(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(a) → b
f(c) → d
f(g(x, y)) → g(f(x), f(y))
f(h(x, y)) → g(h(y, f(x)), h(x, f(y)))
g(x, x) → h(e, 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(g(x, y)) → G(f(x), f(y))
F(g(x, y)) → F(x)
F(g(x, y)) → F(y)
F(h(x, y)) → G(h(y, f(x)), h(x, f(y)))
F(h(x, y)) → F(x)
F(h(x, y)) → F(y)
The TRS R consists of the following rules:
f(a) → b
f(c) → d
f(g(x, y)) → g(f(x), f(y))
f(h(x, y)) → g(h(y, f(x)), h(x, f(y)))
g(x, x) → h(e, 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:
F(g(x, y)) → F(y)
F(g(x, y)) → F(x)
F(h(x, y)) → F(x)
F(h(x, y)) → F(y)
The TRS R consists of the following rules:
f(a) → b
f(c) → d
f(g(x, y)) → g(f(x), f(y))
f(h(x, y)) → g(h(y, f(x)), h(x, f(y)))
g(x, x) → h(e, 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.
F(g(x, y)) → F(y)
F(g(x, y)) → F(x)
F(h(x, y)) → F(x)
F(h(x, y)) → F(y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(
x1) =
x1
g(
x1,
x2) =
g(
x1,
x2)
h(
x1,
x2) =
h(
x1,
x2)
Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented:
none
(6) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(a) → b
f(c) → d
f(g(x, y)) → g(f(x), f(y))
f(h(x, y)) → g(h(y, f(x)), h(x, f(y)))
g(x, x) → h(e, 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