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
          ↳ UsableRulesProof

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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
QDP
              ↳ QDPSizeChangeProof

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

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

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: