(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(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)  =  x1
mark(x1)  =  x1
ok(x1)  =  ok(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)  =  x2
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:
active1 > from1 > [ok1, nil]
active1 > cons2 > take2 > [ok1, nil]
[0, proper1] > from1 > [ok1, nil]
[0, proper1] > cons2 > take2 > [ok1, nil]
top > [ok1, nil]

Status:
ok1: [1]
active1: multiset
from1: [1]
cons2: [1,2]
take2: [1,2]
0: multiset
nil: multiset
proper1: [1]
top: multiset


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(mark(X1), 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(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)
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
ok(x1)  =  x1
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:
SEL1 > [nil, top]
active1 > from1 > [mark1, s1] > [nil, top]
active1 > head1 > [mark1, s1] > [nil, top]
active1 > take2 > cons2 > sel2 > [mark1, s1] > [nil, top]
0 > [mark1, s1] > [nil, top]

Status:
SEL1: multiset
mark1: [1]
active1: [1]
from1: multiset
cons2: [1,2]
s1: [1]
head1: multiset
take2: [1,2]
0: multiset
nil: multiset
sel2: multiset
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))

(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)  =  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)  =  x1
ok(x1)  =  x1
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:
0 > [active1, nil] > cons2 > take2 > mark1
0 > [active1, nil] > cons2 > sel2 > mark1
top > mark1

Status:
mark1: [1]
active1: [1]
cons2: [1,2]
take2: multiset
0: multiset
nil: multiset
sel2: [1,2]
top: multiset


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)  =  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)  =  x1
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:
TAKE1 > [mark1, top]
active1 > [cons2, take2] > sel2 > [mark1, top]
active1 > nil > [mark1, top]
0 > [mark1, top]

Status:
TAKE1: [1]
mark1: [1]
active1: [1]
cons2: multiset
take2: [1,2]
0: multiset
nil: multiset
sel2: [1,2]
top: multiset


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(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)
ok(x1)  =  x1
active(x1)  =  active(x1)
from(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  s(x1)
head(x1)  =  x1
2nd(x1)  =  2nd(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 with status [RPO].
Quasi-Precedence:
[active1, 2nd1] > [cons2, s1, take2, sel2] > [mark1, top]
[active1, 2nd1] > [cons2, s1, take2, sel2] > [0, nil]

Status:
mark1: [1]
active1: [1]
cons2: [1,2]
s1: [1]
2nd1: [1]
take2: [1,2]
0: multiset
nil: multiset
sel2: [2,1]
top: multiset


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(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.

(19) 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, x2)
ok(x1)  =  ok(x1)
active(x1)  =  x1
from(x1)  =  x1
mark(x1)  =  mark
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  x1
head(x1)  =  x1
2nd(x1)  =  x1
take(x1, x2)  =  take(x2)
0  =  0
nil  =  nil
sel(x1, x2)  =  x1
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:
proper1 > take1 > cons2 > ok1 > TAKE2 > mark
proper1 > take1 > nil > mark
proper1 > 0 > ok1 > TAKE2 > mark
proper1 > 0 > nil > mark
top > mark

Status:
TAKE2: multiset
ok1: [1]
mark: multiset
cons2: [2,1]
take1: [1]
0: multiset
nil: multiset
proper1: multiset
top: multiset


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(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)  =  x1
ok(x1)  =  x1
mark(x1)  =  mark(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)  =  x1
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:
active1 > from1 > mark1 > top
active1 > take2 > [cons2, sel2] > mark1 > top
active1 > take2 > nil > top
0 > mark1 > top
0 > nil > top

Status:
mark1: multiset
active1: [1]
from1: [1]
cons2: [1,2]
take2: [2,1]
0: multiset
nil: multiset
sel2: [1,2]
top: multiset


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(ok(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(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)  =  2ND(x1)
ok(x1)  =  ok(x1)
active(x1)  =  x1
from(x1)  =  x1
mark(x1)  =  mark
cons(x1, x2)  =  x2
s(x1)  =  x1
head(x1)  =  x1
2nd(x1)  =  x1
take(x1, x2)  =  x1
0  =  0
nil  =  nil
sel(x1, x2)  =  x1
proper(x1)  =  proper
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:
2ND1 > [mark, nil, top]
proper > 0 > ok1 > [mark, nil, top]

Status:
2ND1: multiset
ok1: [1]
mark: []
0: multiset
nil: multiset
proper: multiset
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))

(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(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)  =  x1
ok(x1)  =  x1
mark(x1)  =  mark(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)  =  x1
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:
active1 > from1 > mark1 > top
active1 > take2 > [cons2, sel2] > mark1 > top
active1 > take2 > nil > top
0 > mark1 > top
0 > nil > top

Status:
mark1: multiset
active1: [1]
from1: [1]
cons2: [1,2]
take2: [2,1]
0: multiset
nil: multiset
sel2: [1,2]
top: multiset


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(ok(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(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)  =  HEAD(x1)
ok(x1)  =  ok(x1)
active(x1)  =  x1
from(x1)  =  x1
mark(x1)  =  mark
cons(x1, x2)  =  x2
s(x1)  =  x1
head(x1)  =  x1
2nd(x1)  =  x1
take(x1, x2)  =  x1
0  =  0
nil  =  nil
sel(x1, x2)  =  x1
proper(x1)  =  proper
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:
HEAD1 > [mark, nil, top]
proper > 0 > ok1 > [mark, nil, top]

Status:
HEAD1: multiset
ok1: [1]
mark: []
0: multiset
nil: multiset
proper: multiset
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))

(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(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)  =  x1
ok(x1)  =  x1
mark(x1)  =  mark(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)  =  x1
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:
active1 > from1 > mark1 > top
active1 > take2 > [cons2, sel2] > mark1 > top
active1 > take2 > nil > top
0 > mark1 > top
0 > nil > top

Status:
mark1: multiset
active1: [1]
from1: [1]
cons2: [1,2]
take2: [2,1]
0: multiset
nil: multiset
sel2: [1,2]
top: multiset


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(ok(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(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)  =  S(x1)
ok(x1)  =  ok(x1)
active(x1)  =  x1
from(x1)  =  x1
mark(x1)  =  mark
cons(x1, x2)  =  x2
s(x1)  =  x1
head(x1)  =  x1
2nd(x1)  =  x1
take(x1, x2)  =  x1
0  =  0
nil  =  nil
sel(x1, x2)  =  x1
proper(x1)  =  proper
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:
S1 > [mark, nil, top]
proper > 0 > ok1 > [mark, nil, top]

Status:
S1: multiset
ok1: [1]
mark: []
0: multiset
nil: multiset
proper: multiset
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))

(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(x2)
ok(x1)  =  ok(x1)
mark(x1)  =  mark
active(x1)  =  x1
from(x1)  =  x1
cons(x1, x2)  =  x1
s(x1)  =  s(x1)
head(x1)  =  x1
2nd(x1)  =  2nd(x1)
take(x1, x2)  =  x1
0  =  0
nil  =  nil
sel(x1, x2)  =  x2
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:
[nil, proper1] > s1 > [ok1, top] > [CONS1, mark]
[nil, proper1] > 2nd1 > [ok1, top] > [CONS1, mark]
[nil, proper1] > 0 > [CONS1, mark]

Status:
CONS1: multiset
ok1: [1]
mark: multiset
s1: [1]
2nd1: multiset
0: multiset
nil: multiset
proper1: [1]
top: multiset


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)  =  CONS(x1, x2)
mark(x1)  =  mark(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)  =  take(x1, x2)
0  =  0
nil  =  nil
sel(x1, x2)  =  sel(x1, x2)
proper(x1)  =  proper(x1)
ok(x1)  =  ok(x1)
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:
active1 > cons2 > take2 > [mark1, s1, ok1] > CONS2
active1 > cons2 > take2 > [mark1, s1, ok1] > top
active1 > cons2 > sel2 > [mark1, s1, ok1] > CONS2
active1 > cons2 > sel2 > [mark1, s1, ok1] > top
active1 > [0, nil] > [mark1, s1, ok1] > CONS2
active1 > [0, nil] > [mark1, s1, ok1] > top
proper1 > cons2 > take2 > [mark1, s1, ok1] > CONS2
proper1 > cons2 > take2 > [mark1, s1, ok1] > top
proper1 > cons2 > sel2 > [mark1, s1, ok1] > CONS2
proper1 > cons2 > sel2 > [mark1, s1, ok1] > top
proper1 > [0, nil] > [mark1, s1, ok1] > CONS2
proper1 > [0, nil] > [mark1, s1, ok1] > top

Status:
CONS2: multiset
mark1: [1]
active1: [1]
cons2: [2,1]
s1: [1]
take2: [2,1]
0: multiset
nil: multiset
sel2: [2,1]
proper1: [1]
ok1: [1]
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(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)  =  x1
ok(x1)  =  x1
mark(x1)  =  mark(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)  =  x1
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:
active1 > from1 > mark1 > top
active1 > take2 > [cons2, sel2] > mark1 > top
active1 > take2 > nil > top
0 > mark1 > top
0 > nil > top

Status:
mark1: multiset
active1: [1]
from1: [1]
cons2: [1,2]
take2: [2,1]
0: multiset
nil: multiset
sel2: [1,2]
top: multiset


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(ok(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(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)  =  FROM(x1)
ok(x1)  =  ok(x1)
active(x1)  =  x1
from(x1)  =  x1
mark(x1)  =  mark
cons(x1, x2)  =  x2
s(x1)  =  x1
head(x1)  =  x1
2nd(x1)  =  x1
take(x1, x2)  =  x1
0  =  0
nil  =  nil
sel(x1, x2)  =  x1
proper(x1)  =  proper
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:
FROM1 > [mark, nil, top]
proper > 0 > ok1 > [mark, nil, top]

Status:
FROM1: multiset
ok1: [1]
mark: []
0: multiset
nil: multiset
proper: multiset
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))

(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)  =  x1
mark(x1)  =  mark
0  =  0
nil  =  nil
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:
cons2 > [PROPER1, take2] > mark
sel2 > [PROPER1, take2] > mark
[0, nil] > mark
top > mark

Status:
PROPER1: multiset
cons2: multiset
take2: multiset
sel2: [2,1]
mark: []
0: multiset
nil: multiset
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))

(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(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)  =  x1
from(x1)  =  x1
s(x1)  =  x1
head(x1)  =  x1
2nd(x1)  =  2nd(x1)
active(x1)  =  x1
mark(x1)  =  mark
cons(x1, x2)  =  x2
take(x1, x2)  =  take
0  =  0
nil  =  nil
sel(x1, x2)  =  sel
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:
[0, nil] > [mark, take, ok]
sel > [2nd1, proper1] > [mark, take, ok]
top > [2nd1, proper1] > [mark, take, ok]

Status:
2nd1: [1]
mark: []
take: []
0: multiset
nil: multiset
sel: []
proper1: [1]
ok: []
top: multiset


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(from(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)
PROPER(head(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(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)  =  x1
s(x1)  =  s(x1)
head(x1)  =  x1
active(x1)  =  x1
mark(x1)  =  mark
cons(x1, x2)  =  cons
2nd(x1)  =  x1
take(x1, x2)  =  take
0  =  0
nil  =  nil
sel(x1, x2)  =  x1
proper(x1)  =  x1
ok(x1)  =  ok
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:
PROPER1 > [mark, cons, ok]
s1 > [mark, cons, ok]
take > nil > [mark, cons, ok]
0 > [mark, cons, ok]
top > [mark, cons, ok]

Status:
PROPER1: multiset
s1: multiset
mark: []
cons: []
take: multiset
0: multiset
nil: multiset
ok: []
top: multiset


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(from(X)) → PROPER(X)
PROPER(head(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(from(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)
head(x1)  =  x1
active(x1)  =  x1
mark(x1)  =  mark
cons(x1, x2)  =  cons
s(x1)  =  s(x1)
2nd(x1)  =  2nd
take(x1, x2)  =  take
0  =  0
nil  =  nil
sel(x1, x2)  =  x2
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:
PROPER1 > [mark, cons, top]
from1 > [mark, cons, top]
s1 > [mark, cons, top]
2nd > [mark, cons, top]
take > nil > [mark, cons, top]
0 > [mark, cons, top]

Status:
PROPER1: multiset
from1: multiset
mark: []
cons: []
s1: [1]
2nd: []
take: []
0: multiset
nil: multiset
top: multiset


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:
The TRS P consists of the following rules:

PROPER(head(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.

(67) 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)  =  PROPER(x1)
head(x1)  =  head(x1)
active(x1)  =  active(x1)
from(x1)  =  x1
mark(x1)  =  mark
cons(x1, x2)  =  cons
s(x1)  =  s
2nd(x1)  =  x1
take(x1, x2)  =  take
0  =  0
nil  =  nil
sel(x1, x2)  =  sel(x1)
proper(x1)  =  x1
ok(x1)  =  ok
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:
PROPER1 > ok
active1 > [s, take, nil] > [head1, mark, cons, 0] > ok
active1 > sel1 > [head1, mark, cons, 0] > ok
top > ok

Status:
PROPER1: multiset
head1: [1]
active1: [1]
mark: []
cons: []
s: multiset
take: multiset
0: multiset
nil: multiset
sel1: [1]
ok: []
top: multiset


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))

(68) 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.

(69) PisEmptyProof (EQUIVALENT transformation)

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

(70) TRUE

(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)
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.

(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)
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)  =  ACTIVE(x1)
cons(x1, x2)  =  cons(x1)
from(x1)  =  x1
s(x1)  =  x1
head(x1)  =  x1
2nd(x1)  =  2nd(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)  =  x1
ok(x1)  =  ok
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:
active1 > sel2 > mark > 2nd1 > [0, ok] > cons1
active1 > sel2 > mark > take2 > ACTIVE1 > cons1
active1 > sel2 > mark > take2 > [0, ok] > cons1
active1 > sel2 > mark > top > cons1
active1 > nil > [0, ok] > cons1

Status:
ACTIVE1: multiset
cons1: multiset
2nd1: [1]
take2: [2,1]
sel2: multiset
active1: [1]
mark: []
0: multiset
nil: multiset
ok: []
top: multiset


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(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)  =  ACTIVE(x1)
from(x1)  =  x1
s(x1)  =  s(x1)
head(x1)  =  x1
active(x1)  =  x1
mark(x1)  =  mark
cons(x1, x2)  =  cons
2nd(x1)  =  x1
take(x1, x2)  =  take
0  =  0
nil  =  nil
sel(x1, x2)  =  x1
proper(x1)  =  x1
ok(x1)  =  ok
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:
ACTIVE1 > [mark, cons, ok]
s1 > [mark, cons, ok]
take > nil > [mark, cons, ok]
0 > [mark, cons, ok]
top > [mark, cons, ok]

Status:
ACTIVE1: multiset
s1: multiset
mark: []
cons: []
take: multiset
0: multiset
nil: multiset
ok: []
top: multiset


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(from(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.

(76) 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)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  ACTIVE(x1)
from(x1)  =  from(x1)
head(x1)  =  x1
active(x1)  =  x1
mark(x1)  =  mark
cons(x1, x2)  =  cons
s(x1)  =  s(x1)
2nd(x1)  =  2nd
take(x1, x2)  =  take
0  =  0
nil  =  nil
sel(x1, x2)  =  x2
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:
ACTIVE1 > [mark, cons, top]
from1 > [mark, cons, top]
s1 > [mark, cons, top]
2nd > [mark, cons, top]
take > nil > [mark, cons, top]
0 > [mark, cons, top]

Status:
ACTIVE1: multiset
from1: multiset
mark: []
cons: []
s1: [1]
2nd: []
take: []
0: multiset
nil: multiset
top: multiset


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:
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.

(78) 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)  =  x1
mark(x1)  =  mark
cons(x1, x2)  =  cons
s(x1)  =  s
2nd(x1)  =  x1
take(x1, x2)  =  take
0  =  0
nil  =  nil
sel(x1, x2)  =  sel(x1)
proper(x1)  =  x1
ok(x1)  =  ok
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:
ACTIVE1 > ok
active1 > [s, take, nil] > [head1, mark, cons, 0] > ok
active1 > sel1 > [head1, mark, cons, 0] > ok
top > ok

Status:
ACTIVE1: multiset
head1: [1]
active1: [1]
mark: []
cons: []
s: multiset
take: multiset
0: multiset
nil: multiset
sel1: [1]
ok: []
top: multiset


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))

(79) 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.

(80) PisEmptyProof (EQUIVALENT transformation)

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

(81) TRUE

(82) 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.