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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

app2(f, app2(g, x)) -> app2(g, app2(g, app2(f, x)))
app2(f, app2(g, x)) -> app2(g, app2(g, app2(g, 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(f, app2(g, x)) -> APP2(f, x)
Used argument filtering: APP2(x1, x2)  =  x2
app2(x1, x2)  =  app1(x2)
g  =  g
Used ordering: Quasi Precedence: trivial 


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

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

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