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:

c → f(g(c))
f(g(X)) → g(X)

The set Q consists of the following terms:

f(g(x0))
c



↳ QTRS
  ↳ DependencyPairsProof

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

c → f(g(c))
f(g(X)) → g(X)

The set Q consists of the following terms:

f(g(x0))
c


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

C → C
C → F(g(c))

The TRS R consists of the following rules:

c → f(g(c))
f(g(X)) → g(X)

The set Q consists of the following terms:

f(g(x0))
c

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:

C → C
C → F(g(c))

The TRS R consists of the following rules:

c → f(g(c))
f(g(X)) → g(X)

The set Q consists of the following terms:

f(g(x0))
c

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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
          ↳ UsableRulesProof

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

C → C

The TRS R consists of the following rules:

c → f(g(c))
f(g(X)) → g(X)

The set Q consists of the following terms:

f(g(x0))
c

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
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
          ↳ UsableRulesProof

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

C → C

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

f(g(x0))
c

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(g(x0))
c



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

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

C → C

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:

C → C

The TRS R consists of the following rules:none


s = C evaluates to t =C

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 C to C.




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
        ↳ QDP
          ↳ UsableRulesProof
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof

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

C → C

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

f(g(x0))
c

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(g(x0))
c



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

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

C → C

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