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:

ap(ap(g, x), y) → y
ap(f, x) → ap(f, app(g, x))

Q is empty.


↳ QTRS
  ↳ RRRPoloQTRSProof

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

ap(ap(g, x), y) → y
ap(f, x) → ap(f, app(g, x))

Q is empty.

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

ap(ap(g, x), y) → y
ap(f, x) → ap(f, app(g, x))

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

ap(ap(g, x), y) → y
Used ordering:
Polynomial interpretation [25]:

POL(ap(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(app(x1, x2)) = x1 + x2   
POL(f) = 0   
POL(g) = 0   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ AAECC Innermost

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

ap(f, x) → ap(f, app(g, x))

Q is empty.

We have applied [19,8] to switch to innermost. The TRS R 1 is none

The TRS R 2 is

ap(f, x) → ap(f, app(g, x))

The signature Sigma is {ap}

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ AAECC Innermost
        ↳ QTRS
          ↳ DependencyPairsProof

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

ap(f, x) → ap(f, app(g, x))

The set Q consists of the following terms:

ap(f, 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:

AP(f, x) → AP(f, app(g, x))

The TRS R consists of the following rules:

ap(f, x) → ap(f, app(g, x))

The set Q consists of the following terms:

ap(f, x0)

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

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ AAECC Innermost
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ UsableRulesProof

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

AP(f, x) → AP(f, app(g, x))

The TRS R consists of the following rules:

ap(f, x) → ap(f, app(g, x))

The set Q consists of the following terms:

ap(f, 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
      ↳ AAECC Innermost
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof

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

AP(f, x) → AP(f, app(g, x))

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

ap(f, 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.

ap(f, x0)



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ AAECC Innermost
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
                    ↳ QDP
                      ↳ ATransformationProof

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

AP(f, x) → AP(f, app(g, x))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We have applied the A-Transformation [17] to get from an applicative problem to a standard problem. 

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ AAECC Innermost
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
                    ↳ QDP
                      ↳ ATransformationProof
                        ↳ QDP
                          ↳ Instantiation

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

f(x) → f(g(x))

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

f(g(z0)) → f(g(g(z0)))



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ AAECC Innermost
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
                    ↳ QDP
                      ↳ ATransformationProof
                        ↳ QDP
                          ↳ Instantiation
                            ↳ QDP
                              ↳ Instantiation

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

f(g(z0)) → f(g(g(z0)))

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

f(g(g(z0))) → f(g(g(g(z0))))



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ AAECC Innermost
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
                    ↳ QDP
                      ↳ ATransformationProof
                        ↳ QDP
                          ↳ Instantiation
                            ↳ QDP
                              ↳ Instantiation
                                ↳ QDP
                                  ↳ NonTerminationProof

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

f(g(g(z0))) → f(g(g(g(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 semiunifying a rule from P directly.

The TRS P consists of the following rules:

f(g(g(z0))) → f(g(g(g(z0))))

The TRS R consists of the following rules:none


s = f(g(g(z0))) evaluates to t =f(g(g(g(z0))))

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(g(g(z0))) to f(g(g(g(z0)))).