IF1(lt(x0, x1), s(x0), s(x1))
IF1(false, 0, x0) → MOD(0, x0)
IF1(false, s(x0), x1) → MOD(if(gt(s(x0), x1), x0, x1), x1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Instantiation

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Instantiation

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

IF1(false, s(x0), x1) → MOD(if(gt(s(x0), x1), x0, x1), x1)
MOD(s(x0), s(x1)) → IF1(lt(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [LPAR04] the rule IF1(false, s(x0), x1) → MOD(if(gt(s(x0), x1), x0, x1), x1) we obtained the following new rules [LPAR04]:

IF1(false, s(z0), s(z1)) → MOD(if(gt(s(z0), s(z1)), z0, s(z1)), s(z1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Instantiation

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ Rewriting

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

MOD(s(x0), s(x1)) → IF1(lt(x0, x1), s(x0), s(x1))
IF1(false, s(z0), s(z1)) → MOD(if(gt(s(z0), s(z1)), z0, s(z1)), s(z1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [LPAR04] the rule IF1(false, s(z0), s(z1)) → MOD(if(gt(s(z0), s(z1)), z0, s(z1)), s(z1)) at position [0,0] we obtained the following new rules [LPAR04]:

IF1(false, s(z0), s(z1)) → MOD(if(gt(z0, z1), z0, s(z1)), s(z1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Instantiation

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ Induction-Processor

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

MOD(s(x0), s(x1)) → IF1(lt(x0, x1), s(x0), s(x1))
IF1(false, s(z0), s(z1)) → MOD(if(gt(z0, z1), z0, s(z1)), s(z1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

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

This DP could be deleted by the Induction-Processor:
IF1(false, s(z0'), s(z1')) → MOD(if(gt(z0', z1'), z0', s(z1')), s(z1'))

This order was computed:
Polynomial interpretation [POLO]:

POL(0) = 0
POL(IF1(x₁, x₂, x₃)) = x₁ + x₂
POL(MOD(x₁, x₂)) = 1 + x₁
POL(false) = 1
POL(gt(x₁, x₂)) = 1
POL(if(x₁, x₂, x₃)) = 1 + x₂
POL(lt(x₁, x₂)) = 1
POL(minus(x₁, x₂)) = x₁
POL(s(x₁)) = 1 + x₁
POL(true) = 0

At least one of these decreasing rules is always used after the deleted DP:
if(false, x413, y353) → 0

The following formula is valid:
z0':sort[a0],z1':sort[a0].if'(gt(z0' , z1' ), z0' , s(z1' ))=true

The transformed set:
minus'(0, y3) → false
minus'(s(x13), y11) → if'(gt(s(x13), y11), x13, y11)
if'(true, x32, y27) → minus'(x32, y27)
if'(false, x41, y35) → true
gt(s(x'), 0) → true
minus(0, y3) → 0
minus(s(x13), y11) → if(gt(s(x13), y11), x13, y11)
gt(s(x23), s(y19)) → gt(x23, y19)
if(true, x32, y27) → s(minus(x32, y27))
if(false, x41, y35) → 0
gt(0, y43) → false
lt(0, s(x58)) → true
lt(s(x67), s(y58)) → lt(x67, y58)
lt(x76, 0) → false
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a39](witness_sort[a39], witness_sort[a39]) → true

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Instantiation

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ Induction-Processor

                                                  ↳ AND

                                                    ↳ QDP

                                                      ↳ DependencyGraphProof

                                                    ↳ QTRS

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

MOD(s(x0), s(x1)) → IF1(lt(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Instantiation

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ Induction-Processor

                                                  ↳ AND

                                                    ↳ QDP

                                                    ↳ QTRS

                                                      ↳ QTRSRRRProof

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

minus'(0, y3) → false
minus'(s(x13), y11) → if'(gt(s(x13), y11), x13, y11)
if'(true, x32, y27) → minus'(x32, y27)
if'(false, x41, y35) → true
gt(s(x'), 0) → true
minus(0, y3) → 0
minus(s(x13), y11) → if(gt(s(x13), y11), x13, y11)
gt(s(x23), s(y19)) → gt(x23, y19)
if(true, x32, y27) → s(minus(x32, y27))
if(false, x41, y35) → 0
gt(0, y43) → false
lt(0, s(x58)) → true
lt(s(x67), s(y58)) → lt(x67, y58)
lt(x76, 0) → false
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a39](witness_sort[a39], witness_sort[a39]) → true

Q is empty.

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

minus'(0, y3) → false
minus'(s(x13), y11) → if'(gt(s(x13), y11), x13, y11)
if'(true, x32, y27) → minus'(x32, y27)
if'(false, x41, y35) → true
gt(s(x'), 0) → true
minus(0, y3) → 0
minus(s(x13), y11) → if(gt(s(x13), y11), x13, y11)
gt(s(x23), s(y19)) → gt(x23, y19)
if(true, x32, y27) → s(minus(x32, y27))
if(false, x41, y35) → 0
gt(0, y43) → false
lt(0, s(x58)) → true
lt(s(x67), s(y58)) → lt(x67, y58)
lt(x76, 0) → false
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a39](witness_sort[a39], witness_sort[a39]) → true

Q is empty.
Used ordering:
Recursive path order with status [RPO].
Quasi-Precedence:

[minus'₂, 0, false, if'₃, not₁, isa_false₁] > [true, minus₂, if₃, equal_sort[a0]₂] > s₁
[minus'₂, 0, false, if'₃, not₁, isa_false₁] > [true, minus₂, if₃, equal_sort[a0]₂] > gt₂
equal_sort[a39]₂ > [true, minus₂, if₃, equal_sort[a0]₂] > s₁
equal_sort[a39]₂ > [true, minus₂, if₃, equal_sort[a0]₂] > gt₂

Status:

minus₂: [1,2]
minus'₂: [1,2]
true: multiset
or₂: multiset
and₂: multiset
gt₂: multiset
0: multiset
equal_bool₂: [2,1]
equal_sort[a0]₂: multiset
equal_sort[a39]₂: multiset
if₃: [2,3,1]
if'₃: [2,3,1]
not₁: multiset
witness_sort[a39]: multiset
isa_false₁: multiset
false: multiset
s₁: multiset
lt₂: [2,1]
isa_true₁: multiset

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

minus'(0, y3) → false
minus'(s(x13), y11) → if'(gt(s(x13), y11), x13, y11)
if'(true, x32, y27) → minus'(x32, y27)
if'(false, x41, y35) → true
gt(s(x'), 0) → true
minus(0, y3) → 0
minus(s(x13), y11) → if(gt(s(x13), y11), x13, y11)
gt(s(x23), s(y19)) → gt(x23, y19)
if(true, x32, y27) → s(minus(x32, y27))
if(false, x41, y35) → 0
gt(0, y43) → false
lt(0, s(x58)) → true
lt(s(x67), s(y58)) → lt(x67, y58)
lt(x76, 0) → false
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a39](witness_sort[a39], witness_sort[a39]) → true

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Instantiation

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ Induction-Processor

                                                  ↳ AND

                                                    ↳ QDP

                                                    ↳ QTRS

                                                      ↳ QTRSRRRProof

                                                        ↳ QTRS

                                                          ↳ RisEmptyProof

                                                          ↳ RisEmptyProof

                                                          ↳ RisEmptyProof

Q restricted rewrite system:
R is empty.
Q is empty.

The TRS R is empty. Hence, termination is trivially proven.
The TRS R is empty. Hence, termination is trivially proven.
The TRS R is empty. Hence, termination is trivially proven.