(0) Obligation:

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

from(X) → cons(X, from(s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(XS)
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, take(N, XS))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, XS)

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)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(XS)
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, take(N, XS))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, XS)

The set Q consists of the following terms:

from(x0)
head(cons(x0, x1))
2nd(cons(x0, x1))
take(0, x0)
take(s(x0), cons(x1, x2))
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))
2ND(cons(X, XS)) → HEAD(XS)
TAKE(s(N), cons(X, XS)) → TAKE(N, XS)
SEL(s(N), cons(X, XS)) → SEL(N, XS)

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(XS)
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, take(N, XS))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, XS)

The set Q consists of the following terms:

from(x0)
head(cons(x0, x1))
2nd(cons(x0, x1))
take(0, x0)
take(s(x0), cons(x1, x2))
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 3 SCCs with 1 less node.

(6) Complex Obligation (AND)

(7) Obligation:

Q DP problem:
The TRS P consists of the following rules:

SEL(s(N), cons(X, XS)) → SEL(N, XS)

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(XS)
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, take(N, XS))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, XS)

The set Q consists of the following terms:

from(x0)
head(cons(x0, x1))
2nd(cons(x0, x1))
take(0, x0)
take(s(x0), cons(x1, x2))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))

We have to consider all minimal (P,Q,R)-chains.

(8) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TAKE(s(N), cons(X, XS)) → TAKE(N, XS)

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(XS)
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, take(N, XS))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, XS)

The set Q consists of the following terms:

from(x0)
head(cons(x0, x1))
2nd(cons(x0, x1))
take(0, x0)
take(s(x0), cons(x1, x2))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))

We have to consider all minimal (P,Q,R)-chains.

(9) 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)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(XS)
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, take(N, XS))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, XS)

The set Q consists of the following terms:

from(x0)
head(cons(x0, x1))
2nd(cons(x0, x1))
take(0, x0)
take(s(x0), cons(x1, x2))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))

We have to consider all minimal (P,Q,R)-chains.