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)


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)

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
          ↳ 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)

The set Q consists of the following terms:

app2(app2(apply, x0), x1)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


APP2(app2(apply, f), x) -> APP2(f, x)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  APP1(x1)
app2(x1, x2)  =  app1(x2)
apply  =  apply

Lexicographic Path Order [19].
Precedence:
app1 > APP1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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)

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.