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
↳ QDPOrderProof
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.
We use the reduction pair processor [13].
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.
F(s(s(x))) → F(f(s(x)))
Used ordering: Combined order from the following AFS and order.
F(x1) = F(x1)
s(x1) = s(x1)
f(x1) = f(x1)
0 = 0
Lexicographic Path Order [19].
Precedence:
F1 > [s1, f1, 0]
The following usable rules [14] were oriented:
f(s(0)) → s(0)
f(s(s(x))) → f(f(s(x)))
f(0) → s(0)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
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)))
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.