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.

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(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 [13,14,18] contains 1 SCC with 10 less nodes.

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

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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


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 remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  APP1(x1)
app2(x1, x2)  =  app2(x1, x2)
:  =  :
+  =  +
f  =  f
g  =  g
a  =  a

Lexicographic Path Order [19].
Precedence:
APP1 > [app2, +, f, g, a]
: > [app2, +, f, g, a]


The following usable rules [14] 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:

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.