Termination w.r.t. Q of the given QTRS could be disproven:

(0) Obligation:

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.

(1) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(2) Obligation:

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

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

Q is empty.

(3) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(4) Obligation:

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

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

Q is empty.

(5) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(6) Obligation:

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

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(8) Obligation:

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.

(9) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(10) Obligation:

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.

(11) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

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:

• Semiunifier: [ ]
• Matcher: [ ]

---

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.

(12) FALSE

(13) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(14) Obligation:

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.