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, x), y) -> f2(f2(x, f2(a, y)), a)
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
f2(f2(a, x), y) -> f2(f2(x, f2(a, y)), 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, x), y) -> f2(f2(x, f2(a, y)), a)
The set Q consists of the following terms:
f2(f2(a, x0), x1)
Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
F2(f2(a, x), y) -> F2(x, f2(a, y))
F2(f2(a, x), y) -> F2(f2(x, f2(a, y)), a)
F2(f2(a, x), y) -> F2(a, y)
The TRS R consists of the following rules:
f2(f2(a, x), y) -> f2(f2(x, f2(a, y)), a)
The set Q consists of the following terms:
f2(f2(a, x0), x1)
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, x), y) -> F2(x, f2(a, y))
F2(f2(a, x), y) -> F2(f2(x, f2(a, y)), a)
F2(f2(a, x), y) -> F2(a, y)
The TRS R consists of the following rules:
f2(f2(a, x), y) -> f2(f2(x, f2(a, y)), a)
The set Q consists of the following terms:
f2(f2(a, x0), x1)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] 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:
F2(f2(a, x), y) -> F2(x, f2(a, y))
F2(f2(a, x), y) -> F2(f2(x, f2(a, y)), a)
The TRS R consists of the following rules:
f2(f2(a, x), y) -> f2(f2(x, f2(a, y)), a)
The set Q consists of the following terms:
f2(f2(a, x0), x1)
We have to consider all minimal (P,Q,R)-chains.