(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(0, 1, g(x, y), z) → f(g(x, y), g(x, y), g(x, y), h(x))
g(0, 1) → 0
g(0, 1) → 1
h(g(x, y)) → h(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(0, 1, g(x, y), z) → F(g(x, y), g(x, y), g(x, y), h(x))
F(0, 1, g(x, y), z) → H(x)
H(g(x, y)) → H(x)
The TRS R consists of the following rules:
f(0, 1, g(x, y), z) → f(g(x, y), g(x, y), g(x, y), h(x))
g(0, 1) → 0
g(0, 1) → 1
h(g(x, y)) → h(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 2 SCCs with 1 less node.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
H(g(x, y)) → H(x)
The TRS R consists of the following rules:
f(0, 1, g(x, y), z) → f(g(x, y), g(x, y), g(x, y), h(x))
g(0, 1) → 0
g(0, 1) → 1
h(g(x, y)) → h(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
H(g(x, y)) → H(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
H(
x1) =
x1
g(
x1,
x2) =
g(
x1)
f(
x1,
x2,
x3,
x4) =
f
0 =
0
1 =
1
h(
x1) =
h
Lexicographic Path Order [LPO].
Precedence:
g1 > 1
The following usable rules [FROCOS05] were oriented:
f(0, 1, g(x, y), z) → f(g(x, y), g(x, y), g(x, y), h(x))
g(0, 1) → 0
g(0, 1) → 1
h(g(x, y)) → h(x)
(7) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(0, 1, g(x, y), z) → f(g(x, y), g(x, y), g(x, y), h(x))
g(0, 1) → 0
g(0, 1) → 1
h(g(x, y)) → h(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(9) TRUE
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(0, 1, g(x, y), z) → F(g(x, y), g(x, y), g(x, y), h(x))
The TRS R consists of the following rules:
f(0, 1, g(x, y), z) → f(g(x, y), g(x, y), g(x, y), h(x))
g(0, 1) → 0
g(0, 1) → 1
h(g(x, y)) → h(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.