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.

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(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 [13,14,18] contains 1 SCC with 4 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


APP2(g, app2(g, x)) -> APP2(g, app2(h, app2(g, 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)  =  x2
g  =  g
app2(x1, x2)  =  x1
h  =  h
f  =  f

Lexicographic Path Order [19].
Precedence:
g > h
f > h


The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ 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.