(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(empty, l) → l
f(cons(x, k), l) → g(k, l, cons(x, k))
g(a, b, c) → f(a, cons(b, c))
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(cons(x, k), l) → G(k, l, cons(x, k))
G(a, b, c) → F(a, cons(b, c))
The TRS R consists of the following rules:
f(empty, l) → l
f(cons(x, k), l) → g(k, l, cons(x, k))
g(a, b, c) → f(a, cons(b, c))
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(cons(x, k), l) → G(k, l, cons(x, k))
G(a, b, c) → F(a, cons(b, c))
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
F(
x0,
x1,
x2) =
F(
x1)
G(
x0,
x1,
x2,
x3) =
G(
x1,
x3)
Tags:
F has argument tags [6,2,3] and root tag 1
G has argument tags [7,4,7,2] and root tag 0
Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:
POL(F(x1, x2)) = x2
POL(G(x1, x2, x3)) = x2
POL(cons(x1, x2)) = 1 + x1 + x2
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(empty, l) → l
f(cons(x, k), l) → g(k, l, cons(x, k))
g(a, b, c) → f(a, cons(b, c))
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