(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
from(X) → cons(X, from(s(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
Q is empty.
(1) Overlay + Local Confluence (EQUIVALENT transformation)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
from(X) → cons(X, from(s(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
The set Q consists of the following terms:
from(x0)
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
(3) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FROM(X) → FROM(s(X))
SEL(s(X), cons(Y, Z)) → SEL(X, Z)
The TRS R consists of the following rules:
from(X) → cons(X, from(s(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
The set Q consists of the following terms:
from(x0)
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
SEL(s(X), cons(Y, Z)) → SEL(X, Z)
The TRS R consists of the following rules:
from(X) → cons(X, from(s(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
The set Q consists of the following terms:
from(x0)
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
We have to consider all minimal (P,Q,R)-chains.
(8) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
SEL(s(X), cons(Y, Z)) → SEL(X, Z)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SEL(
x1,
x2) =
SEL(
x1,
x2)
s(
x1) =
s(
x1)
cons(
x1,
x2) =
cons(
x2)
Recursive path order with status [RPO].
Quasi-Precedence:
trivial
Status:
SEL2: multiset
s1: multiset
cons1: multiset
The following usable rules [FROCOS05] were oriented:
none
(9) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
from(X) → cons(X, from(s(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
The set Q consists of the following terms:
from(x0)
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
We have to consider all minimal (P,Q,R)-chains.
(10) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(11) TRUE
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FROM(X) → FROM(s(X))
The TRS R consists of the following rules:
from(X) → cons(X, from(s(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
The set Q consists of the following terms:
from(x0)
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
We have to consider all minimal (P,Q,R)-chains.