Termination w.r.t. Q proof of /tmp/tmpXTMEF

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

:(:(:(:(C, x), y), z), u) → :(:(x, z), :(:(:(x, y), z), u))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

:(:(:(:(C, x), y), z), u) → :(:(x, z), :(:(:(x, y), z), u))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

:¹(:(:(:(C, x), y), z), u) → :¹(x, z)
:¹(:(:(:(C, x), y), z), u) → :¹(:(x, z), :(:(:(x, y), z), u))
:¹(:(:(:(C, x), y), z), u) → :¹(x, y)
:¹(:(:(:(C, x), y), z), u) → :¹(:(:(x, y), z), u)
:¹(:(:(:(C, x), y), z), u) → :¹(:(x, y), z)

The TRS R consists of the following rules:

:(:(:(:(C, x), y), z), u) → :(:(x, z), :(:(:(x, y), z), u))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

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

:¹(:(:(:(C, x), y), z), u) → :¹(x, z)
:¹(:(:(:(C, x), y), z), u) → :¹(:(x, z), :(:(:(x, y), z), u))
:¹(:(:(:(C, x), y), z), u) → :¹(x, y)
:¹(:(:(:(C, x), y), z), u) → :¹(:(:(x, y), z), u)
:¹(:(:(:(C, x), y), z), u) → :¹(:(x, y), z)

The TRS R consists of the following rules:

:(:(:(:(C, x), y), z), u) → :(:(x, z), :(:(:(x, y), z), u))

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