(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(0) → s(0)
f(s(0)) → s(0)
f(s(s(x))) → f(f(s(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(s(x))) → F(f(s(x)))
F(s(s(x))) → F(s(x))
The TRS R consists of the following rules:
f(0) → s(0)
f(s(0)) → s(0)
f(s(s(x))) → f(f(s(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
F(s(s(x))) → F(s(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(0) = 1
POL(F(x1)) = x1
POL(f(x1)) = 1 + x1
POL(s(x1)) = 1 + x1
The following usable rules [FROCOS05] were oriented:
f(s(0)) → s(0)
f(s(s(x))) → f(f(s(x)))
f(0) → s(0)
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(s(s(x))) → F(f(s(x)))
The TRS R consists of the following rules:
f(0) → s(0)
f(s(0)) → s(0)
f(s(s(x))) → f(f(s(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(s(x))) → F(f(s(x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(0) = 0
POL(F(x1)) = x1
POL(f(x1)) = 1
POL(s(x1)) = 1 + x1
The following usable rules [FROCOS05] were oriented:
f(s(0)) → s(0)
f(s(s(x))) → f(f(s(x)))
f(0) → s(0)
(6) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(0) → s(0)
f(s(0)) → s(0)
f(s(s(x))) → f(f(s(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