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:

:2(:2(:2(:2(C, x), y), z), u) -> :2(:2(x, z), :2(:2(:2(x, y), z), u))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

:2(:2(:2(:2(C, x), y), z), u) -> :2(:2(x, z), :2(:2(:2(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:

:2(:2(:2(:2(C, x), y), z), u) -> :2(:2(x, z), :2(:2(:2(x, y), z), u))

The set Q consists of the following terms:

:2(:2(:2(:2(C, x0), x1), x2), x3)


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

:12(:2(:2(:2(C, x), y), z), u) -> :12(x, z)
:12(:2(:2(:2(C, x), y), z), u) -> :12(x, y)
:12(:2(:2(:2(C, x), y), z), u) -> :12(:2(x, z), :2(:2(:2(x, y), z), u))
:12(:2(:2(:2(C, x), y), z), u) -> :12(:2(x, y), z)
:12(:2(:2(:2(C, x), y), z), u) -> :12(:2(:2(x, y), z), u)

The TRS R consists of the following rules:

:2(:2(:2(:2(C, x), y), z), u) -> :2(:2(x, z), :2(:2(:2(x, y), z), u))

The set Q consists of the following terms:

:2(:2(:2(:2(C, x0), x1), x2), x3)

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

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

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

:12(:2(:2(:2(C, x), y), z), u) -> :12(x, z)
:12(:2(:2(:2(C, x), y), z), u) -> :12(x, y)
:12(:2(:2(:2(C, x), y), z), u) -> :12(:2(x, z), :2(:2(:2(x, y), z), u))
:12(:2(:2(:2(C, x), y), z), u) -> :12(:2(x, y), z)
:12(:2(:2(:2(C, x), y), z), u) -> :12(:2(:2(x, y), z), u)

The TRS R consists of the following rules:

:2(:2(:2(:2(C, x), y), z), u) -> :2(:2(x, z), :2(:2(:2(x, y), z), u))

The set Q consists of the following terms:

:2(:2(:2(:2(C, x0), x1), x2), x3)

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.

:12(:2(:2(:2(C, x), y), z), u) -> :12(x, z)
:12(:2(:2(:2(C, x), y), z), u) -> :12(x, y)
:12(:2(:2(:2(C, x), y), z), u) -> :12(:2(x, z), :2(:2(:2(x, y), z), u))
:12(:2(:2(:2(C, x), y), z), u) -> :12(:2(x, y), z)
:12(:2(:2(:2(C, x), y), z), u) -> :12(:2(:2(x, y), z), u)
Used argument filtering: :12(x1, x2)  =  x1
:2(x1, x2)  =  :2(x1, x2)
C  =  C
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPAfsSolverProof
            ↳ QDP
              ↳ PisEmptyProof

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

:2(:2(:2(:2(C, x), y), z), u) -> :2(:2(x, z), :2(:2(:2(x, y), z), u))

The set Q consists of the following terms:

:2(:2(:2(:2(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.