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:
f2(x, g1(x)) -> x
f2(x, h1(y)) -> f2(h1(x), y)
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
f2(x, g1(x)) -> x
f2(x, h1(y)) -> f2(h1(x), y)
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(x, g1(x)) -> x
f2(x, h1(y)) -> f2(h1(x), y)
The set Q consists of the following terms:
f2(x0, g1(x0))
f2(x0, h1(x1))
Q DP problem:
The TRS P consists of the following rules:
F2(x, h1(y)) -> F2(h1(x), y)
The TRS R consists of the following rules:
f2(x, g1(x)) -> x
f2(x, h1(y)) -> f2(h1(x), y)
The set Q consists of the following terms:
f2(x0, g1(x0))
f2(x0, h1(x1))
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:
F2(x, h1(y)) -> F2(h1(x), y)
The TRS R consists of the following rules:
f2(x, g1(x)) -> x
f2(x, h1(y)) -> f2(h1(x), y)
The set Q consists of the following terms:
f2(x0, g1(x0))
f2(x0, h1(x1))
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.
F2(x, h1(y)) -> F2(h1(x), y)
Used argument filtering: F2(x1, x2) = x2
h1(x1) = h1(x1)
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:
f2(x, g1(x)) -> x
f2(x, h1(y)) -> f2(h1(x), y)
The set Q consists of the following terms:
f2(x0, g1(x0))
f2(x0, h1(x1))
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.