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(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> app2(app2(:, app2(app2(:, x), z)), app2(app2(:, app2(app2(:, app2(app2(:, x), y)), z)), u))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

app2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> app2(app2(:, app2(app2(:, x), z)), app2(app2(:, app2(app2(:, app2(app2(:, x), y)), z)), u))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof

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

app2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> app2(app2(:, app2(app2(:, x), z)), app2(app2(:, app2(app2(:, app2(app2(:, x), y)), z)), u))

The set Q consists of the following terms:

app2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x0)), x1)), x2)), x3)


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(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(:, app2(app2(:, x), z))
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, x), y)
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, app2(app2(:, x), z)), app2(app2(:, app2(app2(:, app2(app2(:, x), y)), z)), u))
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, app2(app2(:, x), y)), z)
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, x), z)
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(:, app2(app2(:, app2(app2(:, x), y)), z))
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, app2(app2(:, app2(app2(:, x), y)), z)), u)
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(:, app2(app2(:, x), y))
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(:, x)

The TRS R consists of the following rules:

app2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> app2(app2(:, app2(app2(:, x), z)), app2(app2(:, app2(app2(:, app2(app2(:, x), y)), z)), u))

The set Q consists of the following terms:

app2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x0)), x1)), x2)), x3)

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(:, app2(app2(:, x), z))
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, x), y)
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, app2(app2(:, x), z)), app2(app2(:, app2(app2(:, app2(app2(:, x), y)), z)), u))
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, app2(app2(:, x), y)), z)
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, x), z)
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(:, app2(app2(:, app2(app2(:, x), y)), z))
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, app2(app2(:, app2(app2(:, x), y)), z)), u)
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(:, app2(app2(:, x), y))
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(:, x)

The TRS R consists of the following rules:

app2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> app2(app2(:, app2(app2(:, x), z)), app2(app2(:, app2(app2(:, app2(app2(:, x), y)), z)), u))

The set Q consists of the following terms:

app2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x0)), x1)), x2)), x3)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 4 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof

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

APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, x), y)
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, app2(app2(:, x), z)), app2(app2(:, app2(app2(:, app2(app2(:, x), y)), z)), u))
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, app2(app2(:, x), y)), z)
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, x), z)
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, app2(app2(:, app2(app2(:, x), y)), z)), u)

The TRS R consists of the following rules:

app2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> app2(app2(:, app2(app2(:, x), z)), app2(app2(:, app2(app2(:, app2(app2(:, x), y)), z)), u))

The set Q consists of the following terms:

app2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x0)), x1)), x2)), x3)

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(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, x), y)
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, app2(app2(:, x), z)), app2(app2(:, app2(app2(:, app2(app2(:, x), y)), z)), u))
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, app2(app2(:, x), y)), z)
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, x), z)
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, app2(app2(:, app2(app2(:, x), y)), z)), u)
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)
app2(x1, x2)  =  app2(x1, x2)
:  =  :
C  =  C

Lexicographic Path Order [19].
Precedence:
[APP2, :] > app2
C > app2


The following usable rules [14] were oriented:

app2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> app2(app2(:, app2(app2(:, x), z)), app2(app2(:, app2(app2(:, app2(app2(:, x), y)), z)), u))



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> app2(app2(:, app2(app2(:, x), z)), app2(app2(:, app2(app2(:, app2(app2(:, x), y)), z)), u))

The set Q consists of the following terms:

app2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x0)), x1)), x2)), x3)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.