(0) Obligation:

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

f(x, empty) → x
f(empty, cons(a, k)) → f(cons(a, k), k)
f(cons(a, k), y) → f(y, k)

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:

F(empty, cons(a, k)) → F(cons(a, k), k)
F(cons(a, k), y) → F(y, k)

The TRS R consists of the following rules:

f(x, empty) → x
f(empty, cons(a, k)) → f(cons(a, k), k)
f(cons(a, k), y) → f(y, k)

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.


F(empty, cons(a, k)) → F(cons(a, k), k)
F(cons(a, k), y) → F(y, k)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
F(x1, x2)  =  F(x1, x2)

Tags:
F has tags [0,1]

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(cons(x1, x2)) = 1 + x1 + x2   
POL(empty) = 0   

The following usable rules [FROCOS05] were oriented: none

(4) Obligation:

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

f(x, empty) → x
f(empty, cons(a, k)) → f(cons(a, k), k)
f(cons(a, k), y) → f(y, k)

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