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

The set Q consists of the following terms:

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



↳ 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:

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

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:

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

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

↳ 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:

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

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
  ↳ 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:

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

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

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



↳ 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: 



↳ 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:

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

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



↳ 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
  ↳ 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
  ↳ 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
  ↳ 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
  ↳ 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
  ↳ 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
  ↳ 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
  ↳ 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
  ↳ 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
  ↳ 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:



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).