(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