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(x, x) → f(i(x), g(g(x)))
f(x, y) → x
g(x) → i(x)
f(x, i(x)) → f(x, x)
f(i(x), i(g(x))) → a

The set Q consists of the following terms:

f(x0, x1)
g(x0)



↳ QTRS
  ↳ RRRPoloQTRSProof

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

f(x, x) → f(i(x), g(g(x)))
f(x, y) → x
g(x) → i(x)
f(x, i(x)) → f(x, x)
f(i(x), i(g(x))) → a

The set Q consists of the following terms:

f(x0, x1)
g(x0)


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

f(x, x) → f(i(x), g(g(x)))
f(x, y) → x
g(x) → i(x)
f(x, i(x)) → f(x, x)
f(i(x), i(g(x))) → a

The set Q consists of the following terms:

f(x0, x1)
g(x0)

The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

f(x, y) → x
f(i(x), i(g(x))) → a
Used ordering:
Polynomial interpretation [25]:

POL(a) = 1   
POL(f(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(g(x1)) = x1   
POL(i(x1)) = x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof

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

f(x, x) → f(i(x), g(g(x)))
g(x) → i(x)
f(x, i(x)) → f(x, x)

The set Q consists of the following terms:

f(x0, x1)
g(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:

F(x, x) → G(g(x))
F(x, x) → F(i(x), g(g(x)))
F(x, i(x)) → F(x, x)
F(x, x) → G(x)

The TRS R consists of the following rules:

f(x, x) → f(i(x), g(g(x)))
g(x) → i(x)
f(x, i(x)) → f(x, x)

The set Q consists of the following terms:

f(x0, x1)
g(x0)

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

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

F(x, x) → G(g(x))
F(x, x) → F(i(x), g(g(x)))
F(x, i(x)) → F(x, x)
F(x, x) → G(x)

The TRS R consists of the following rules:

f(x, x) → f(i(x), g(g(x)))
g(x) → i(x)
f(x, i(x)) → f(x, x)

The set Q consists of the following terms:

f(x0, x1)
g(x0)

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

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof

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

F(x, x) → F(i(x), g(g(x)))
F(x, i(x)) → F(x, x)

The TRS R consists of the following rules:

f(x, x) → f(i(x), g(g(x)))
g(x) → i(x)
f(x, i(x)) → f(x, x)

The set Q consists of the following terms:

f(x0, x1)
g(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
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof

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

F(x, x) → F(i(x), g(g(x)))
F(x, i(x)) → F(x, x)

The TRS R consists of the following rules:

g(x) → i(x)

The set Q consists of the following terms:

f(x0, x1)
g(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(x0, x1)



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
                    ↳ QDP
                      ↳ Rewriting

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

F(x, x) → F(i(x), g(g(x)))
F(x, i(x)) → F(x, x)

The TRS R consists of the following rules:

g(x) → i(x)

The set Q consists of the following terms:

g(x0)

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

F(x, x) → F(i(x), i(g(x)))



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
                    ↳ QDP
                      ↳ Rewriting
                        ↳ QDP
                          ↳ Rewriting

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

F(x, i(x)) → F(x, x)
F(x, x) → F(i(x), i(g(x)))

The TRS R consists of the following rules:

g(x) → i(x)

The set Q consists of the following terms:

g(x0)

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

F(x, x) → F(i(x), i(i(x)))



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
                    ↳ QDP
                      ↳ Rewriting
                        ↳ QDP
                          ↳ Rewriting
                            ↳ QDP
                              ↳ UsableRulesProof

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

F(x, i(x)) → F(x, x)
F(x, x) → F(i(x), i(i(x)))

The TRS R consists of the following rules:

g(x) → i(x)

The set Q consists of the following terms:

g(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
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
                    ↳ QDP
                      ↳ Rewriting
                        ↳ QDP
                          ↳ Rewriting
                            ↳ QDP
                              ↳ UsableRulesProof
                                ↳ QDP
                                  ↳ QReductionProof

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

F(x, i(x)) → F(x, x)
F(x, x) → F(i(x), i(i(x)))

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

g(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.

g(x0)



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
                    ↳ QDP
                      ↳ Rewriting
                        ↳ QDP
                          ↳ Rewriting
                            ↳ QDP
                              ↳ UsableRulesProof
                                ↳ QDP
                                  ↳ QReductionProof
                                    ↳ QDP
                                      ↳ Instantiation

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

F(x, i(x)) → F(x, x)
F(x, x) → F(i(x), i(i(x)))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule F(x, i(x)) → F(x, x) we obtained the following new rules:

F(i(z0), i(i(z0))) → F(i(z0), i(z0))



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
                    ↳ QDP
                      ↳ Rewriting
                        ↳ QDP
                          ↳ Rewriting
                            ↳ QDP
                              ↳ UsableRulesProof
                                ↳ QDP
                                  ↳ QReductionProof
                                    ↳ QDP
                                      ↳ Instantiation
                                        ↳ QDP
                                          ↳ Instantiation

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

F(i(z0), i(i(z0))) → F(i(z0), i(z0))
F(x, x) → F(i(x), i(i(x)))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule F(x, x) → F(i(x), i(i(x))) we obtained the following new rules:

F(i(z0), i(z0)) → F(i(i(z0)), i(i(i(z0))))



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
                    ↳ QDP
                      ↳ Rewriting
                        ↳ QDP
                          ↳ Rewriting
                            ↳ QDP
                              ↳ UsableRulesProof
                                ↳ QDP
                                  ↳ QReductionProof
                                    ↳ QDP
                                      ↳ Instantiation
                                        ↳ QDP
                                          ↳ Instantiation
                                            ↳ QDP
                                              ↳ Instantiation

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

F(i(z0), i(i(z0))) → F(i(z0), i(z0))
F(i(z0), i(z0)) → F(i(i(z0)), i(i(i(z0))))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule F(i(z0), i(i(z0))) → F(i(z0), i(z0)) we obtained the following new rules:

F(i(i(z0)), i(i(i(z0)))) → F(i(i(z0)), i(i(z0)))



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
                    ↳ QDP
                      ↳ Rewriting
                        ↳ QDP
                          ↳ Rewriting
                            ↳ QDP
                              ↳ UsableRulesProof
                                ↳ QDP
                                  ↳ QReductionProof
                                    ↳ QDP
                                      ↳ Instantiation
                                        ↳ QDP
                                          ↳ Instantiation
                                            ↳ QDP
                                              ↳ Instantiation
                                                ↳ QDP
                                                  ↳ Instantiation

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

F(i(z0), i(z0)) → F(i(i(z0)), i(i(i(z0))))
F(i(i(z0)), i(i(i(z0)))) → F(i(i(z0)), i(i(z0)))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule F(i(z0), i(z0)) → F(i(i(z0)), i(i(i(z0)))) we obtained the following new rules:

F(i(i(z0)), i(i(z0))) → F(i(i(i(z0))), i(i(i(i(z0)))))



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
                    ↳ QDP
                      ↳ Rewriting
                        ↳ QDP
                          ↳ Rewriting
                            ↳ QDP
                              ↳ UsableRulesProof
                                ↳ QDP
                                  ↳ QReductionProof
                                    ↳ QDP
                                      ↳ Instantiation
                                        ↳ QDP
                                          ↳ Instantiation
                                            ↳ QDP
                                              ↳ Instantiation
                                                ↳ QDP
                                                  ↳ Instantiation
                                                    ↳ QDP
                                                      ↳ NonTerminationProof

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

F(i(i(z0)), i(i(i(z0)))) → F(i(i(z0)), i(i(z0)))
F(i(i(z0)), i(i(z0))) → F(i(i(i(z0))), i(i(i(i(z0)))))

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

The TRS P consists of the following rules:

F(i(i(z0)), i(i(i(z0)))) → F(i(i(z0)), i(i(z0)))
F(i(i(z0)), i(i(z0))) → F(i(i(i(z0))), i(i(i(i(z0)))))

The TRS R consists of the following rules:none


s = F(i(i(z0')), i(i(z0'))) evaluates to t =F(i(i(i(z0'))), i(i(i(z0'))))

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



Rewriting sequence

F(i(i(z0')), i(i(z0'))) → F(i(i(i(z0'))), i(i(i(i(z0')))))
with rule F(i(i(z0'')), i(i(z0''))) → F(i(i(i(z0''))), i(i(i(i(z0''))))) at position [] and matcher [z0'' / z0']

F(i(i(i(z0'))), i(i(i(i(z0'))))) → F(i(i(i(z0'))), i(i(i(z0'))))
with rule F(i(i(z0)), i(i(i(z0)))) → F(i(i(z0)), i(i(z0)))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.