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(h1(x)) -> f1(i1(x))
g1(i1(x)) -> g1(h1(x))
h1(a) -> b
i1(a) -> b
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(h1(x)) -> f1(i1(x))
g1(i1(x)) -> g1(h1(x))
h1(a) -> b
i1(a) -> b
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
G1(i1(x)) -> H1(x)
F1(h1(x)) -> F1(i1(x))
F1(h1(x)) -> I1(x)
G1(i1(x)) -> G1(h1(x))
The TRS R consists of the following rules:
f1(h1(x)) -> f1(i1(x))
g1(i1(x)) -> g1(h1(x))
h1(a) -> b
i1(a) -> b
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(i1(x)) -> H1(x)
F1(h1(x)) -> F1(i1(x))
F1(h1(x)) -> I1(x)
G1(i1(x)) -> G1(h1(x))
The TRS R consists of the following rules:
f1(h1(x)) -> f1(i1(x))
g1(i1(x)) -> g1(h1(x))
h1(a) -> b
i1(a) -> b
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 4 less nodes.