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:
f2(f2(a, a), x) -> f2(f2(a, x), f2(a, f2(a, a)))
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
f2(f2(a, a), x) -> f2(f2(a, x), f2(a, f2(a, a)))
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:
f2(f2(a, a), x) -> f2(f2(a, x), f2(a, f2(a, a)))
The set Q consists of the following terms:
f2(f2(a, a), x0)
Q DP problem:
The TRS P consists of the following rules:
F2(f2(a, a), x) -> F2(f2(a, x), f2(a, f2(a, a)))
F2(f2(a, a), x) -> F2(a, f2(a, a))
F2(f2(a, a), x) -> F2(a, x)
The TRS R consists of the following rules:
f2(f2(a, a), x) -> f2(f2(a, x), f2(a, f2(a, a)))
The set Q consists of the following terms:
f2(f2(a, a), 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:
F2(f2(a, a), x) -> F2(f2(a, x), f2(a, f2(a, a)))
F2(f2(a, a), x) -> F2(a, f2(a, a))
F2(f2(a, a), x) -> F2(a, x)
The TRS R consists of the following rules:
f2(f2(a, a), x) -> f2(f2(a, x), f2(a, f2(a, a)))
The set Q consists of the following terms:
f2(f2(a, a), x0)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 2 less nodes.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F2(f2(a, a), x) -> F2(f2(a, x), f2(a, f2(a, a)))
The TRS R consists of the following rules:
f2(f2(a, a), x) -> f2(f2(a, x), f2(a, f2(a, a)))
The set Q consists of the following terms:
f2(f2(a, a), x0)
We have to consider all minimal (P,Q,R)-chains.