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:
app2(app2(apply, f), x) -> app2(f, x)
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
app2(app2(apply, f), x) -> app2(f, 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:
app2(app2(apply, f), x) -> app2(f, x)
The set Q consists of the following terms:
app2(app2(apply, x0), x1)
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(apply, f), x) -> APP2(f, x)
The TRS R consists of the following rules:
app2(app2(apply, f), x) -> app2(f, x)
The set Q consists of the following terms:
app2(app2(apply, x0), 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:
APP2(app2(apply, f), x) -> APP2(f, x)
The TRS R consists of the following rules:
app2(app2(apply, f), x) -> app2(f, x)
The set Q consists of the following terms:
app2(app2(apply, x0), 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.
APP2(app2(apply, f), x) -> APP2(f, x)
Used argument filtering: APP2(x1, x2) = x1
app2(x1, x2) = app1(x2)
apply = apply
Used ordering: 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:
app2(app2(apply, f), x) -> app2(f, x)
The set Q consists of the following terms:
app2(app2(apply, x0), x1)
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.