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)


Using Dependency Pairs [1,13] we result in the following initial DP problem:
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
          ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


: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 remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
:12(x1, x2)  =  :11(x1)
:2(x1, x2)  =  :2(x1, x2)
C  =  C

Lexicographic Path Order [19].
Precedence:
:^11 > :2
C > :2

The following usable rules [14] were oriented:

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



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