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) → g(f(x))
g(f(x)) → x
g(x) → a

The set Q consists of the following terms:

g(x0)
f(x0)



↳ QTRS
  ↳ RRRPoloQTRSProof

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

f(x) → g(f(x))
g(f(x)) → x
g(x) → a

The set Q consists of the following terms:

g(x0)
f(x0)


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

f(x) → g(f(x))
g(f(x)) → x
g(x) → a

The set Q consists of the following terms:

g(x0)
f(x0)

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

g(f(x)) → x
Used ordering:
Polynomial interpretation [25]:

POL(a) = 0   
POL(f(x1)) = 1 + 2·x1   
POL(g(x1)) = x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof

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

f(x) → g(f(x))
g(x) → a

The set Q consists of the following terms:

g(x0)
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:

F(x) → G(f(x))
F(x) → F(x)

The TRS R consists of the following rules:

f(x) → g(f(x))
g(x) → a

The set Q consists of the following terms:

g(x0)
f(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) → G(f(x))
F(x) → F(x)

The TRS R consists of the following rules:

f(x) → g(f(x))
g(x) → a

The set Q consists of the following terms:

g(x0)
f(x0)

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

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

F(x) → F(x)

The TRS R consists of the following rules:

f(x) → g(f(x))
g(x) → a

The set Q consists of the following terms:

g(x0)
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
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
              ↳ UsableRulesProof

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

F(x) → F(x)

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

g(x0)
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.

g(x0)
f(x0)



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

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

F(x) → F(x)

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(x) → F(x)

The TRS R consists of the following rules:none


s = F(x) evaluates to t =F(x)

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(x) to F(x).




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
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof

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

F(x) → F(x)

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

g(x0)
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.

g(x0)
f(x0)



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

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

F(x) → F(x)

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