(0) Obligation:

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

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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(2nd(cons(X, X1))) → 2ND(cons1(X, X1))
ACTIVE(2nd(cons(X, X1))) → CONS1(X, X1)
ACTIVE(from(X)) → CONS(X, from(s(X)))
ACTIVE(from(X)) → FROM(s(X))
ACTIVE(from(X)) → S(X)
ACTIVE(2nd(X)) → 2ND(active(X))
ACTIVE(2nd(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(from(X)) → FROM(active(X))
ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(s(X)) → S(active(X))
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(cons1(X1, X2)) → CONS1(active(X1), X2)
ACTIVE(cons1(X1, X2)) → ACTIVE(X1)
ACTIVE(cons1(X1, X2)) → CONS1(X1, active(X2))
ACTIVE(cons1(X1, X2)) → ACTIVE(X2)
2ND(mark(X)) → 2ND(X)
CONS(mark(X1), X2) → CONS(X1, X2)
FROM(mark(X)) → FROM(X)
S(mark(X)) → S(X)
CONS1(mark(X1), X2) → CONS1(X1, X2)
CONS1(X1, mark(X2)) → CONS1(X1, X2)
PROPER(2nd(X)) → 2ND(proper(X))
PROPER(2nd(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(from(X)) → FROM(proper(X))
PROPER(from(X)) → PROPER(X)
PROPER(s(X)) → S(proper(X))
PROPER(s(X)) → PROPER(X)
PROPER(cons1(X1, X2)) → CONS1(proper(X1), proper(X2))
PROPER(cons1(X1, X2)) → PROPER(X1)
PROPER(cons1(X1, X2)) → PROPER(X2)
2ND(ok(X)) → 2ND(X)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
FROM(ok(X)) → FROM(X)
S(ok(X)) → S(X)
CONS1(ok(X1), ok(X2)) → CONS1(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(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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 18 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

CONS1(X1, mark(X2)) → CONS1(X1, X2)
CONS1(mark(X1), X2) → CONS1(X1, X2)
CONS1(ok(X1), ok(X2)) → CONS1(X1, X2)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.


CONS1(X1, mark(X2)) → CONS1(X1, X2)
CONS1(mark(X1), X2) → CONS1(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS1(x1, x2)  =  CONS1(x1, x2)
mark(x1)  =  mark(x1)
ok(x1)  =  x1
active(x1)  =  active(x1)
2nd(x1)  =  x1
cons1(x1, x2)  =  cons1(x1, x2)
cons(x1, x2)  =  cons(x1, x2)
from(x1)  =  from(x1)
s(x1)  =  x1
proper(x1)  =  proper(x1)
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
proper1 > cons12 > mark1 > CONS12
proper1 > cons2 > mark1 > CONS12
proper1 > from1 > mark1 > CONS12
top > active1 > cons12 > mark1 > CONS12
top > active1 > cons2 > mark1 > CONS12
top > active1 > from1 > mark1 > CONS12

Status:
CONS12: [1,2]
mark1: [1]
active1: [1]
cons12: [1,2]
cons2: [2,1]
from1: [1]
proper1: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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:

CONS1(ok(X1), ok(X2)) → CONS1(X1, X2)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.


CONS1(ok(X1), ok(X2)) → CONS1(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS1(x1, x2)  =  CONS1(x1)
ok(x1)  =  ok(x1)
active(x1)  =  active(x1)
2nd(x1)  =  2nd(x1)
cons1(x1, x2)  =  cons1(x1, x2)
cons(x1, x2)  =  cons(x1, x2)
mark(x1)  =  x1
from(x1)  =  from(x1)
s(x1)  =  s(x1)
proper(x1)  =  x1
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
CONS11 > ok1
active1 > cons2 > 2nd1 > cons12 > ok1
active1 > from1 > ok1
active1 > s1 > ok1
top > ok1

Status:
CONS11: [1]
ok1: [1]
active1: [1]
2nd1: [1]
cons12: [1,2]
cons2: [1,2]
from1: [1]
s1: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(9) Obligation:

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

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

S(ok(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

(13) 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)
2nd(x1)  =  x1
cons1(x1, x2)  =  cons1(x1, x2)
cons(x1, x2)  =  cons(x1, x2)
from(x1)  =  from(x1)
s(x1)  =  x1
proper(x1)  =  x1
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
active1 > cons2 > cons12 > mark1
active1 > from1 > mark1

Status:
S1: [1]
mark1: [1]
active1: [1]
cons12: [2,1]
cons2: [1,2]
from1: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(14) 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(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.


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)  =  x1
2nd(x1)  =  x1
cons1(x1, x2)  =  cons1(x2)
cons(x1, x2)  =  cons(x2)
mark(x1)  =  mark
from(x1)  =  from(x1)
s(x1)  =  x1
proper(x1)  =  x1
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
cons11 > ok1
cons11 > mark
from1 > cons1 > ok1
from1 > cons1 > mark

Status:
ok1: [1]
cons11: [1]
cons1: [1]
mark: []
from1: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(16) Obligation:

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

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE

(19) 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(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

(20) 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)
2nd(x1)  =  x1
cons1(x1, x2)  =  cons1(x1, x2)
cons(x1, x2)  =  cons(x1, x2)
from(x1)  =  from(x1)
s(x1)  =  x1
proper(x1)  =  x1
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
active1 > cons2 > cons12 > mark1
active1 > from1 > mark1

Status:
FROM1: [1]
mark1: [1]
active1: [1]
cons12: [2,1]
cons2: [1,2]
from1: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(21) 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(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

(22) 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)  =  x1
2nd(x1)  =  x1
cons1(x1, x2)  =  cons1(x2)
cons(x1, x2)  =  cons(x2)
mark(x1)  =  mark
from(x1)  =  from(x1)
s(x1)  =  x1
proper(x1)  =  x1
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
cons11 > ok1
cons11 > mark
from1 > cons1 > ok1
from1 > cons1 > mark

Status:
ok1: [1]
cons11: [1]
cons1: [1]
mark: []
from1: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(23) Obligation:

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

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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) PisEmptyProof (EQUIVALENT transformation)

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

(25) TRUE

(26) 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(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

(27) 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)
2nd(x1)  =  2nd(x1)
cons1(x1, x2)  =  cons1(x1, x2)
cons(x1, x2)  =  cons(x1, x2)
from(x1)  =  from(x1)
s(x1)  =  s(x1)
proper(x1)  =  proper(x1)
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
active1 > 2nd1 > mark1 > CONS2
active1 > cons12 > mark1 > CONS2
active1 > cons2 > mark1 > CONS2
active1 > from1 > mark1 > CONS2
active1 > s1 > mark1 > CONS2
proper1 > 2nd1 > mark1 > CONS2
proper1 > cons12 > mark1 > CONS2
proper1 > cons2 > mark1 > CONS2
proper1 > from1 > mark1 > CONS2
proper1 > s1 > mark1 > CONS2

Status:
CONS2: [2,1]
mark1: [1]
active1: [1]
2nd1: [1]
cons12: [1,2]
cons2: [2,1]
from1: [1]
s1: [1]
proper1: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(28) 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(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

(29) 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)
2nd(x1)  =  2nd(x1)
cons1(x1, x2)  =  cons1(x1, x2)
cons(x1, x2)  =  cons(x1, x2)
mark(x1)  =  x1
from(x1)  =  from(x1)
s(x1)  =  s(x1)
proper(x1)  =  x1
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
CONS1 > ok1
active1 > cons2 > 2nd1 > cons12 > ok1
active1 > from1 > ok1
active1 > s1 > ok1
top > ok1

Status:
CONS1: [1]
ok1: [1]
active1: [1]
2nd1: [1]
cons12: [1,2]
cons2: [1,2]
from1: [1]
s1: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(30) Obligation:

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

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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) PisEmptyProof (EQUIVALENT transformation)

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

(32) TRUE

(33) 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(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

(34) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


2ND(mark(X)) → 2ND(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
2ND(x1)  =  2ND(x1)
ok(x1)  =  x1
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
2nd(x1)  =  x1
cons1(x1, x2)  =  cons1(x1, x2)
cons(x1, x2)  =  cons(x1, x2)
from(x1)  =  from(x1)
s(x1)  =  x1
proper(x1)  =  x1
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
active1 > cons2 > cons12 > mark1
active1 > from1 > mark1

Status:
2ND1: [1]
mark1: [1]
active1: [1]
cons12: [2,1]
cons2: [1,2]
from1: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(35) 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(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

(36) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


2ND(ok(X)) → 2ND(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
2ND(x1)  =  x1
ok(x1)  =  ok(x1)
active(x1)  =  x1
2nd(x1)  =  x1
cons1(x1, x2)  =  cons1(x2)
cons(x1, x2)  =  cons(x2)
mark(x1)  =  mark
from(x1)  =  from(x1)
s(x1)  =  x1
proper(x1)  =  x1
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
cons11 > ok1
cons11 > mark
from1 > cons1 > ok1
from1 > cons1 > mark

Status:
ok1: [1]
cons11: [1]
cons1: [1]
mark: []
from1: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(37) Obligation:

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

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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) PisEmptyProof (EQUIVALENT transformation)

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

(39) TRUE

(40) Obligation:

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

PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(2nd(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(from(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)
PROPER(cons1(X1, X2)) → PROPER(X1)
PROPER(cons1(X1, X2)) → PROPER(X2)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

(41) 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(from(X)) → PROPER(X)
PROPER(cons1(X1, X2)) → PROPER(X1)
PROPER(cons1(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)
2nd(x1)  =  x1
from(x1)  =  from(x1)
s(x1)  =  x1
cons1(x1, x2)  =  cons1(x1, x2)
active(x1)  =  active(x1)
mark(x1)  =  x1
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
active1 > cons2 > cons12 > PROPER1 > top
active1 > from1 > PROPER1 > top

Status:
PROPER1: [1]
cons2: [2,1]
from1: [1]
cons12: [2,1]
active1: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(42) Obligation:

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

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

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

(43) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(2nd(X)) → PROPER(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  PROPER(x1)
2nd(x1)  =  2nd(x1)
s(x1)  =  x1
active(x1)  =  active(x1)
cons1(x1, x2)  =  cons1(x1, x2)
cons(x1, x2)  =  cons(x1, x2)
mark(x1)  =  mark
from(x1)  =  from(x1)
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
proper1 > 2nd1 > PROPER1
proper1 > 2nd1 > cons12
proper1 > cons2 > cons12
proper1 > from1
ok > active1 > 2nd1 > PROPER1
ok > active1 > 2nd1 > cons12
ok > active1 > cons2 > cons12
ok > active1 > mark > cons12
ok > active1 > mark > from1
top > active1 > 2nd1 > PROPER1
top > active1 > 2nd1 > cons12
top > active1 > cons2 > cons12
top > active1 > mark > cons12
top > active1 > mark > from1

Status:
PROPER1: [1]
2nd1: [1]
active1: [1]
cons12: [1,2]
cons2: [1,2]
mark: []
from1: [1]
proper1: [1]
ok: []
top: []

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(44) Obligation:

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

PROPER(s(X)) → PROPER(X)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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(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
s(x1)  =  s(x1)
active(x1)  =  active(x1)
2nd(x1)  =  x1
cons1(x1, x2)  =  cons1(x1, x2)
cons(x1, x2)  =  cons(x1, x2)
mark(x1)  =  mark
from(x1)  =  from(x1)
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
proper1 > s1 > from1
proper1 > cons12 > from1
proper1 > cons2 > from1
top > active1 > s1 > from1
top > active1 > cons12 > from1
top > active1 > cons2 > from1
top > active1 > mark > from1

Status:
s1: [1]
active1: [1]
cons12: [1,2]
cons2: [2,1]
mark: []
from1: [1]
proper1: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(46) Obligation:

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

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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) PisEmptyProof (EQUIVALENT transformation)

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

(48) TRUE

(49) Obligation:

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

ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(2nd(X)) → ACTIVE(X)
ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(cons1(X1, X2)) → ACTIVE(X1)
ACTIVE(cons1(X1, X2)) → ACTIVE(X2)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

(50) 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(cons1(X1, X2)) → ACTIVE(X1)
ACTIVE(cons1(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)
2nd(x1)  =  x1
from(x1)  =  x1
s(x1)  =  x1
cons1(x1, x2)  =  cons1(x1, x2)
active(x1)  =  active(x1)
mark(x1)  =  x1
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
top > active1 > cons2
top > active1 > cons12

Status:
ACTIVE1: [1]
cons2: [1,2]
cons12: [1,2]
active1: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(51) Obligation:

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

ACTIVE(2nd(X)) → ACTIVE(X)
ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.


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)
2nd(x1)  =  x1
from(x1)  =  x1
s(x1)  =  s(x1)
active(x1)  =  active(x1)
cons1(x1, x2)  =  cons1(x2)
cons(x1, x2)  =  cons(x1, x2)
mark(x1)  =  mark
proper(x1)  =  proper(x1)
ok(x1)  =  ok(x1)
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
active1 > s1 > ACTIVE1
active1 > s1 > mark
active1 > s1 > ok1
active1 > cons11 > mark
active1 > cons11 > ok1
active1 > cons2 > mark
active1 > cons2 > ok1
proper1 > s1 > ACTIVE1
proper1 > s1 > mark
proper1 > s1 > ok1
proper1 > cons11 > mark
proper1 > cons11 > ok1
proper1 > cons2 > mark
proper1 > cons2 > ok1

Status:
ACTIVE1: [1]
s1: [1]
active1: [1]
cons11: [1]
cons2: [1,2]
mark: []
proper1: [1]
ok1: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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:

ACTIVE(2nd(X)) → ACTIVE(X)
ACTIVE(from(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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(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)
2nd(x1)  =  x1
from(x1)  =  from(x1)
active(x1)  =  x1
cons1(x1, x2)  =  cons1(x1)
cons(x1, x2)  =  cons
mark(x1)  =  mark
s(x1)  =  x1
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
proper1 > from1 > mark
proper1 > cons11 > mark
proper1 > cons > mark

Status:
ACTIVE1: [1]
from1: [1]
cons11: [1]
cons: []
mark: []
proper1: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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(2nd(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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(2nd(X)) → ACTIVE(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  x1
2nd(x1)  =  2nd(x1)
active(x1)  =  active(x1)
cons1(x1, x2)  =  cons1(x1, x2)
cons(x1, x2)  =  cons(x1, x2)
mark(x1)  =  x1
from(x1)  =  from(x1)
s(x1)  =  x1
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
top > active1 > cons2 > 2nd1
top > active1 > cons2 > cons12
top > active1 > from1
top > proper1 > cons2 > 2nd1
top > proper1 > cons2 > cons12
top > proper1 > from1

Status:
2nd1: [1]
active1: [1]
cons12: [1,2]
cons2: [2,1]
from1: [1]
proper1: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(57) Obligation:

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

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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) PisEmptyProof (EQUIVALENT transformation)

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

(59) TRUE

(60) 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(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.