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:

app(app(F, app(app(F, f), x)), x) → app(app(F, app(G, app(app(F, f), x))), app(f, x))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

app(app(F, app(app(F, f), x)), x) → app(app(F, app(G, app(app(F, f), x))), app(f, x))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

APP(app(F, app(app(F, f), x)), x) → APP(F, app(G, app(app(F, f), x)))
APP(app(F, app(app(F, f), x)), x) → APP(app(F, app(G, app(app(F, f), x))), app(f, x))
APP(app(F, app(app(F, f), x)), x) → APP(f, x)
APP(app(F, app(app(F, f), x)), x) → APP(G, app(app(F, f), x))

The TRS R consists of the following rules:

app(app(F, app(app(F, f), x)), x) → app(app(F, app(G, app(app(F, f), x))), app(f, 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:

APP(app(F, app(app(F, f), x)), x) → APP(F, app(G, app(app(F, f), x)))
APP(app(F, app(app(F, f), x)), x) → APP(app(F, app(G, app(app(F, f), x))), app(f, x))
APP(app(F, app(app(F, f), x)), x) → APP(f, x)
APP(app(F, app(app(F, f), x)), x) → APP(G, app(app(F, f), x))

The TRS R consists of the following rules:

app(app(F, app(app(F, f), x)), x) → app(app(F, app(G, app(app(F, f), x))), app(f, x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP
          ↳ QDPOrderProof

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

APP(app(F, app(app(F, f), x)), x) → APP(f, x)

The TRS R consists of the following rules:

app(app(F, app(app(F, f), x)), x) → app(app(F, app(G, app(app(F, f), x))), app(f, x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


APP(app(F, app(app(F, f), x)), x) → APP(f, x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(APP(x1, x2)) = x_1   
POL(app(x1, x2)) = (2)x_1 + x_2   
POL(F) = 1   
The value of delta used in the strict ordering is 6.
The following usable rules [17] were oriented: none



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

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

app(app(F, app(app(F, f), x)), x) → app(app(F, app(G, app(app(F, f), x))), app(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.