(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:
- Matcher: [ ]
- Semiunifier: [ ]
Rewriting sequenceH(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.