(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(a) → g(h(a))
h(g(x)) → g(h(f(x)))
k(x, h(x), a) → h(x)
k(f(x), y, x) → f(x)
Q is empty.
(1) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(2) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(a) → H(a)
H(g(x)) → H(f(x))
H(g(x)) → F(x)
The TRS R consists of the following rules:
f(a) → g(h(a))
h(g(x)) → g(h(f(x)))
k(x, h(x), a) → h(x)
k(f(x), y, x) → f(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
H(g(x)) → H(f(x))
The TRS R consists of the following rules:
f(a) → g(h(a))
h(g(x)) → g(h(f(x)))
k(x, h(x), a) → h(x)
k(f(x), y, x) → f(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
H(g(x)) → H(f(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
H(
x1) =
H(
x1)
g(
x1) =
g(
x1)
f(
x1) =
f(
x1)
a =
a
h(
x1) =
h
Recursive Path Order [RPO].
Precedence:
H1 > f1
a > g1 > f1
a > h > f1
The following usable rules [FROCOS05] were oriented:
f(a) → g(h(a))
(6) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(a) → g(h(a))
h(g(x)) → g(h(f(x)))
k(x, h(x), a) → h(x)
k(f(x), y, x) → f(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(7) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(8) TRUE