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(h(f(X)))

The set Q consists of the following terms:

f(x0)



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof

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

f(X) → g(h(f(X)))

The set Q consists of the following terms:

f(x0)


We have reversed the following QTRS:
The set of rules R is

f(X) → g(h(f(X)))

The set Q is {f(x0)}.
We have obtained the following QTRS:

f(x) → f(h(g(x)))

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
    ↳ QTRS
  ↳ DependencyPairsProof

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

f(x) → f(h(g(x)))

Q is empty.

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) → F(X)

The TRS R consists of the following rules:

f(X) → g(h(f(X)))

The set Q consists of the following terms:

f(x0)

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ 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(h(f(X)))

The set Q consists of the following terms:

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
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ 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:

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.

f(x0)



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ 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
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ 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:

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.

f(x0)



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ 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.