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:
f1(s1(x)) -> s1(s1(f1(p1(s1(x)))))
f1(0) -> 0
p1(s1(x)) -> x
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(s1(x)) -> s1(s1(f1(p1(s1(x)))))
f1(0) -> 0
p1(s1(x)) -> x
Q is empty.
The TRS is non-overlapping. Hence, we can switch to innermost.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(s1(x)) -> s1(s1(f1(p1(s1(x)))))
f1(0) -> 0
p1(s1(x)) -> x
The set Q consists of the following terms:
f1(s1(x0))
f1(0)
p1(s1(x0))
Q DP problem:
The TRS P consists of the following rules:
F1(s1(x)) -> P1(s1(x))
F1(s1(x)) -> F1(p1(s1(x)))
The TRS R consists of the following rules:
f1(s1(x)) -> s1(s1(f1(p1(s1(x)))))
f1(0) -> 0
p1(s1(x)) -> x
The set Q consists of the following terms:
f1(s1(x0))
f1(0)
p1(s1(x0))
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F1(s1(x)) -> P1(s1(x))
F1(s1(x)) -> F1(p1(s1(x)))
The TRS R consists of the following rules:
f1(s1(x)) -> s1(s1(f1(p1(s1(x)))))
f1(0) -> 0
p1(s1(x)) -> x
The set Q consists of the following terms:
f1(s1(x0))
f1(0)
p1(s1(x0))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F1(s1(x)) -> F1(p1(s1(x)))
The TRS R consists of the following rules:
f1(s1(x)) -> s1(s1(f1(p1(s1(x)))))
f1(0) -> 0
p1(s1(x)) -> x
The set Q consists of the following terms:
f1(s1(x0))
f1(0)
p1(s1(x0))
We have to consider all minimal (P,Q,R)-chains.