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(g1(X)) -> g1(f1(f1(X)))
f1(h1(X)) -> h1(g1(X))
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(g1(X)) -> g1(f1(f1(X)))
f1(h1(X)) -> h1(g1(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:
f1(g1(X)) -> g1(f1(f1(X)))
f1(h1(X)) -> h1(g1(X))
The set Q consists of the following terms:
f1(g1(x0))
f1(h1(x0))
Q DP problem:
The TRS P consists of the following rules:
F1(g1(X)) -> F1(f1(X))
F1(g1(X)) -> F1(X)
The TRS R consists of the following rules:
f1(g1(X)) -> g1(f1(f1(X)))
f1(h1(X)) -> h1(g1(X))
The set Q consists of the following terms:
f1(g1(x0))
f1(h1(x0))
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
F1(g1(X)) -> F1(f1(X))
F1(g1(X)) -> F1(X)
The TRS R consists of the following rules:
f1(g1(X)) -> g1(f1(f1(X)))
f1(h1(X)) -> h1(g1(X))
The set Q consists of the following terms:
f1(g1(x0))
f1(h1(x0))
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
F1(g1(X)) -> F1(f1(X))
F1(g1(X)) -> F1(X)
Used argument filtering: F1(x1) = x1
g1(x1) = g1(x1)
f1(x1) = x1
h1(x1) = h
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f1(g1(X)) -> g1(f1(f1(X)))
f1(h1(X)) -> h1(g1(X))
The set Q consists of the following terms:
f1(g1(x0))
f1(h1(x0))
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.