Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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

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.

(3) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
:¹(x1, x2)  =  :¹(x1)
:(x1, x2)  =  :(x1, x2)
C  =  C

Lexicographic path order with status [LPO].
Quasi-Precedence:

C > :^1₁ > :₂

Status:

:^1₁: [1]
:₂: [1,2]
C: []

The following usable rules [FROCOS05] were oriented:

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

(4) Obligation:

Q DP problem:
P is empty.
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.

(5) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(6) TRUE