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) → s(0)
f(s(0)) → s(0)
f(s(s(x))) → f(f(s(x)))
Q is empty.
↳ QTRS
↳ Overlay + Local Confluence
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.
The TRS is overlay and locally confluent. By [15] we can switch to innermost.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
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)))
The set Q consists of the following terms:
f(0)
f(s(0))
f(s(s(x0)))
Using Dependency Pairs [1,13] we result in the following initial DP problem:
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)))
The set Q consists of the following terms:
f(0)
f(s(0))
f(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
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)))
The set Q consists of the following terms:
f(0)
f(s(0))
f(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.