(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