(0) Obligation:

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

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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(first(s(X), cons(Y, Z))) → CONS(Y, first(X, Z))
ACTIVE(first(s(X), cons(Y, Z))) → FIRST(X, Z)
ACTIVE(sel(s(X), cons(Y, Z))) → SEL(X, Z)
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(first(X1, X2)) → FIRST(active(X1), X2)
ACTIVE(first(X1, X2)) → ACTIVE(X1)
ACTIVE(first(X1, X2)) → FIRST(X1, active(X2))
ACTIVE(first(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)
FIRST(mark(X1), X2) → FIRST(X1, X2)
FIRST(X1, mark(X2)) → FIRST(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(first(X1, X2)) → FIRST(proper(X1), proper(X2))
PROPER(first(X1, X2)) → PROPER(X1)
PROPER(first(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)
FIRST(ok(X1), ok(X2)) → FIRST(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(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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 8 SCCs with 20 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(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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)  =  SEL(x1, x2)
mark(x1)  =  x1
ok(x1)  =  ok(x1)
active(x1)  =  active(x1)
from(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  s(x1)
first(x1, x2)  =  first(x2)
0  =  0
nil  =  nil
sel(x1, x2)  =  sel(x1, x2)
proper(x1)  =  proper(x1)
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
top > active1 > cons2 > first1 > ok1 > SEL2
top > active1 > cons2 > sel2 > ok1 > SEL2
top > active1 > s1 > ok1 > SEL2
top > active1 > nil > ok1 > SEL2
top > proper1 > cons2 > first1 > ok1 > SEL2
top > proper1 > cons2 > sel2 > ok1 > SEL2
top > proper1 > s1 > ok1 > SEL2
top > proper1 > 0
top > proper1 > nil > ok1 > SEL2

Status:
sel2: [1,2]
active1: [1]
cons2: [1,2]
ok1: [1]
s1: [1]
SEL2: [2,1]
proper1: [1]
top: []
0: []
first1: [1]
nil: []

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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)  =  x1
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
from(x1)  =  from(x1)
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  s(x1)
first(x1, x2)  =  first(x1, x2)
0  =  0
nil  =  nil
sel(x1, x2)  =  sel(x1, x2)
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
active1 > from1 > cons2 > sel2 > mark1
active1 > from1 > s1 > mark1
active1 > first2 > cons2 > sel2 > mark1
active1 > nil > mark1
0 > mark1
top > proper1 > from1 > cons2 > sel2 > mark1
top > proper1 > from1 > s1 > mark1
top > proper1 > first2 > cons2 > sel2 > mark1

Status:
sel2: [1,2]
active1: [1]
from1: [1]
cons2: [2,1]
mark1: [1]
s1: [1]
proper1: [1]
top: []
first2: [2,1]
0: []
nil: []

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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)  =  SEL(x2)
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
from(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  x1
first(x1, x2)  =  first(x1, x2)
0  =  0
nil  =  nil
sel(x1, x2)  =  sel(x1, x2)
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
0 > mark1
top > active1 > cons2 > mark1
top > active1 > first2 > mark1
top > active1 > first2 > nil
top > active1 > sel2 > mark1

Status:
sel2: [2,1]
active1: [1]
cons2: [2,1]
mark1: [1]
SEL1: [1]
top: []
first2: [1,2]
0: []
nil: []

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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:

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

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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.


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

Lexicographic path order with status [LPO].
Precedence:
top > active1 > cons2 > first1 > ok1 > FIRST2
top > active1 > cons2 > sel2 > ok1 > FIRST2
top > active1 > s1 > ok1 > FIRST2
top > active1 > nil > ok1 > FIRST2
top > proper1 > cons2 > first1 > ok1 > FIRST2
top > proper1 > cons2 > sel2 > ok1 > FIRST2
top > proper1 > s1 > ok1 > FIRST2
top > proper1 > 0
top > proper1 > nil > ok1 > FIRST2

Status:
sel2: [1,2]
FIRST2: [2,1]
active1: [1]
cons2: [1,2]
ok1: [1]
s1: [1]
proper1: [1]
top: []
0: []
first1: [1]
nil: []

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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:

FIRST(X1, mark(X2)) → FIRST(X1, X2)
FIRST(mark(X1), X2) → FIRST(X1, X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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.


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

Lexicographic path order with status [LPO].
Precedence:
active1 > from1 > cons2 > sel2 > mark1
active1 > from1 > s1 > mark1
active1 > first2 > cons2 > sel2 > mark1
active1 > nil > mark1
0 > mark1
top > proper1 > from1 > cons2 > sel2 > mark1
top > proper1 > from1 > s1 > mark1
top > proper1 > first2 > cons2 > sel2 > mark1

Status:
sel2: [1,2]
active1: [1]
from1: [1]
cons2: [2,1]
mark1: [1]
s1: [1]
proper1: [1]
top: []
first2: [2,1]
0: []
nil: []

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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:

FIRST(X1, mark(X2)) → FIRST(X1, X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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.


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

Lexicographic path order with status [LPO].
Precedence:
0 > mark1
top > active1 > cons2 > mark1
top > active1 > first2 > mark1
top > active1 > first2 > nil
top > active1 > sel2 > mark1

Status:
FIRST1: [1]
sel2: [2,1]
active1: [1]
cons2: [2,1]
mark1: [1]
top: []
first2: [1,2]
0: []
nil: []

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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:

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(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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.


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

Lexicographic path order with status [LPO].
Precedence:
S1 > mark1
active1 > first2 > cons2 > mark1
active1 > first2 > nil > mark1
active1 > sel2 > mark1
proper1 > first2 > cons2 > mark1
proper1 > first2 > nil > mark1
proper1 > 0 > mark1
proper1 > sel2 > mark1
top > mark1

Status:
sel2: [2,1]
active1: [1]
cons2: [1,2]
mark1: [1]
proper1: [1]
top: []
first2: [1,2]
0: []
nil: []
S1: [1]

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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:

S(ok(X)) → S(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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.


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

Lexicographic path order with status [LPO].
Precedence:
top > active1 > cons2 > ok1
top > active1 > s1 > ok1
top > active1 > first2 > ok1
top > active1 > nil > ok1
top > proper1 > cons2 > ok1
top > proper1 > s1 > ok1
top > proper1 > first2 > ok1
top > proper1 > 0 > ok1
top > proper1 > nil > ok1

Status:
active1: [1]
cons2: [2,1]
ok1: [1]
s1: [1]
proper1: [1]
top: []
first2: [1,2]
0: []
nil: []

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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:

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(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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.


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)
ok(x1)  =  x1
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
from(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  s(x1)
first(x1, x2)  =  first(x1, x2)
0  =  0
nil  =  nil
sel(x1, x2)  =  sel(x1, x2)
proper(x1)  =  x1
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
active1 > s1 > cons2 > first2 > mark1
active1 > nil
active1 > sel2 > mark1

Status:
sel2: [1,2]
active1: [1]
cons2: [2,1]
CONS2: [2,1]
mark1: [1]
s1: [1]
top: []
first2: [2,1]
0: []
nil: []

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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:

CONS(ok(X1), ok(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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.


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

Lexicographic path order with status [LPO].
Precedence:
CONS1 > ok1
top > active1 > mark > proper1 > from1 > ok1
top > active1 > mark > proper1 > first2 > ok1
top > active1 > mark > proper1 > 0 > ok1
top > active1 > mark > proper1 > sel2 > ok1
top > active1 > nil > ok1

Status:
sel2: [2,1]
active1: [1]
from1: [1]
CONS1: [1]
mark: []
ok1: [1]
proper1: [1]
top: []
first2: [2,1]
0: []
nil: []

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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:

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(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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.


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

Lexicographic path order with status [LPO].
Precedence:
FROM1 > mark1
active1 > first2 > cons2 > mark1
active1 > first2 > nil > mark1
active1 > sel2 > mark1
proper1 > first2 > cons2 > mark1
proper1 > first2 > nil > mark1
proper1 > 0 > mark1
proper1 > sel2 > mark1
top > mark1

Status:
sel2: [2,1]
active1: [1]
cons2: [1,2]
mark1: [1]
proper1: [1]
top: []
first2: [1,2]
0: []
nil: []
FROM1: [1]

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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:

FROM(ok(X)) → FROM(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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.


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

Lexicographic path order with status [LPO].
Precedence:
top > active1 > cons2 > ok1
top > active1 > s1 > ok1
top > active1 > first2 > ok1
top > active1 > nil > ok1
top > proper1 > cons2 > ok1
top > proper1 > s1 > ok1
top > proper1 > first2 > ok1
top > proper1 > 0 > ok1
top > proper1 > nil > ok1

Status:
active1: [1]
cons2: [2,1]
ok1: [1]
s1: [1]
proper1: [1]
top: []
first2: [1,2]
0: []
nil: []

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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:

PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(from(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(s(X)) → PROPER(X)
PROPER(first(X1, X2)) → PROPER(X1)
PROPER(first(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(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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.


PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(first(X1, X2)) → PROPER(X1)
PROPER(first(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)  =  x1
cons(x1, x2)  =  cons(x1, x2)
from(x1)  =  x1
s(x1)  =  x1
first(x1, x2)  =  first(x1, x2)
sel(x1, x2)  =  sel(x1, x2)
active(x1)  =  active(x1)
mark(x1)  =  mark(x1)
0  =  0
nil  =  nil
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
proper1 > cons2 > mark1
proper1 > cons2 > ok
proper1 > first2 > mark1
proper1 > first2 > ok
proper1 > sel2 > mark1
proper1 > sel2 > ok
proper1 > 0
proper1 > nil > ok
top > active1 > cons2 > mark1
top > active1 > cons2 > ok
top > active1 > first2 > mark1
top > active1 > first2 > ok
top > active1 > sel2 > mark1
top > active1 > sel2 > ok
top > active1 > nil > ok

Status:
sel2: [1,2]
cons2: [1,2]
active1: [1]
mark1: [1]
ok: []
proper1: [1]
top: []
first2: [2,1]
0: []
nil: []

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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:

PROPER(from(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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.


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)  =  x1
from(x1)  =  x1
s(x1)  =  s(x1)
active(x1)  =  active(x1)
mark(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)
first(x1, x2)  =  first(x2)
0  =  0
nil  =  nil
sel(x1, x2)  =  x2
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
active1 > s1 > first1 > cons2
active1 > s1 > ok > cons2
active1 > nil > ok > cons2
0 > nil > ok > cons2
proper1 > s1 > first1 > cons2
proper1 > s1 > ok > cons2
proper1 > nil > ok > cons2

Status:
active1: [1]
cons2: [1,2]
s1: [1]
ok: []
proper1: [1]
top: []
0: []
first1: [1]
nil: []

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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:
The TRS P consists of the following rules:

PROPER(from(X)) → PROPER(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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) 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)
active(x1)  =  x1
mark(x1)  =  mark
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  x1
first(x1, x2)  =  first(x1, x2)
0  =  0
nil  =  nil
sel(x1, x2)  =  sel(x1)
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
proper1 > first2 > ok > from1 > mark
proper1 > first2 > ok > from1 > cons2
proper1 > 0 > mark
proper1 > nil > ok > from1 > mark
proper1 > nil > ok > from1 > cons2
proper1 > sel1 > ok > from1 > mark
proper1 > sel1 > ok > from1 > cons2

Status:
PROPER1: [1]
from1: [1]
cons2: [2,1]
sel1: [1]
mark: []
ok: []
proper1: [1]
top: []
first2: [2,1]
0: []
nil: []

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(50) 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(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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.

(51) PisEmptyProof (EQUIVALENT transformation)

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

(52) TRUE

(53) 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(first(X1, X2)) → ACTIVE(X1)
ACTIVE(first(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(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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.


ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(first(X1, X2)) → ACTIVE(X1)
ACTIVE(first(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, x2)
from(x1)  =  x1
s(x1)  =  x1
first(x1, x2)  =  first(x1, x2)
sel(x1, x2)  =  sel(x1, x2)
active(x1)  =  active(x1)
mark(x1)  =  mark(x1)
0  =  0
nil  =  nil
proper(x1)  =  x1
ok(x1)  =  ok
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
active1 > first2 > cons2 > mark1 > top
active1 > first2 > cons2 > ok
active1 > sel2 > mark1 > top
active1 > sel2 > ok
active1 > nil > ok
0 > mark1 > top
0 > ok

Status:
sel2: [2,1]
cons2: [2,1]
active1: [1]
mark1: [1]
ok: []
top: []
first2: [1,2]
0: []
ACTIVE1: [1]
nil: []

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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) 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)  =  x1
from(x1)  =  x1
s(x1)  =  s(x1)
active(x1)  =  active(x1)
mark(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)
first(x1, x2)  =  first(x2)
0  =  0
nil  =  nil
sel(x1, x2)  =  x2
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
active1 > s1 > first1 > cons2
active1 > s1 > ok > cons2
active1 > nil > ok > cons2
0 > nil > ok > cons2
proper1 > s1 > first1 > cons2
proper1 > s1 > ok > cons2
proper1 > nil > ok > cons2

Status:
active1: [1]
cons2: [1,2]
s1: [1]
ok: []
proper1: [1]
top: []
0: []
first1: [1]
nil: []

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(57) Obligation:

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

ACTIVE(from(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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.

(58) 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)
active(x1)  =  x1
mark(x1)  =  mark
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  x1
first(x1, x2)  =  first(x1, x2)
0  =  0
nil  =  nil
sel(x1, x2)  =  sel(x1)
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
proper1 > first2 > ok > from1 > mark
proper1 > first2 > ok > from1 > cons2
proper1 > 0 > mark
proper1 > nil > ok > from1 > mark
proper1 > nil > ok > from1 > cons2
proper1 > sel1 > ok > from1 > mark
proper1 > sel1 > ok > from1 > cons2

Status:
from1: [1]
cons2: [2,1]
sel1: [1]
mark: []
ok: []
proper1: [1]
top: []
first2: [2,1]
0: []
ACTIVE1: [1]
nil: []

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(59) 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(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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.

(60) PisEmptyProof (EQUIVALENT transformation)

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

(61) TRUE

(62) 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(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(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))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(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(first(X1, X2)) → first(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))
first(ok(X1), ok(X2)) → ok(first(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.