Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(0) → cons(0, f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(0) → cons(0, f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
F(0) → F(s(0))
F(s(0)) → F(p(s(0)))
F(s(0)) → P(s(0))
The TRS R consists of the following rules:
f(0) → cons(0, f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F(0) → F(s(0))
F(s(0)) → F(p(s(0)))
F(s(0)) → P(s(0))
The TRS R consists of the following rules:
f(0) → cons(0, f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F(0) → F(s(0))
F(s(0)) → F(p(s(0)))
The TRS R consists of the following rules:
f(0) → cons(0, f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
Q is empty.
We have to consider all minimal (P,Q,R)-chains.