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(app2(:, app2(app2(:, x), y)), z) -> app2(app2(:, x), app2(app2(:, y), z))
app2(app2(:, app2(app2(+, x), y)), z) -> app2(app2(+, app2(app2(:, x), z)), app2(app2(:, y), z))
app2(app2(:, z), app2(app2(+, x), app2(f, y))) -> app2(app2(:, app2(app2(g, z), y)), app2(app2(+, x), a))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

app2(app2(:, app2(app2(:, x), y)), z) -> app2(app2(:, x), app2(app2(:, y), z))
app2(app2(:, app2(app2(+, x), y)), z) -> app2(app2(+, app2(app2(:, x), z)), app2(app2(:, y), z))
app2(app2(:, z), app2(app2(+, x), app2(f, y))) -> app2(app2(:, app2(app2(g, z), y)), app2(app2(+, x), a))

Q is empty.

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

APP2(app2(:, z), app2(app2(+, x), app2(f, y))) -> APP2(app2(:, app2(app2(g, z), y)), app2(app2(+, x), a))
APP2(app2(:, app2(app2(:, x), y)), z) -> APP2(app2(:, y), z)
APP2(app2(:, app2(app2(+, x), y)), z) -> APP2(app2(+, app2(app2(:, x), z)), app2(app2(:, y), z))
APP2(app2(:, app2(app2(+, x), y)), z) -> APP2(app2(:, y), z)
APP2(app2(:, z), app2(app2(+, x), app2(f, y))) -> APP2(app2(g, z), y)
APP2(app2(:, z), app2(app2(+, x), app2(f, y))) -> APP2(app2(+, x), a)
APP2(app2(:, app2(app2(+, x), y)), z) -> APP2(+, app2(app2(:, x), z))
APP2(app2(:, z), app2(app2(+, x), app2(f, y))) -> APP2(:, app2(app2(g, z), y))
APP2(app2(:, app2(app2(+, x), y)), z) -> APP2(:, y)
APP2(app2(:, z), app2(app2(+, x), app2(f, y))) -> APP2(g, z)
APP2(app2(:, app2(app2(:, x), y)), z) -> APP2(app2(:, x), app2(app2(:, y), z))
APP2(app2(:, app2(app2(+, x), y)), z) -> APP2(app2(:, x), z)
APP2(app2(:, app2(app2(:, x), y)), z) -> APP2(:, y)
APP2(app2(:, app2(app2(+, x), y)), z) -> APP2(:, x)

The TRS R consists of the following rules:

app2(app2(:, app2(app2(:, x), y)), z) -> app2(app2(:, x), app2(app2(:, y), z))
app2(app2(:, app2(app2(+, x), y)), z) -> app2(app2(+, app2(app2(:, x), z)), app2(app2(:, y), z))
app2(app2(:, z), app2(app2(+, x), app2(f, y))) -> app2(app2(:, app2(app2(g, z), y)), app2(app2(+, x), a))

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(app2(:, z), app2(app2(+, x), app2(f, y))) -> APP2(app2(:, app2(app2(g, z), y)), app2(app2(+, x), a))
APP2(app2(:, app2(app2(:, x), y)), z) -> APP2(app2(:, y), z)
APP2(app2(:, app2(app2(+, x), y)), z) -> APP2(app2(+, app2(app2(:, x), z)), app2(app2(:, y), z))
APP2(app2(:, app2(app2(+, x), y)), z) -> APP2(app2(:, y), z)
APP2(app2(:, z), app2(app2(+, x), app2(f, y))) -> APP2(app2(g, z), y)
APP2(app2(:, z), app2(app2(+, x), app2(f, y))) -> APP2(app2(+, x), a)
APP2(app2(:, app2(app2(+, x), y)), z) -> APP2(+, app2(app2(:, x), z))
APP2(app2(:, z), app2(app2(+, x), app2(f, y))) -> APP2(:, app2(app2(g, z), y))
APP2(app2(:, app2(app2(+, x), y)), z) -> APP2(:, y)
APP2(app2(:, z), app2(app2(+, x), app2(f, y))) -> APP2(g, z)
APP2(app2(:, app2(app2(:, x), y)), z) -> APP2(app2(:, x), app2(app2(:, y), z))
APP2(app2(:, app2(app2(+, x), y)), z) -> APP2(app2(:, x), z)
APP2(app2(:, app2(app2(:, x), y)), z) -> APP2(:, y)
APP2(app2(:, app2(app2(+, x), y)), z) -> APP2(:, x)

The TRS R consists of the following rules:

app2(app2(:, app2(app2(:, x), y)), z) -> app2(app2(:, x), app2(app2(:, y), z))
app2(app2(:, app2(app2(+, x), y)), z) -> app2(app2(+, app2(app2(:, x), z)), app2(app2(:, y), z))
app2(app2(:, z), app2(app2(+, x), app2(f, y))) -> app2(app2(:, app2(app2(g, z), y)), app2(app2(+, x), a))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 10 less nodes.

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

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

APP2(app2(:, app2(app2(:, x), y)), z) -> APP2(app2(:, x), app2(app2(:, y), z))
APP2(app2(:, app2(app2(+, x), y)), z) -> APP2(app2(:, x), z)
APP2(app2(:, app2(app2(:, x), y)), z) -> APP2(app2(:, y), z)
APP2(app2(:, app2(app2(+, x), y)), z) -> APP2(app2(:, y), z)

The TRS R consists of the following rules:

app2(app2(:, app2(app2(:, x), y)), z) -> app2(app2(:, x), app2(app2(:, y), z))
app2(app2(:, app2(app2(+, x), y)), z) -> app2(app2(+, app2(app2(:, x), z)), app2(app2(:, y), z))
app2(app2(:, z), app2(app2(+, x), app2(f, y))) -> app2(app2(:, app2(app2(g, z), y)), app2(app2(+, x), a))

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(app2(:, app2(app2(:, x), y)), z) -> APP2(app2(:, x), app2(app2(:, y), z))
APP2(app2(:, app2(app2(+, x), y)), z) -> APP2(app2(:, x), z)
APP2(app2(:, app2(app2(:, x), y)), z) -> APP2(app2(:, y), z)
APP2(app2(:, app2(app2(+, x), y)), z) -> APP2(app2(:, y), z)
Used argument filtering: APP2(x1, x2)  =  x1
app2(x1, x2)  =  app2(x1, x2)
:  =  :
+  =  +
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(app2(:, app2(app2(:, x), y)), z) -> app2(app2(:, x), app2(app2(:, y), z))
app2(app2(:, app2(app2(+, x), y)), z) -> app2(app2(+, app2(app2(:, x), z)), app2(app2(:, y), z))
app2(app2(:, z), app2(app2(+, x), app2(f, y))) -> app2(app2(:, app2(app2(g, z), y)), app2(app2(+, x), a))

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.