Termination w.r.t. Q proof of /tmp/tmpJS1ddF/tpa3.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
f(s(x)) → f(-(*(s(s(0)), s(x)), s(s(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)
+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
f(s(x)) → f(-(*(s(s(0)), s(x)), s(s(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)
+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))

The TRS R 2 is

f(s(x)) → f(-(*(s(s(0)), s(x)), s(s(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)
+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
f(s(x)) → f(-(*(s(s(0)), s(x)), s(s(x))))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
f(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:

-¹(s(x), s(y)) → -¹(x, y)
F(s(x)) → -¹(*(s(s(0)), s(x)), s(s(x)))
+¹(s(x), y) → +¹(x, y)
F(s(x)) → *¹(s(s(0)), s(x))
*¹(x, s(y)) → *¹(x, y)
F(s(x)) → F(-(*(s(s(0)), s(x)), s(s(x))))
*¹(x, s(y)) → +¹(x, *(x, y))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
f(s(x)) → f(-(*(s(s(0)), s(x)), s(s(x))))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
f(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:

-¹(s(x), s(y)) → -¹(x, y)
F(s(x)) → -¹(*(s(s(0)), s(x)), s(s(x)))
+¹(s(x), y) → +¹(x, y)
F(s(x)) → *¹(s(s(0)), s(x))
*¹(x, s(y)) → *¹(x, y)
F(s(x)) → F(-(*(s(s(0)), s(x)), s(s(x))))
*¹(x, s(y)) → +¹(x, *(x, y))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
f(s(x)) → f(-(*(s(s(0)), s(x)), s(s(x))))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
f(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 3 less nodes.

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

+¹(s(x), y) → +¹(x, y)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
f(s(x)) → f(-(*(s(s(0)), s(x)), s(s(x))))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
f(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

              ↳ QDP

              ↳ QDP

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

+¹(s(x), y) → +¹(x, y)

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

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
f(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.

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
f(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ 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:

+¹(s(x), 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), 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

              ↳ QDP

              ↳ QDP

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

*¹(x, s(y)) → *¹(x, y)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
f(s(x)) → f(-(*(s(s(0)), s(x)), s(s(x))))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
f(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

              ↳ QDP

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

*¹(x, s(y)) → *¹(x, y)

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

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
f(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.

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
f(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ 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:

*¹(x, s(y)) → *¹(x, y)

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

*¹(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

              ↳ 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)
+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
f(s(x)) → f(-(*(s(s(0)), s(x)), s(s(x))))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
f(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ 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))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
f(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.

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
f(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ 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:

-¹(s(x), s(y)) → -¹(x, y)

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

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                ↳ UsableRulesProof

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

F(s(x)) → F(-(*(s(s(0)), s(x)), s(s(x))))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
+(0, y) → y
+(s(x), y) → s(+(x, y))
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
f(s(x)) → f(-(*(s(s(0)), s(x)), s(s(x))))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
f(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                ↳ UsableRulesProof

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

F(s(x)) → F(-(*(s(s(0)), s(x)), s(s(x))))

The TRS R consists of the following rules:

*(x, s(y)) → +(x, *(x, y))
-(s(x), s(y)) → -(x, y)
-(x, 0) → x
*(x, 0) → 0
+(0, y) → y
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
f(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.

f(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                ↳ UsableRulesProof

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

F(s(x)) → F(-(*(s(s(0)), s(x)), s(s(x))))

The TRS R consists of the following rules:

*(x, s(y)) → +(x, *(x, y))
-(s(x), s(y)) → -(x, y)
-(x, 0) → x
*(x, 0) → 0
+(0, y) → y
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

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

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

F(s(x)) → F(-(+(s(s(0)), *(s(s(0)), x)), s(s(x))))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                ↳ UsableRulesProof

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

F(s(x)) → F(-(+(s(s(0)), *(s(s(0)), x)), s(s(x))))

The TRS R consists of the following rules:

*(x, s(y)) → +(x, *(x, y))
-(s(x), s(y)) → -(x, y)
-(x, 0) → x
*(x, 0) → 0
+(0, y) → y
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

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

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

F(s(x)) → F(-(s(+(s(0), *(s(s(0)), x))), s(s(x))))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                ↳ UsableRulesProof

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

F(s(x)) → F(-(s(+(s(0), *(s(s(0)), x))), s(s(x))))

The TRS R consists of the following rules:

*(x, s(y)) → +(x, *(x, y))
-(s(x), s(y)) → -(x, y)
-(x, 0) → x
*(x, 0) → 0
+(0, y) → y
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

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

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

F(s(x)) → F(-(+(s(0), *(s(s(0)), x)), s(x)))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                ↳ UsableRulesProof

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

F(s(x)) → F(-(+(s(0), *(s(s(0)), x)), s(x)))

The TRS R consists of the following rules:

*(x, s(y)) → +(x, *(x, y))
-(s(x), s(y)) → -(x, y)
-(x, 0) → x
*(x, 0) → 0
+(0, y) → y
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

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

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

F(s(x)) → F(-(s(+(0, *(s(s(0)), x))), s(x)))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Rewriting

                ↳ UsableRulesProof

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

F(s(x)) → F(-(s(+(0, *(s(s(0)), x))), s(x)))

The TRS R consists of the following rules:

*(x, s(y)) → +(x, *(x, y))
-(s(x), s(y)) → -(x, y)
-(x, 0) → x
*(x, 0) → 0
+(0, y) → y
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

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

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

F(s(x)) → F(-(+(0, *(s(s(0)), x)), x))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ Rewriting

                ↳ UsableRulesProof

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

F(s(x)) → F(-(+(0, *(s(s(0)), x)), x))

The TRS R consists of the following rules:

*(x, s(y)) → +(x, *(x, y))
-(s(x), s(y)) → -(x, y)
-(x, 0) → x
*(x, 0) → 0
+(0, y) → y
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

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

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

F(s(x)) → F(-(*(s(s(0)), x), x))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ Narrowing

                ↳ UsableRulesProof

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

F(s(x)) → F(-(*(s(s(0)), x), x))

The TRS R consists of the following rules:

*(x, s(y)) → +(x, *(x, y))
-(s(x), s(y)) → -(x, y)
-(x, 0) → x
*(x, 0) → 0
+(0, y) → y
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

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

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

F(s(s(x1))) → F(-(+(s(s(0)), *(s(s(0)), x1)), s(x1)))
F(s(0)) → F(-(0, 0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ Rewriting

                ↳ UsableRulesProof

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

F(s(0)) → F(-(0, 0))
F(s(s(x1))) → F(-(+(s(s(0)), *(s(s(0)), x1)), s(x1)))

The TRS R consists of the following rules:

*(x, s(y)) → +(x, *(x, y))
-(s(x), s(y)) → -(x, y)
-(x, 0) → x
*(x, 0) → 0
+(0, y) → y
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

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

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

F(s(s(x1))) → F(-(s(+(s(0), *(s(s(0)), x1))), s(x1)))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ Rewriting

                                                      ↳ QDP

                                                        ↳ Rewriting

                ↳ UsableRulesProof

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

F(s(s(x1))) → F(-(s(+(s(0), *(s(s(0)), x1))), s(x1)))
F(s(0)) → F(-(0, 0))

The TRS R consists of the following rules:

*(x, s(y)) → +(x, *(x, y))
-(s(x), s(y)) → -(x, y)
-(x, 0) → x
*(x, 0) → 0
+(0, y) → y
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

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

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

F(s(0)) → F(0)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ Rewriting

                                                      ↳ QDP

                                                        ↳ Rewriting

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                ↳ UsableRulesProof

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

F(s(s(x1))) → F(-(s(+(s(0), *(s(s(0)), x1))), s(x1)))
F(s(0)) → F(0)

The TRS R consists of the following rules:

*(x, s(y)) → +(x, *(x, y))
-(s(x), s(y)) → -(x, y)
-(x, 0) → x
*(x, 0) → 0
+(0, y) → y
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

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

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

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ Rewriting

                                                      ↳ QDP

                                                        ↳ Rewriting

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ Rewriting

                ↳ UsableRulesProof

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

F(s(s(x1))) → F(-(s(+(s(0), *(s(s(0)), x1))), s(x1)))

The TRS R consists of the following rules:

*(x, s(y)) → +(x, *(x, y))
-(s(x), s(y)) → -(x, y)
-(x, 0) → x
*(x, 0) → 0
+(0, y) → y
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

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

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

F(s(s(x1))) → F(-(+(s(0), *(s(s(0)), x1)), x1))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ Rewriting

                                                      ↳ QDP

                                                        ↳ Rewriting

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

                ↳ UsableRulesProof

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

F(s(s(x1))) → F(-(+(s(0), *(s(s(0)), x1)), x1))

The TRS R consists of the following rules:

*(x, s(y)) → +(x, *(x, y))
-(s(x), s(y)) → -(x, y)
-(x, 0) → x
*(x, 0) → 0
+(0, y) → y
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

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

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

F(s(s(x1))) → F(-(s(+(0, *(s(s(0)), x1))), x1))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ Rewriting

                                                      ↳ QDP

                                                        ↳ Rewriting

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

                                                                      ↳ QDP

                                                                        ↳ Rewriting

                ↳ UsableRulesProof

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

F(s(s(x1))) → F(-(s(+(0, *(s(s(0)), x1))), x1))

The TRS R consists of the following rules:

*(x, s(y)) → +(x, *(x, y))
-(s(x), s(y)) → -(x, y)
-(x, 0) → x
*(x, 0) → 0
+(0, y) → y
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

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

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

F(s(s(x1))) → F(-(s(*(s(s(0)), x1)), x1))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ Rewriting

                                                      ↳ QDP

                                                        ↳ Rewriting

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

                                                                      ↳ QDP

                                                                        ↳ Rewriting

                                                                          ↳ QDP

                                                                            ↳ MNOCProof

                ↳ UsableRulesProof

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

F(s(s(x1))) → F(-(s(*(s(s(0)), x1)), x1))

The TRS R consists of the following rules:

*(x, s(y)) → +(x, *(x, y))
-(s(x), s(y)) → -(x, y)
-(x, 0) → x
*(x, 0) → 0
+(0, y) → y
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ Rewriting

                                                      ↳ QDP

                                                        ↳ Rewriting

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

                                                                      ↳ QDP

                                                                        ↳ Rewriting

                                                                          ↳ QDP

                                                                            ↳ MNOCProof

                                                                              ↳ QDP

                ↳ UsableRulesProof

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

F(s(s(x1))) → F(-(s(*(s(s(0)), x1)), x1))

The TRS R consists of the following rules:

*(x, s(y)) → +(x, *(x, y))
-(s(x), s(y)) → -(x, y)
-(x, 0) → x
*(x, 0) → 0
+(0, y) → y
+(s(x), y) → s(+(x, y))

Q is empty.
We have to consider all (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

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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

F(s(x)) → F(-(*(s(s(0)), s(x)), s(s(x))))

The TRS R consists of the following rules:

*(x, s(y)) → +(x, *(x, y))
-(s(x), s(y)) → -(x, y)
-(x, 0) → x
*(x, 0) → 0
+(0, y) → y
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
f(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.

f(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ MNOCProof

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

F(s(x)) → F(-(*(s(s(0)), s(x)), s(s(x))))

The TRS R consists of the following rules:

*(x, s(y)) → +(x, *(x, y))
-(s(x), s(y)) → -(x, y)
-(x, 0) → x
*(x, 0) → 0
+(0, y) → y
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ MNOCProof

                          ↳ QDP

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

F(s(x)) → F(-(*(s(s(0)), s(x)), s(s(x))))

The TRS R consists of the following rules:

*(x, s(y)) → +(x, *(x, y))
-(s(x), s(y)) → -(x, y)
-(x, 0) → x
*(x, 0) → 0
+(0, y) → y
+(s(x), y) → s(+(x, y))

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