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(g, app2(h, app2(g, x))) -> app2(g, x)
app2(g, app2(g, x)) -> app2(g, app2(h, app2(g, x)))
app2(h, app2(h, x)) -> app2(h, app2(app2(f, app2(h, x)), x))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

app2(g, app2(h, app2(g, x))) -> app2(g, x)
app2(g, app2(g, x)) -> app2(g, app2(h, app2(g, x)))
app2(h, app2(h, x)) -> app2(h, app2(app2(f, app2(h, x)), x))

Q is empty.

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

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

The TRS R consists of the following rules:

app2(g, app2(h, app2(g, x))) -> app2(g, x)
app2(g, app2(g, x)) -> app2(g, app2(h, app2(g, x)))
app2(h, app2(h, x)) -> app2(h, app2(app2(f, app2(h, x)), x))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

app2(g, app2(h, app2(g, x))) -> app2(g, x)
app2(g, app2(g, x)) -> app2(g, app2(h, app2(g, x)))
app2(h, app2(h, x)) -> app2(h, app2(app2(f, app2(h, x)), x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 4 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPAfsSolverProof

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

APP2(g, app2(g, x)) -> APP2(g, app2(h, app2(g, x)))

The TRS R consists of the following rules:

app2(g, app2(h, app2(g, x))) -> app2(g, x)
app2(g, app2(g, x)) -> app2(g, app2(h, app2(g, x)))
app2(h, app2(h, x)) -> app2(h, app2(app2(f, app2(h, x)), x))

Q is empty.
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(g, app2(g, x)) -> APP2(g, app2(h, app2(g, x)))
Used argument filtering: APP2(x1, x2)  =  x2
app2(x1, x2)  =  x1
g  =  g
h  =  h
Used ordering: Quasi Precedence: g > h 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPAfsSolverProof
            ↳ QDP
              ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

app2(g, app2(h, app2(g, x))) -> app2(g, x)
app2(g, app2(g, x)) -> app2(g, app2(h, app2(g, x)))
app2(h, app2(h, x)) -> app2(h, app2(app2(f, app2(h, x)), 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.