(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(0) → 1
f(s(x)) → g(f(x))
g(x) → +(x, s(x))
f(s(x)) → +(f(x), s(f(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(x)) → G(f(x))
F(s(x)) → F(x)
The TRS R consists of the following rules:
f(0) → 1
f(s(x)) → g(f(x))
g(x) → +(x, s(x))
f(s(x)) → +(f(x), s(f(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 1 less node.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(s(x)) → F(x)
The TRS R consists of the following rules:
f(0) → 1
f(s(x)) → g(f(x))
g(x) → +(x, s(x))
f(s(x)) → +(f(x), s(f(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) QDPSizeChangeProof (EQUIVALENT transformation)
We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.
Order:Homeomorphic Embedding Order
AFS:
s(x1) = s(x1)
From the DPs we obtained the following set of size-change graphs:
- F(s(x)) → F(x) (allowed arguments on rhs = {1})
The graph contains the following edges 1 > 1
We oriented the following set of usable rules [AAECC05,FROCOS05].
none
(6) TRUE