Termination w.r.t. Q proof of /tmp/aprove_benchmark_example

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:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

Q is empty.

↳ QTRS

  ↳ AAECC Innermost

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

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

Q is empty.

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

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x

The TRS R 2 is

f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

The signature Sigma is {f}

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

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

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(x1))

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

-¹(s(x), s(y)) → -¹(x, y)
F(x, s(y)) → P(-(s(y), x))
F(x, s(y)) → P(-(x, s(y)))
F(x, s(y)) → F(p(-(x, s(y))), p(-(s(y), x)))
F(s(x), y) → -¹(y, s(x))
F(s(x), y) → -¹(s(x), y)
F(x, s(y)) → -¹(s(y), x)
F(x, s(y)) → -¹(x, s(y))
F(s(x), y) → F(p(-(s(x), y)), p(-(y, s(x))))
F(s(x), y) → P(-(y, s(x)))
F(s(x), y) → P(-(s(x), y))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(x1))

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:

-¹(s(x), s(y)) → -¹(x, y)
F(x, s(y)) → P(-(s(y), x))
F(x, s(y)) → P(-(x, s(y)))
F(x, s(y)) → F(p(-(x, s(y))), p(-(s(y), x)))
F(s(x), y) → -¹(y, s(x))
F(s(x), y) → -¹(s(x), y)
F(x, s(y)) → -¹(s(y), x)
F(x, s(y)) → -¹(x, s(y))
F(s(x), y) → F(p(-(s(x), y)), p(-(y, s(x))))
F(s(x), y) → P(-(y, s(x)))
F(s(x), y) → P(-(s(x), y))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 8 less nodes.

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

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

-¹(s(x), s(y)) → -¹(x, y)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(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 [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:

-¹(s(x), s(y)) → -¹(x, y)

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

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(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.

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(x1))

↳ 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:

-¹(s(x), s(y)) → -¹(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:

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

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

F(x, s(y)) → F(p(-(x, s(y))), p(-(s(y), x)))
F(s(x), y) → F(p(-(s(x), y)), p(-(y, s(x))))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(x1))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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

F(x, s(y)) → F(p(-(x, s(y))), p(-(s(y), x)))
F(s(x), y) → F(p(-(s(x), y)), p(-(y, s(x))))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(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.

f(s(x0), x1)
f(x0, s(x1))

↳ 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:

F(x, s(y)) → F(p(-(x, s(y))), p(-(s(y), x)))
F(s(x), y) → F(p(-(s(x), y)), p(-(y, s(x))))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x

The set Q consists of the following terms:

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

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

F(s(x1), s(x0)) → F(p(-(s(x1), s(x0))), p(-(x0, x1)))
F(0, s(y1)) → F(p(-(0, s(y1))), p(s(y1)))
F(s(x0), s(x1)) → F(p(-(x0, x1)), p(-(s(x1), s(x0))))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

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

F(s(x0), s(x1)) → F(p(-(x0, x1)), p(-(s(x1), s(x0))))
F(s(x1), s(x0)) → F(p(-(s(x1), s(x0))), p(-(x0, x1)))
F(s(x), y) → F(p(-(s(x), y)), p(-(y, s(x))))
F(0, s(y1)) → F(p(-(0, s(y1))), p(s(y1)))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
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

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Rewriting

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

F(s(x0), s(x1)) → F(p(-(x0, x1)), p(-(s(x1), s(x0))))
F(s(x1), s(x0)) → F(p(-(s(x1), s(x0))), p(-(x0, x1)))
F(s(x), y) → F(p(-(s(x), y)), p(-(y, s(x))))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule F(s(x1), s(x0)) → F(p(-(s(x1), s(x0))), p(-(x0, x1))) at position [0,0] we obtained the following new rules:

F(s(x1), s(x0)) → F(p(-(x1, x0)), p(-(x0, x1)))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

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

F(s(x0), s(x1)) → F(p(-(x0, x1)), p(-(s(x1), s(x0))))
F(s(x), y) → F(p(-(s(x), y)), p(-(y, s(x))))
F(s(x1), s(x0)) → F(p(-(x1, x0)), p(-(x0, x1)))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule F(s(x0), s(x1)) → F(p(-(x0, x1)), p(-(s(x1), s(x0)))) at position [1,0] we obtained the following new rules:

F(s(x0), s(x1)) → F(p(-(x0, x1)), p(-(x1, x0)))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Narrowing

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

F(s(x), y) → F(p(-(s(x), y)), p(-(y, s(x))))
F(s(x1), s(x0)) → F(p(-(x1, x0)), p(-(x0, x1)))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x

The set Q consists of the following terms:

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

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

F(s(x0), s(x1)) → F(p(-(x0, x1)), p(-(s(x1), s(x0))))
F(s(y0), 0) → F(p(s(y0)), p(-(0, s(y0))))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

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

F(s(x0), s(x1)) → F(p(-(x0, x1)), p(-(s(x1), s(x0))))
F(s(x1), s(x0)) → F(p(-(x1, x0)), p(-(x0, x1)))
F(s(y0), 0) → F(p(s(y0)), p(-(0, s(y0))))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
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

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Rewriting

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

F(s(x0), s(x1)) → F(p(-(x0, x1)), p(-(s(x1), s(x0))))
F(s(x1), s(x0)) → F(p(-(x1, x0)), p(-(x0, x1)))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule F(s(x0), s(x1)) → F(p(-(x0, x1)), p(-(s(x1), s(x0)))) at position [1,0] we obtained the following new rules:

F(s(x0), s(x1)) → F(p(-(x0, x1)), p(-(x1, x0)))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ UsableRulesReductionPairsProof

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

F(s(x1), s(x0)) → F(p(-(x1, x0)), p(-(x0, x1)))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

F(s(x1), s(x0)) → F(p(-(x1, x0)), p(-(x0, x1)))

The following rules are removed from R:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x

Used ordering: POLO with Polynomial interpretation [25]:

POL(-(x₁, x₂)) = x₁ + x₂
POL(0) = 0
POL(F(x₁, x₂)) = 2·x₁ + 2·x₂
POL(p(x₁)) = 1 + x₁
POL(s(x₁)) = 2 + 2·x₁

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ UsableRulesReductionPairsProof

                                                      ↳ QDP

                                                        ↳ PisEmptyProof

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

-(x0, 0)
-(s(x0), s(x1))
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.