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:
f3(s1(x), y, y) -> f3(y, x, s1(x))
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
f3(s1(x), y, y) -> f3(y, x, 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:
f3(s1(x), y, y) -> f3(y, x, s1(x))
The set Q consists of the following terms:
f3(s1(x0), x1, x1)
Q DP problem:
The TRS P consists of the following rules:
F3(s1(x), y, y) -> F3(y, x, s1(x))
The TRS R consists of the following rules:
f3(s1(x), y, y) -> f3(y, x, s1(x))
The set Q consists of the following terms:
f3(s1(x0), x1, 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:
F3(s1(x), y, y) -> F3(y, x, s1(x))
The TRS R consists of the following rules:
f3(s1(x), y, y) -> f3(y, x, s1(x))
The set Q consists of the following terms:
f3(s1(x0), x1, x1)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.