Termination w.r.t. Q of the given QTRS could not be shown:

(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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SEL(s(N), cons(X, XS)) → SEL(N, XS)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SEL(x1, x2)  =  SEL(x2)
s(x1)  =  s
cons(x1, x2)  =  cons(x1, x2)

Recursive Path Order [RPO].
Precedence:

s > SEL₁
cons₂ > SEL₁

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)))
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.

(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:

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.

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

TAKE(s(N), cons(X, XS)) → TAKE(N, XS)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TAKE(x1, x2)  =  TAKE(x2)
s(x1)  =  s
cons(x1, x2)  =  cons(x1, x2)

Recursive Path Order [RPO].
Precedence:

s > TAKE₁
cons₂ > TAKE₁

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

Q DP problem:
P is empty.
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.

(15) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(16) TRUE

(17) 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.