(0) Obligation:

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

f(0) → cons(0, n__f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
activate(n__f(X)) → f(X)
activate(X) → 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(s(0)) → F(p(s(0)))
F(s(0)) → P(s(0))
ACTIVATE(n__f(X)) → F(X)

The TRS R consists of the following rules:

f(0) → cons(0, n__f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
activate(n__f(X)) → f(X)
activate(X) → 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(s(0)) → F(p(s(0)))

The TRS R consists of the following rules:

f(0) → cons(0, n__f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
activate(n__f(X)) → f(X)
activate(X) → 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(s(0)) → F(p(s(0)))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1)  =  F(x1)
s(x1)  =  s(x1)
0  =  0
p(x1)  =  p
f(x1)  =  x1
cons(x1, x2)  =  x1
n__f(x1)  =  x1
activate(x1)  =  x1

Recursive Path Order [RPO].
Precedence:
s1 > F1 > 0
s1 > p > 0

The following usable rules [FROCOS05] were oriented:

f(0) → cons(0, n__f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
activate(n__f(X)) → f(X)
activate(X) → X

(6) Obligation:

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

f(0) → cons(0, n__f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
activate(n__f(X)) → f(X)
activate(X) → 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