Termination w.r.t. Q proof of /tmp/tmpT-VlAJ/nonterm.trs

Termination w.r.t. Q of the following Term Rewriting System could be disproven:

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

f(s(s(s(s(s(s(s(s(x)))))))), y, y) → f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)
id(s(x)) → s(id(x))
id(0) → 0

Q is empty.

↳ QTRS

  ↳ AAECC Innermost

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

f(s(s(s(s(s(s(s(s(x)))))))), y, y) → f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)
id(s(x)) → s(id(x))
id(0) → 0

Q is empty.

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

id(s(x)) → s(id(x))
id(0) → 0

The TRS R 2 is

f(s(s(s(s(s(s(s(s(x)))))))), y, y) → f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)

The signature Sigma is {f}

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

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

f(s(s(s(s(s(s(s(s(x)))))))), y, y) → f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)
id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1)
id(s(x0))
id(0)

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → ID(s(s(s(s(s(s(s(s(x)))))))))
ID(s(x)) → ID(x)
F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(id(s(s(s(s(s(s(s(s(x))))))))), y, y)

The TRS R consists of the following rules:

f(s(s(s(s(s(s(s(s(x)))))))), y, y) → f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)
id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1)
id(s(x0))
id(0)

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:

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → ID(s(s(s(s(s(s(s(s(x)))))))))
ID(s(x)) → ID(x)
F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(id(s(s(s(s(s(s(s(s(x))))))))), y, y)

The TRS R consists of the following rules:

f(s(s(s(s(s(s(s(s(x)))))))), y, y) → f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)
id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1)
id(s(x0))
id(0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 1 less node.

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

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

ID(s(x)) → ID(x)

The TRS R consists of the following rules:

f(s(s(s(s(s(s(s(s(x)))))))), y, y) → f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)
id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1)
id(s(x0))
id(0)

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:

ID(s(x)) → ID(x)

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

f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1)
id(s(x0))
id(0)

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(s(s(s(s(s(s(s(x0)))))))), x1, x1)
id(s(x0))
id(0)

↳ 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:

ID(s(x)) → ID(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

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

ID(s(x)) → ID(x)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(id(s(s(s(s(s(s(s(s(x))))))))), y, y)

The TRS R consists of the following rules:

f(s(s(s(s(s(s(s(s(x)))))))), y, y) → f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)
id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1)
id(s(x0))
id(0)

↳ 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(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(id(s(s(s(s(s(s(s(s(x))))))))), y, y)

The TRS R consists of the following rules:

id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1)
id(s(x0))
id(0)

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(id(s(s(s(s(s(s(s(s(x))))))))), y, y)

The TRS R consists of the following rules:

id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

id(s(x0))
id(0)

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(id(s(s(s(s(s(s(s(x))))))))), y, y)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(id(s(s(s(s(s(s(s(x))))))))), y, y)

The TRS R consists of the following rules:

id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

id(s(x0))
id(0)

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(id(s(s(s(s(s(s(x))))))))), y, y)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(id(s(s(s(s(s(s(x))))))))), y, y)

The TRS R consists of the following rules:

id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

id(s(x0))
id(0)

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(s(id(s(s(s(s(s(x))))))))), y, y)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(s(id(s(s(s(s(s(x))))))))), y, y)

The TRS R consists of the following rules:

id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

id(s(x0))
id(0)

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(s(s(id(s(s(s(s(x))))))))), y, y)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Rewriting

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(s(s(id(s(s(s(s(x))))))))), y, y)

The TRS R consists of the following rules:

id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

id(s(x0))
id(0)

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(s(s(s(id(s(s(s(x))))))))), y, y)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ Rewriting

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(s(s(s(id(s(s(s(x))))))))), y, y)

The TRS R consists of the following rules:

id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

id(s(x0))
id(0)

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(s(s(s(s(id(s(s(x))))))))), y, y)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ Rewriting

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(s(s(s(s(id(s(s(x))))))))), y, y)

The TRS R consists of the following rules:

id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

id(s(x0))
id(0)

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(s(s(s(s(s(id(s(x))))))))), y, y)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ Rewriting

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(s(s(s(s(s(id(s(x))))))))), y, y)

The TRS R consists of the following rules:

id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

id(s(x0))
id(0)

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(s(s(s(s(s(s(id(x))))))))), y, y)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ Rewriting

                                                      ↳ QDP

                                                        ↳ MNOCProof

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(s(s(s(s(s(s(id(x))))))))), y, y)

The TRS R consists of the following rules:

id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

id(s(x0))
id(0)

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

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ Rewriting

                                                      ↳ QDP

                                                        ↳ MNOCProof

                                                          ↳ QDP

                                                            ↳ NonTerminationProof

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(s(s(s(s(s(s(id(x))))))))), y, y)

The TRS R consists of the following rules:

id(s(x)) → s(id(x))
id(0) → 0

Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(s(s(s(s(s(s(id(x))))))))), y, y)

The TRS R consists of the following rules:

id(s(x)) → s(id(x))
id(0) → 0

s = F(s(s(s(s(s(s(s(s(x)))))))), y, y) evaluates to t =F(s(s(s(s(s(s(s(s(id(x))))))))), y, y)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

Semiunifier: [ ]
Matcher: [x / id(x)]

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from F(s(s(s(s(s(s(s(s(x)))))))), y, y) to F(s(s(s(s(s(s(s(s(id(x))))))))), y, y).