Termination w.r.t. Q of the following Term Rewriting System could be proven:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x) → s(x)
f(s(s(x))) → s(f(f(x)))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x) → s(x)
f(s(s(x))) → s(f(f(x)))
Q is empty.
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(x)
F(s(s(x))) → F(f(x))
The TRS R consists of the following rules:
f(x) → s(x)
f(s(s(x))) → s(f(f(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
F(s(s(x))) → F(x)
F(s(s(x))) → F(f(x))
The TRS R consists of the following rules:
f(x) → s(x)
f(s(s(x))) → s(f(f(x)))
Q is empty.
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(x)
F(s(s(x))) → F(f(x))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
F(x1) = x1
s(x1) = s(x1)
f(x1) = f(x1)
Lexicographic Path Order [19].
Precedence:
[s1, f1]
The following usable rules [14] were oriented:
f(s(s(x))) → s(f(f(x)))
f(x) → s(x)
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(x) → s(x)
f(s(s(x))) → s(f(f(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.