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

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

app2(f, app2(f, x)) -> app2(g, app2(f, x))
app2(g, app2(g, x)) -> app2(f, 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(f, x)
APP2(f, app2(f, x)) -> APP2(g, 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)  =  APP2(x1, x2)
g  =  g
app2(x1, x2)  =  app1(x2)
f  =  f

Lexicographic Path Order [19].
Precedence:
[app1, f] > APP2 > g


The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
QDP
          ↳ PisEmptyProof

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

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