Termination w.r.t. Q of the following Term Rewriting System could not be shown:

Q restricted rewrite system:
The TRS R consists of the following rules:

app2(app2(app2(comp, f), g), x) -> app2(f, app2(g, x))
app2(twice, f) -> app2(app2(comp, f), f)

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

Q restricted rewrite system:
The TRS R consists of the following rules:

app2(app2(app2(comp, f), g), x) -> app2(f, app2(g, x))
app2(twice, f) -> app2(app2(comp, f), f)

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(app2(comp, f), g), x) -> app2(f, app2(g, x))
app2(twice, f) -> app2(app2(comp, f), f)

The set Q consists of the following terms:

app2(app2(app2(comp, x0), x1), x2)
app2(twice, x0)


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(twice, f) -> APP2(comp, f)
APP2(twice, f) -> APP2(app2(comp, f), f)
APP2(app2(app2(comp, f), g), x) -> APP2(f, app2(g, x))
APP2(app2(app2(comp, f), g), x) -> APP2(g, x)

The TRS R consists of the following rules:

app2(app2(app2(comp, f), g), x) -> app2(f, app2(g, x))
app2(twice, f) -> app2(app2(comp, f), f)

The set Q consists of the following terms:

app2(app2(app2(comp, x0), x1), x2)
app2(twice, x0)

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

APP2(twice, f) -> APP2(comp, f)
APP2(twice, f) -> APP2(app2(comp, f), f)
APP2(app2(app2(comp, f), g), x) -> APP2(f, app2(g, x))
APP2(app2(app2(comp, f), g), x) -> APP2(g, x)

The TRS R consists of the following rules:

app2(app2(app2(comp, f), g), x) -> app2(f, app2(g, x))
app2(twice, f) -> app2(app2(comp, f), f)

The set Q consists of the following terms:

app2(app2(app2(comp, x0), x1), x2)
app2(twice, x0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 2 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

APP2(app2(app2(comp, f), g), x) -> APP2(f, app2(g, x))
APP2(app2(app2(comp, f), g), x) -> APP2(g, x)

The TRS R consists of the following rules:

app2(app2(app2(comp, f), g), x) -> app2(f, app2(g, x))
app2(twice, f) -> app2(app2(comp, f), f)

The set Q consists of the following terms:

app2(app2(app2(comp, x0), x1), x2)
app2(twice, x0)

We have to consider all minimal (P,Q,R)-chains.