(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(g(x), s(0), y) → f(g(s(0)), y, g(x))
g(s(x)) → s(g(x))
g(0) → 0
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), s(0), y) → F(g(s(0)), y, g(x))
F(g(x), s(0), y) → G(s(0))
G(s(x)) → G(x)
The TRS R consists of the following rules:
f(g(x), s(0), y) → f(g(s(0)), y, g(x))
g(s(x)) → s(g(x))
g(0) → 0
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:
G(s(x)) → G(x)
The TRS R consists of the following rules:
f(g(x), s(0), y) → f(g(s(0)), y, g(x))
g(s(x)) → s(g(x))
g(0) → 0
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.
G(s(x)) → G(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
G(
x1) =
x1
s(
x1) =
s(
x1)
Recursive path order with status [RPO].
Precedence:
trivial
Status:
trivial
The following usable rules [FROCOS05] were oriented:
none
(7) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(g(x), s(0), y) → f(g(s(0)), y, g(x))
g(s(x)) → s(g(x))
g(0) → 0
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(g(x), s(0), y) → F(g(s(0)), y, g(x))
The TRS R consists of the following rules:
f(g(x), s(0), y) → f(g(s(0)), y, g(x))
g(s(x)) → s(g(x))
g(0) → 0
Q is empty.
We have to consider all minimal (P,Q,R)-chains.