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(0) -> cons1(0)
f1(s1(0)) -> f1(p1(s1(0)))
p1(s1(0)) -> 0
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(0) -> cons1(0)
f1(s1(0)) -> f1(p1(s1(0)))
p1(s1(0)) -> 0
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(0) -> cons1(0)
f1(s1(0)) -> f1(p1(s1(0)))
p1(s1(0)) -> 0
The set Q consists of the following terms:
f1(0)
f1(s1(0))
p1(s1(0))
Q DP problem:
The TRS P consists of the following rules:
F1(s1(0)) -> P1(s1(0))
F1(s1(0)) -> F1(p1(s1(0)))
The TRS R consists of the following rules:
f1(0) -> cons1(0)
f1(s1(0)) -> f1(p1(s1(0)))
p1(s1(0)) -> 0
The set Q consists of the following terms:
f1(0)
f1(s1(0))
p1(s1(0))
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(0)) -> P1(s1(0))
F1(s1(0)) -> F1(p1(s1(0)))
The TRS R consists of the following rules:
f1(0) -> cons1(0)
f1(s1(0)) -> f1(p1(s1(0)))
p1(s1(0)) -> 0
The set Q consists of the following terms:
f1(0)
f1(s1(0))
p1(s1(0))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 2 less nodes.