(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) =
x1
s(
x1) =
s
0 =
0
p(
x1) =
p
f(
x1) =
f
cons(
x1,
x2) =
cons
n__f(
x1) =
n__f
activate(
x1) =
x1
Lexicographic path order with status [LPO].
Quasi-Precedence:
s > p > 0
[f, nf] > cons
Status:
trivial
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