(0) Obligation:

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

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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:

ACTIVE(from(X)) → CONS(X, from(s(X)))
ACTIVE(from(X)) → FROM(s(X))
ACTIVE(from(X)) → S(X)
ACTIVE(2nd(cons(X, XS))) → HEAD(XS)
ACTIVE(take(s(N), cons(X, XS))) → CONS(X, take(N, XS))
ACTIVE(take(s(N), cons(X, XS))) → TAKE(N, XS)
ACTIVE(sel(s(N), cons(X, XS))) → SEL(N, XS)
ACTIVE(from(X)) → FROM(active(X))
ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(s(X)) → S(active(X))
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(head(X)) → HEAD(active(X))
ACTIVE(head(X)) → ACTIVE(X)
ACTIVE(2nd(X)) → 2ND(active(X))
ACTIVE(2nd(X)) → ACTIVE(X)
ACTIVE(take(X1, X2)) → TAKE(active(X1), X2)
ACTIVE(take(X1, X2)) → ACTIVE(X1)
ACTIVE(take(X1, X2)) → TAKE(X1, active(X2))
ACTIVE(take(X1, X2)) → ACTIVE(X2)
ACTIVE(sel(X1, X2)) → SEL(active(X1), X2)
ACTIVE(sel(X1, X2)) → ACTIVE(X1)
ACTIVE(sel(X1, X2)) → SEL(X1, active(X2))
ACTIVE(sel(X1, X2)) → ACTIVE(X2)
FROM(mark(X)) → FROM(X)
CONS(mark(X1), X2) → CONS(X1, X2)
S(mark(X)) → S(X)
HEAD(mark(X)) → HEAD(X)
2ND(mark(X)) → 2ND(X)
TAKE(mark(X1), X2) → TAKE(X1, X2)
TAKE(X1, mark(X2)) → TAKE(X1, X2)
SEL(mark(X1), X2) → SEL(X1, X2)
SEL(X1, mark(X2)) → SEL(X1, X2)
PROPER(from(X)) → FROM(proper(X))
PROPER(from(X)) → PROPER(X)
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(s(X)) → S(proper(X))
PROPER(s(X)) → PROPER(X)
PROPER(head(X)) → HEAD(proper(X))
PROPER(head(X)) → PROPER(X)
PROPER(2nd(X)) → 2ND(proper(X))
PROPER(2nd(X)) → PROPER(X)
PROPER(take(X1, X2)) → TAKE(proper(X1), proper(X2))
PROPER(take(X1, X2)) → PROPER(X1)
PROPER(take(X1, X2)) → PROPER(X2)
PROPER(sel(X1, X2)) → SEL(proper(X1), proper(X2))
PROPER(sel(X1, X2)) → PROPER(X1)
PROPER(sel(X1, X2)) → PROPER(X2)
FROM(ok(X)) → FROM(X)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
S(ok(X)) → S(X)
HEAD(ok(X)) → HEAD(X)
2ND(ok(X)) → 2ND(X)
TAKE(ok(X1), ok(X2)) → TAKE(X1, X2)
SEL(ok(X1), ok(X2)) → SEL(X1, X2)
TOP(mark(X)) → TOP(proper(X))
TOP(mark(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
TOP(ok(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 10 SCCs with 25 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

SEL(X1, mark(X2)) → SEL(X1, X2)
SEL(mark(X1), X2) → SEL(X1, X2)
SEL(ok(X1), ok(X2)) → SEL(X1, X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


SEL(mark(X1), X2) → SEL(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SEL(x1, x2)  =  SEL(x1)
mark(x1)  =  mark(x1)
ok(x1)  =  x1
active(x1)  =  active(x1)
from(x1)  =  from(x1)
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  s(x1)
head(x1)  =  head(x1)
2nd(x1)  =  x1
take(x1, x2)  =  take(x1, x2)
0  =  0
nil  =  nil
sel(x1, x2)  =  sel(x1, x2)
proper(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
0 > mark1 > SEL1
0 > nil > SEL1
top > active1 > from1 > mark1 > SEL1
top > active1 > cons2 > head1 > mark1 > SEL1
top > active1 > s1 > mark1 > SEL1
top > active1 > take2 > mark1 > SEL1
top > active1 > sel2 > mark1 > SEL1

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(7) Obligation:

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

SEL(X1, mark(X2)) → SEL(X1, X2)
SEL(ok(X1), ok(X2)) → SEL(X1, X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
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(ok(X1), ok(X2)) → SEL(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SEL(x1, x2)  =  SEL(x1)
mark(x1)  =  mark
ok(x1)  =  ok(x1)
active(x1)  =  active(x1)
from(x1)  =  from(x1)
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  x1
head(x1)  =  head(x1)
2nd(x1)  =  2nd(x1)
take(x1, x2)  =  x1
0  =  0
nil  =  nil
sel(x1, x2)  =  sel(x1)
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > from1 > ok1 > SEL1 > mark
active1 > cons2 > head1 > ok1 > SEL1 > mark
active1 > cons2 > sel1 > ok1 > SEL1 > mark
active1 > 2nd1 > head1 > ok1 > SEL1 > mark
0 > ok1 > SEL1 > mark
nil > ok1 > SEL1 > mark
proper1 > from1 > ok1 > SEL1 > mark
proper1 > cons2 > head1 > ok1 > SEL1 > mark
proper1 > cons2 > sel1 > ok1 > SEL1 > mark
proper1 > 2nd1 > head1 > ok1 > SEL1 > mark
top > mark

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(9) Obligation:

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

SEL(X1, mark(X2)) → SEL(X1, X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(10) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


SEL(X1, mark(X2)) → SEL(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SEL(x1, x2)  =  x2
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
from(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  s(x1)
head(x1)  =  head(x1)
2nd(x1)  =  x1
take(x1, x2)  =  take(x1, x2)
0  =  0
nil  =  nil
sel(x1, x2)  =  sel(x1, x2)
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > nil > ok > s1 > cons2 > head1 > mark1 > top
active1 > nil > ok > s1 > cons2 > take2 > mark1 > top
active1 > nil > ok > s1 > sel2 > mark1 > top
0 > nil > ok > s1 > cons2 > head1 > mark1 > top
0 > nil > ok > s1 > cons2 > take2 > mark1 > top
0 > nil > ok > s1 > sel2 > mark1 > top
proper1 > s1 > cons2 > head1 > mark1 > top
proper1 > s1 > cons2 > take2 > mark1 > top
proper1 > s1 > sel2 > mark1 > top

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(11) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(12) PisEmptyProof (EQUIVALENT transformation)

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

(13) TRUE

(14) Obligation:

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

TAKE(X1, mark(X2)) → TAKE(X1, X2)
TAKE(mark(X1), X2) → TAKE(X1, X2)
TAKE(ok(X1), ok(X2)) → TAKE(X1, X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


TAKE(mark(X1), X2) → TAKE(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TAKE(x1, x2)  =  TAKE(x1)
mark(x1)  =  mark(x1)
ok(x1)  =  x1
active(x1)  =  active(x1)
from(x1)  =  from(x1)
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  s(x1)
head(x1)  =  head(x1)
2nd(x1)  =  x1
take(x1, x2)  =  take(x1, x2)
0  =  0
nil  =  nil
sel(x1, x2)  =  sel(x1, x2)
proper(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
0 > mark1 > TAKE1
0 > nil > TAKE1
top > active1 > from1 > mark1 > TAKE1
top > active1 > cons2 > head1 > mark1 > TAKE1
top > active1 > s1 > mark1 > TAKE1
top > active1 > take2 > mark1 > TAKE1
top > active1 > sel2 > mark1 > TAKE1

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(16) Obligation:

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

TAKE(X1, mark(X2)) → TAKE(X1, X2)
TAKE(ok(X1), ok(X2)) → TAKE(X1, X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


TAKE(ok(X1), ok(X2)) → TAKE(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TAKE(x1, x2)  =  TAKE(x1)
mark(x1)  =  mark
ok(x1)  =  ok(x1)
active(x1)  =  active(x1)
from(x1)  =  from(x1)
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  x1
head(x1)  =  head(x1)
2nd(x1)  =  2nd(x1)
take(x1, x2)  =  x1
0  =  0
nil  =  nil
sel(x1, x2)  =  sel(x1)
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > from1 > ok1 > TAKE1 > mark
active1 > cons2 > head1 > ok1 > TAKE1 > mark
active1 > cons2 > sel1 > ok1 > TAKE1 > mark
active1 > 2nd1 > head1 > ok1 > TAKE1 > mark
0 > ok1 > TAKE1 > mark
nil > ok1 > TAKE1 > mark
proper1 > from1 > ok1 > TAKE1 > mark
proper1 > cons2 > head1 > ok1 > TAKE1 > mark
proper1 > cons2 > sel1 > ok1 > TAKE1 > mark
proper1 > 2nd1 > head1 > ok1 > TAKE1 > mark
top > mark

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(18) Obligation:

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

TAKE(X1, mark(X2)) → TAKE(X1, X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(19) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


TAKE(X1, mark(X2)) → TAKE(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TAKE(x1, x2)  =  x2
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
from(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  s(x1)
head(x1)  =  head(x1)
2nd(x1)  =  x1
take(x1, x2)  =  take(x1, x2)
0  =  0
nil  =  nil
sel(x1, x2)  =  sel(x1, x2)
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > nil > ok > s1 > cons2 > head1 > mark1 > top
active1 > nil > ok > s1 > cons2 > take2 > mark1 > top
active1 > nil > ok > s1 > sel2 > mark1 > top
0 > nil > ok > s1 > cons2 > head1 > mark1 > top
0 > nil > ok > s1 > cons2 > take2 > mark1 > top
0 > nil > ok > s1 > sel2 > mark1 > top
proper1 > s1 > cons2 > head1 > mark1 > top
proper1 > s1 > cons2 > take2 > mark1 > top
proper1 > s1 > sel2 > mark1 > top

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(20) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(21) PisEmptyProof (EQUIVALENT transformation)

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

(22) TRUE

(23) Obligation:

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

2ND(ok(X)) → 2ND(X)
2ND(mark(X)) → 2ND(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(24) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


2ND(ok(X)) → 2ND(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
2ND(x1)  =  x1
ok(x1)  =  ok(x1)
mark(x1)  =  x1
active(x1)  =  active(x1)
from(x1)  =  from(x1)
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  x1
head(x1)  =  x1
2nd(x1)  =  x1
take(x1, x2)  =  take(x1, x2)
0  =  0
nil  =  nil
sel(x1, x2)  =  sel(x1, x2)
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > from1 > ok1
active1 > cons2 > ok1
active1 > take2 > ok1
active1 > nil > ok1
active1 > sel2 > ok1
0 > nil > ok1
proper1 > from1 > ok1
proper1 > cons2 > ok1
proper1 > take2 > ok1
proper1 > nil > ok1
proper1 > sel2 > ok1

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(25) Obligation:

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

2ND(mark(X)) → 2ND(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(26) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


2ND(mark(X)) → 2ND(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
2ND(x1)  =  2ND(x1)
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
from(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  x1
head(x1)  =  head(x1)
2nd(x1)  =  2nd(x1)
take(x1, x2)  =  take(x1, x2)
0  =  0
nil  =  nil
sel(x1, x2)  =  sel(x1, x2)
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > head1 > mark1
active1 > 2nd1 > mark1
active1 > take2 > cons2 > mark1
active1 > sel2 > mark1
0 > mark1
0 > nil

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(27) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(28) PisEmptyProof (EQUIVALENT transformation)

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

(29) TRUE

(30) Obligation:

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

HEAD(ok(X)) → HEAD(X)
HEAD(mark(X)) → HEAD(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(31) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


HEAD(ok(X)) → HEAD(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
HEAD(x1)  =  x1
ok(x1)  =  ok(x1)
mark(x1)  =  x1
active(x1)  =  active(x1)
from(x1)  =  from(x1)
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  x1
head(x1)  =  x1
2nd(x1)  =  x1
take(x1, x2)  =  take(x1, x2)
0  =  0
nil  =  nil
sel(x1, x2)  =  sel(x1, x2)
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > from1 > ok1
active1 > cons2 > ok1
active1 > take2 > ok1
active1 > nil > ok1
active1 > sel2 > ok1
0 > nil > ok1
proper1 > from1 > ok1
proper1 > cons2 > ok1
proper1 > take2 > ok1
proper1 > nil > ok1
proper1 > sel2 > ok1

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(32) Obligation:

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

HEAD(mark(X)) → HEAD(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(33) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


HEAD(mark(X)) → HEAD(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
HEAD(x1)  =  HEAD(x1)
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
from(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  x1
head(x1)  =  head(x1)
2nd(x1)  =  2nd(x1)
take(x1, x2)  =  take(x1, x2)
0  =  0
nil  =  nil
sel(x1, x2)  =  sel(x1, x2)
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > head1 > mark1
active1 > 2nd1 > mark1
active1 > take2 > cons2 > mark1
active1 > sel2 > mark1
0 > mark1
0 > nil

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(34) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(35) PisEmptyProof (EQUIVALENT transformation)

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

(36) TRUE

(37) Obligation:

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

S(ok(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(38) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


S(ok(X)) → S(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
S(x1)  =  x1
ok(x1)  =  ok(x1)
mark(x1)  =  x1
active(x1)  =  active(x1)
from(x1)  =  from(x1)
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  x1
head(x1)  =  x1
2nd(x1)  =  x1
take(x1, x2)  =  take(x1, x2)
0  =  0
nil  =  nil
sel(x1, x2)  =  sel(x1, x2)
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > from1 > ok1
active1 > cons2 > ok1
active1 > take2 > ok1
active1 > nil > ok1
active1 > sel2 > ok1
0 > nil > ok1
proper1 > from1 > ok1
proper1 > cons2 > ok1
proper1 > take2 > ok1
proper1 > nil > ok1
proper1 > sel2 > ok1

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(39) Obligation:

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

S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(40) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


S(mark(X)) → S(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
S(x1)  =  S(x1)
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
from(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  x1
head(x1)  =  head(x1)
2nd(x1)  =  2nd(x1)
take(x1, x2)  =  take(x1, x2)
0  =  0
nil  =  nil
sel(x1, x2)  =  sel(x1, x2)
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > head1 > mark1
active1 > 2nd1 > mark1
active1 > take2 > cons2 > mark1
active1 > sel2 > mark1
0 > mark1
0 > nil

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(41) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(42) PisEmptyProof (EQUIVALENT transformation)

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

(43) TRUE

(44) Obligation:

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

CONS(ok(X1), ok(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(45) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


CONS(ok(X1), ok(X2)) → CONS(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2)  =  CONS(x1)
ok(x1)  =  ok(x1)
mark(x1)  =  x1
active(x1)  =  active(x1)
from(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  s(x1)
head(x1)  =  x1
2nd(x1)  =  x1
take(x1, x2)  =  x2
0  =  0
nil  =  nil
sel(x1, x2)  =  sel(x2)
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
0 > nil > ok1 > CONS1
proper1 > s1 > cons2 > ok1 > CONS1
proper1 > nil > ok1 > CONS1
proper1 > sel1 > ok1 > CONS1
top > active1 > s1 > cons2 > ok1 > CONS1
top > active1 > nil > ok1 > CONS1
top > active1 > sel1 > ok1 > CONS1

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(46) Obligation:

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

CONS(mark(X1), X2) → CONS(X1, X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(47) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


CONS(mark(X1), X2) → CONS(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2)  =  x1
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
from(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  s(x1)
head(x1)  =  head(x1)
2nd(x1)  =  x1
take(x1, x2)  =  take(x1, x2)
0  =  0
nil  =  nil
sel(x1, x2)  =  sel(x1, x2)
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > nil > ok > cons2 > take2 > mark1 > top
active1 > nil > ok > s1 > mark1 > top
active1 > nil > ok > head1 > mark1 > top
active1 > nil > ok > sel2 > mark1 > top
0 > nil > ok > cons2 > take2 > mark1 > top
0 > nil > ok > s1 > mark1 > top
0 > nil > ok > head1 > mark1 > top
0 > nil > ok > sel2 > mark1 > top
proper1 > cons2 > take2 > mark1 > top
proper1 > s1 > mark1 > top
proper1 > head1 > mark1 > top
proper1 > sel2 > mark1 > top

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(48) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(49) PisEmptyProof (EQUIVALENT transformation)

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

(50) TRUE

(51) Obligation:

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

FROM(ok(X)) → FROM(X)
FROM(mark(X)) → FROM(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(52) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


FROM(ok(X)) → FROM(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FROM(x1)  =  x1
ok(x1)  =  ok(x1)
mark(x1)  =  x1
active(x1)  =  active(x1)
from(x1)  =  from(x1)
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  x1
head(x1)  =  x1
2nd(x1)  =  x1
take(x1, x2)  =  take(x1, x2)
0  =  0
nil  =  nil
sel(x1, x2)  =  sel(x1, x2)
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > from1 > ok1
active1 > cons2 > ok1
active1 > take2 > ok1
active1 > nil > ok1
active1 > sel2 > ok1
0 > nil > ok1
proper1 > from1 > ok1
proper1 > cons2 > ok1
proper1 > take2 > ok1
proper1 > nil > ok1
proper1 > sel2 > ok1

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(53) Obligation:

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

FROM(mark(X)) → FROM(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(54) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


FROM(mark(X)) → FROM(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FROM(x1)  =  FROM(x1)
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
from(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  x1
head(x1)  =  head(x1)
2nd(x1)  =  2nd(x1)
take(x1, x2)  =  take(x1, x2)
0  =  0
nil  =  nil
sel(x1, x2)  =  sel(x1, x2)
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > head1 > mark1
active1 > 2nd1 > mark1
active1 > take2 > cons2 > mark1
active1 > sel2 > mark1
0 > mark1
0 > nil

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(55) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(56) PisEmptyProof (EQUIVALENT transformation)

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

(57) TRUE

(58) Obligation:

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

PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(from(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(s(X)) → PROPER(X)
PROPER(head(X)) → PROPER(X)
PROPER(2nd(X)) → PROPER(X)
PROPER(take(X1, X2)) → PROPER(X1)
PROPER(take(X1, X2)) → PROPER(X2)
PROPER(sel(X1, X2)) → PROPER(X1)
PROPER(sel(X1, X2)) → PROPER(X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(59) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(take(X1, X2)) → PROPER(X1)
PROPER(take(X1, X2)) → PROPER(X2)
PROPER(sel(X1, X2)) → PROPER(X1)
PROPER(sel(X1, X2)) → PROPER(X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  PROPER(x1)
cons(x1, x2)  =  cons(x1, x2)
from(x1)  =  x1
s(x1)  =  x1
head(x1)  =  x1
2nd(x1)  =  x1
take(x1, x2)  =  take(x1, x2)
sel(x1, x2)  =  sel(x1, x2)
active(x1)  =  active(x1)
mark(x1)  =  mark
0  =  0
nil  =  nil
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
top > active1 > take2 > cons2 > mark > PROPER1
top > active1 > take2 > nil > PROPER1
top > active1 > sel2 > mark > PROPER1
top > proper1 > take2 > cons2 > mark > PROPER1
top > proper1 > take2 > nil > PROPER1
top > proper1 > sel2 > mark > PROPER1
top > proper1 > 0 > mark > PROPER1
top > proper1 > 0 > nil > PROPER1

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(60) Obligation:

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

PROPER(from(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)
PROPER(head(X)) → PROPER(X)
PROPER(2nd(X)) → PROPER(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(61) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(from(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  PROPER(x1)
from(x1)  =  from(x1)
s(x1)  =  s(x1)
head(x1)  =  x1
2nd(x1)  =  x1
active(x1)  =  active(x1)
mark(x1)  =  mark(x1)
cons(x1, x2)  =  cons(x1, x2)
take(x1, x2)  =  take(x1, x2)
0  =  0
nil  =  nil
sel(x1, x2)  =  sel(x1, x2)
proper(x1)  =  x1
ok(x1)  =  ok
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > from1 > PROPER1
active1 > from1 > mark1 > top
active1 > from1 > ok > top
active1 > s1 > PROPER1
active1 > s1 > cons2 > mark1 > top
active1 > s1 > cons2 > ok > top
active1 > s1 > take2 > mark1 > top
active1 > s1 > take2 > nil > ok > top
active1 > sel2 > mark1 > top
active1 > sel2 > ok > top
0 > nil > ok > top

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(62) Obligation:

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

PROPER(head(X)) → PROPER(X)
PROPER(2nd(X)) → PROPER(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(63) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(head(X)) → PROPER(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  x1
head(x1)  =  head(x1)
2nd(x1)  =  x1
active(x1)  =  x1
from(x1)  =  x1
mark(x1)  =  mark
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  s
take(x1, x2)  =  take(x1)
0  =  0
nil  =  nil
sel(x1, x2)  =  x1
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
top > proper1 > head1 > mark
top > proper1 > s > cons2 > mark
top > proper1 > take1 > cons2 > mark
top > proper1 > take1 > nil > mark
top > proper1 > 0 > mark

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(64) Obligation:

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

PROPER(2nd(X)) → PROPER(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(65) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(2nd(X)) → PROPER(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  PROPER(x1)
2nd(x1)  =  2nd(x1)
active(x1)  =  active(x1)
from(x1)  =  x1
mark(x1)  =  mark
cons(x1, x2)  =  x1
s(x1)  =  x1
head(x1)  =  x1
take(x1, x2)  =  x1
0  =  0
nil  =  nil
sel(x1, x2)  =  sel
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
PROPER1 > mark
active1 > 2nd1 > mark
proper1 > 2nd1 > mark
proper1 > 0 > mark
proper1 > nil > mark
proper1 > sel > mark
top > mark

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(66) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(67) PisEmptyProof (EQUIVALENT transformation)

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

(68) TRUE

(69) Obligation:

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

ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(head(X)) → ACTIVE(X)
ACTIVE(2nd(X)) → ACTIVE(X)
ACTIVE(take(X1, X2)) → ACTIVE(X1)
ACTIVE(take(X1, X2)) → ACTIVE(X2)
ACTIVE(sel(X1, X2)) → ACTIVE(X1)
ACTIVE(sel(X1, X2)) → ACTIVE(X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(70) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(take(X1, X2)) → ACTIVE(X1)
ACTIVE(take(X1, X2)) → ACTIVE(X2)
ACTIVE(sel(X1, X2)) → ACTIVE(X1)
ACTIVE(sel(X1, X2)) → ACTIVE(X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  x1
cons(x1, x2)  =  x1
from(x1)  =  x1
s(x1)  =  x1
head(x1)  =  x1
2nd(x1)  =  x1
take(x1, x2)  =  take(x1, x2)
sel(x1, x2)  =  sel(x1, x2)
active(x1)  =  active(x1)
mark(x1)  =  mark
0  =  0
nil  =  nil
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
top > active1 > take2 > nil > mark
top > active1 > sel2 > mark
top > proper1 > take2 > nil > mark
top > proper1 > sel2 > mark
top > proper1 > 0 > nil > mark

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(71) Obligation:

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

ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(head(X)) → ACTIVE(X)
ACTIVE(2nd(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(72) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(2nd(X)) → ACTIVE(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  ACTIVE(x1)
cons(x1, x2)  =  cons(x1)
from(x1)  =  x1
s(x1)  =  x1
head(x1)  =  x1
2nd(x1)  =  2nd(x1)
active(x1)  =  x1
mark(x1)  =  mark
take(x1, x2)  =  take
0  =  0
nil  =  nil
sel(x1, x2)  =  sel(x2)
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
take > cons1 > ACTIVE1 > mark
take > cons1 > sel1 > mark
take > nil > mark
top > proper1 > cons1 > ACTIVE1 > mark
top > proper1 > cons1 > sel1 > mark
top > proper1 > 2nd1 > ACTIVE1 > mark
top > proper1 > 0 > mark
top > proper1 > nil > mark

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(73) Obligation:

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

ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(head(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(74) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  x1
from(x1)  =  from(x1)
s(x1)  =  s(x1)
head(x1)  =  x1
active(x1)  =  active(x1)
mark(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)
2nd(x1)  =  x1
take(x1, x2)  =  take(x2)
0  =  0
nil  =  nil
sel(x1, x2)  =  x2
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > from1 > s1
active1 > cons2
active1 > take1
active1 > nil
proper1 > from1 > s1
proper1 > cons2
proper1 > take1
proper1 > 0

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(75) Obligation:

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

ACTIVE(head(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(76) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(head(X)) → ACTIVE(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  ACTIVE(x1)
head(x1)  =  head(x1)
active(x1)  =  active(x1)
from(x1)  =  from(x1)
mark(x1)  =  mark(x1)
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  x1
2nd(x1)  =  x1
take(x1, x2)  =  take(x1, x2)
0  =  0
nil  =  nil
sel(x1, x2)  =  sel(x1, x2)
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
ACTIVE1 > mark1
proper1 > 0 > nil > ok > from1 > mark1
proper1 > 0 > nil > ok > cons2 > head1 > mark1
proper1 > 0 > nil > ok > cons2 > take2 > mark1
proper1 > 0 > nil > ok > cons2 > sel2 > mark1
top > active1 > from1 > mark1
top > active1 > cons2 > head1 > mark1
top > active1 > cons2 > take2 > mark1
top > active1 > cons2 > sel2 > mark1

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(77) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(78) PisEmptyProof (EQUIVALENT transformation)

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

(79) TRUE

(80) Obligation:

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

TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(X))

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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