(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

f(g(x), s(0)) → f(g(x), 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)) → F(g(x), g(x))
G(s(x)) → G(x)

The TRS R consists of the following rules:

f(g(x), s(0)) → f(g(x), 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.

(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)) → f(g(x), 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)
f(x1, x2)  =  f
g(x1)  =  x1
0  =  0

Recursive path order with status [RPO].
Precedence:
trivial

Status:
trivial

The following usable rules [FROCOS05] were oriented:

f(g(x), s(0)) → f(g(x), g(x))
g(s(x)) → s(g(x))
g(0) → 0

(7) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f(g(x), s(0)) → f(g(x), 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)) → F(g(x), g(x))

The TRS R consists of the following rules:

f(g(x), s(0)) → f(g(x), g(x))
g(s(x)) → s(g(x))
g(0) → 0

Q is empty.
We have to consider all minimal (P,Q,R)-chains.