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:

h(f(f(x))) → h(f(g(f(x))))
f(g(f(x))) → f(f(x))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

h(f(f(x))) → h(f(g(f(x))))
f(g(f(x))) → f(f(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(g(f(x))) → F(f(x))
H(f(f(x))) → H(f(g(f(x))))
H(f(f(x))) → F(g(f(x)))

The TRS R consists of the following rules:

h(f(f(x))) → h(f(g(f(x))))
f(g(f(x))) → f(f(x))

Q is empty.
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:

F(g(f(x))) → F(f(x))
H(f(f(x))) → H(f(g(f(x))))
H(f(f(x))) → F(g(f(x)))

The TRS R consists of the following rules:

h(f(f(x))) → h(f(g(f(x))))
f(g(f(x))) → f(f(x))

Q is empty.
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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
          ↳ UsableRulesProof

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

H(f(f(x))) → H(f(g(f(x))))

The TRS R consists of the following rules:

h(f(f(x))) → h(f(g(f(x))))
f(g(f(x))) → f(f(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

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

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

H(f(f(x))) → H(f(g(f(x))))

The TRS R consists of the following rules:

f(g(f(x))) → f(f(x))

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:

H(f(f(x))) → H(f(g(f(x))))

The TRS R consists of the following rules:

f(g(f(x))) → f(f(x))


s = H(f(g(f(x')))) evaluates to t =H(f(g(f(x'))))

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



Rewriting sequence

H(f(g(f(x')))) → H(f(f(x')))
with rule f(g(f(x''))) → f(f(x'')) at position [0] and matcher [x'' / x']

H(f(f(x'))) → H(f(g(f(x'))))
with rule H(f(f(x))) → H(f(g(f(x))))

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.




We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

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

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

H(f(f(x))) → H(f(g(f(x))))

The TRS R consists of the following rules:

f(g(f(x))) → f(f(x))

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