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(a) -> f1(b)
g1(b) -> g1(a)
f1(x) -> g1(x)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(a) -> f1(b)
g1(b) -> g1(a)
f1(x) -> g1(x)
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
G1(b) -> G1(a)
F1(x) -> G1(x)
F1(a) -> F1(b)
The TRS R consists of the following rules:
f1(a) -> f1(b)
g1(b) -> g1(a)
f1(x) -> g1(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
G1(b) -> G1(a)
F1(x) -> G1(x)
F1(a) -> F1(b)
The TRS R consists of the following rules:
f1(a) -> f1(b)
g1(b) -> g1(a)
f1(x) -> g1(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 3 less nodes.