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(0) -> s1(0)
f1(s1(0)) -> s1(0)
f1(s1(s1(x))) -> f1(f1(s1(x)))
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(0) -> s1(0)
f1(s1(0)) -> s1(0)
f1(s1(s1(x))) -> f1(f1(s1(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(0) -> s1(0)
f1(s1(0)) -> s1(0)
f1(s1(s1(x))) -> f1(f1(s1(x)))
The set Q consists of the following terms:
f1(0)
f1(s1(0))
f1(s1(s1(x0)))
Q DP problem:
The TRS P consists of the following rules:
F1(s1(s1(x))) -> F1(f1(s1(x)))
F1(s1(s1(x))) -> F1(s1(x))
The TRS R consists of the following rules:
f1(0) -> s1(0)
f1(s1(0)) -> s1(0)
f1(s1(s1(x))) -> f1(f1(s1(x)))
The set Q consists of the following terms:
f1(0)
f1(s1(0))
f1(s1(s1(x0)))
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
F1(s1(s1(x))) -> F1(f1(s1(x)))
F1(s1(s1(x))) -> F1(s1(x))
The TRS R consists of the following rules:
f1(0) -> s1(0)
f1(s1(0)) -> s1(0)
f1(s1(s1(x))) -> f1(f1(s1(x)))
The set Q consists of the following terms:
f1(0)
f1(s1(0))
f1(s1(s1(x0)))
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
F1(s1(s1(x))) -> F1(s1(x))
Used argument filtering: F1(x1) = x1
s1(x1) = s1(x1)
f1(x1) = f1(x1)
0 = 0
Used ordering: Quasi Precedence:
[s_1, f_1]
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F1(s1(s1(x))) -> F1(f1(s1(x)))
The TRS R consists of the following rules:
f1(0) -> s1(0)
f1(s1(0)) -> s1(0)
f1(s1(s1(x))) -> f1(f1(s1(x)))
The set Q consists of the following terms:
f1(0)
f1(s1(0))
f1(s1(s1(x0)))
We have to consider all minimal (P,Q,R)-chains.