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
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
app2(app2(apply, f), x) -> app2(f, x)
Q is empty.
Using Dependency Pairs [1,13] we result in the following initial DP problem:
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)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
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)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].
The following pairs can be oriented strictly and are deleted.
APP2(app2(apply, f), x) -> APP2(f, x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:
POL(APP2(x1, x2)) = x1
POL(app2(x1, x2)) = 2 + 2·x1·x2 + x2
POL(apply) = 2
The following usable rules [14] were oriented:
none
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
app2(app2(apply, f), x) -> app2(f, x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.