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:
f1(x) -> s1(x)
f1(s1(s1(x))) -> s1(f1(f1(x)))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(x) -> s1(x)
f1(s1(s1(x))) -> s1(f1(f1(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:
F1(s1(s1(x))) -> F1(x)
F1(s1(s1(x))) -> F1(f1(x))
The TRS R consists of the following rules:
f1(x) -> s1(x)
f1(s1(s1(x))) -> s1(f1(f1(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:
F1(s1(s1(x))) -> F1(x)
F1(s1(s1(x))) -> F1(f1(x))
The TRS R consists of the following rules:
f1(x) -> s1(x)
f1(s1(s1(x))) -> s1(f1(f1(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 strictly oriented and are deleted.
F1(s1(s1(x))) -> F1(x)
F1(s1(s1(x))) -> F1(f1(x))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
F1(x1) = F1(x1)
s1(x1) = s1(x1)
f1(x1) = f1(x1)
Lexicographic Path Order [19].
Precedence:
F1 > [s1, f1]
The following usable rules [14] were oriented:
f1(x) -> s1(x)
f1(s1(s1(x))) -> s1(f1(f1(x)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f1(x) -> s1(x)
f1(s1(s1(x))) -> s1(f1(f1(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.