(0) Obligation:

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

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
active(__(X1, X2)) → __(active(X1), X2)
active(__(X1, X2)) → __(X1, active(X2))
active(and(X1, X2)) → and(active(X1), X2)
active(isNePal(X)) → isNePal(active(X))
__(mark(X1), X2) → mark(__(X1, X2))
__(X1, mark(X2)) → mark(__(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
isNePal(mark(X)) → mark(isNePal(X))
proper(__(X1, X2)) → __(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(isNePal(X)) → isNePal(proper(X))
__(ok(X1), ok(X2)) → ok(__(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNePal(ok(X)) → ok(isNePal(X))
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(__(__(X, Y), Z)) → __1(X, __(Y, Z))
ACTIVE(__(__(X, Y), Z)) → __1(Y, Z)
ACTIVE(__(X1, X2)) → __1(active(X1), X2)
ACTIVE(__(X1, X2)) → ACTIVE(X1)
ACTIVE(__(X1, X2)) → __1(X1, active(X2))
ACTIVE(__(X1, X2)) → ACTIVE(X2)
ACTIVE(and(X1, X2)) → AND(active(X1), X2)
ACTIVE(and(X1, X2)) → ACTIVE(X1)
ACTIVE(isNePal(X)) → ISNEPAL(active(X))
ACTIVE(isNePal(X)) → ACTIVE(X)
__1(mark(X1), X2) → __1(X1, X2)
__1(X1, mark(X2)) → __1(X1, X2)
AND(mark(X1), X2) → AND(X1, X2)
ISNEPAL(mark(X)) → ISNEPAL(X)
PROPER(__(X1, X2)) → __1(proper(X1), proper(X2))
PROPER(__(X1, X2)) → PROPER(X1)
PROPER(__(X1, X2)) → PROPER(X2)
PROPER(and(X1, X2)) → AND(proper(X1), proper(X2))
PROPER(and(X1, X2)) → PROPER(X1)
PROPER(and(X1, X2)) → PROPER(X2)
PROPER(isNePal(X)) → ISNEPAL(proper(X))
PROPER(isNePal(X)) → PROPER(X)
__1(ok(X1), ok(X2)) → __1(X1, X2)
AND(ok(X1), ok(X2)) → AND(X1, X2)
ISNEPAL(ok(X)) → ISNEPAL(X)
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(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
active(__(X1, X2)) → __(active(X1), X2)
active(__(X1, X2)) → __(X1, active(X2))
active(and(X1, X2)) → and(active(X1), X2)
active(isNePal(X)) → isNePal(active(X))
__(mark(X1), X2) → mark(__(X1, X2))
__(X1, mark(X2)) → mark(__(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
isNePal(mark(X)) → mark(isNePal(X))
proper(__(X1, X2)) → __(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(isNePal(X)) → isNePal(proper(X))
__(ok(X1), ok(X2)) → ok(__(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNePal(ok(X)) → ok(isNePal(X))
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 6 SCCs with 11 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

ISNEPAL(ok(X)) → ISNEPAL(X)
ISNEPAL(mark(X)) → ISNEPAL(X)

The TRS R consists of the following rules:

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
active(__(X1, X2)) → __(active(X1), X2)
active(__(X1, X2)) → __(X1, active(X2))
active(and(X1, X2)) → and(active(X1), X2)
active(isNePal(X)) → isNePal(active(X))
__(mark(X1), X2) → mark(__(X1, X2))
__(X1, mark(X2)) → mark(__(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
isNePal(mark(X)) → mark(isNePal(X))
proper(__(X1, X2)) → __(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(isNePal(X)) → isNePal(proper(X))
__(ok(X1), ok(X2)) → ok(__(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNePal(ok(X)) → ok(isNePal(X))
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.


ISNEPAL(ok(X)) → ISNEPAL(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ISNEPAL(x1)  =  ISNEPAL(x1)
ok(x1)  =  ok(x1)
mark(x1)  =  x1
active(x1)  =  x1
__(x1, x2)  =  __(x1, x2)
nil  =  nil
and(x1, x2)  =  and(x2)
tt  =  tt
isNePal(x1)  =  isNePal(x1)
proper(x1)  =  proper(x1)
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
nil > [ISNEPAL1, ok1, isNePal1]
top > [2, and1, tt, proper1] > [ISNEPAL1, ok1, isNePal1]


The following usable rules [FROCOS05] were oriented:

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
active(__(X1, X2)) → __(active(X1), X2)
active(__(X1, X2)) → __(X1, active(X2))
active(and(X1, X2)) → and(active(X1), X2)
active(isNePal(X)) → isNePal(active(X))
__(mark(X1), X2) → mark(__(X1, X2))
__(X1, mark(X2)) → mark(__(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
isNePal(mark(X)) → mark(isNePal(X))
proper(__(X1, X2)) → __(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(isNePal(X)) → isNePal(proper(X))
__(ok(X1), ok(X2)) → ok(__(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNePal(ok(X)) → ok(isNePal(X))
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:

ISNEPAL(mark(X)) → ISNEPAL(X)

The TRS R consists of the following rules:

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
active(__(X1, X2)) → __(active(X1), X2)
active(__(X1, X2)) → __(X1, active(X2))
active(and(X1, X2)) → and(active(X1), X2)
active(isNePal(X)) → isNePal(active(X))
__(mark(X1), X2) → mark(__(X1, X2))
__(X1, mark(X2)) → mark(__(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
isNePal(mark(X)) → mark(isNePal(X))
proper(__(X1, X2)) → __(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(isNePal(X)) → isNePal(proper(X))
__(ok(X1), ok(X2)) → ok(__(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNePal(ok(X)) → ok(isNePal(X))
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.


ISNEPAL(mark(X)) → ISNEPAL(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ISNEPAL(x1)  =  ISNEPAL(x1)
mark(x1)  =  mark(x1)
active(x1)  =  x1
__(x1, x2)  =  __(x1, x2)
nil  =  nil
and(x1, x2)  =  and(x1, x2)
tt  =  tt
isNePal(x1)  =  isNePal(x1)
proper(x1)  =  x1
ok(x1)  =  ok
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
_2 > [ISNEPAL1, mark1, nil] > [ok, top]
and2 > [ISNEPAL1, mark1, nil] > [ok, top]
[tt, isNePal1] > [ISNEPAL1, mark1, nil] > [ok, top]


The following usable rules [FROCOS05] were oriented:

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
active(__(X1, X2)) → __(active(X1), X2)
active(__(X1, X2)) → __(X1, active(X2))
active(and(X1, X2)) → and(active(X1), X2)
active(isNePal(X)) → isNePal(active(X))
__(mark(X1), X2) → mark(__(X1, X2))
__(X1, mark(X2)) → mark(__(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
isNePal(mark(X)) → mark(isNePal(X))
proper(__(X1, X2)) → __(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(isNePal(X)) → isNePal(proper(X))
__(ok(X1), ok(X2)) → ok(__(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNePal(ok(X)) → ok(isNePal(X))
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(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
active(__(X1, X2)) → __(active(X1), X2)
active(__(X1, X2)) → __(X1, active(X2))
active(and(X1, X2)) → and(active(X1), X2)
active(isNePal(X)) → isNePal(active(X))
__(mark(X1), X2) → mark(__(X1, X2))
__(X1, mark(X2)) → mark(__(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
isNePal(mark(X)) → mark(isNePal(X))
proper(__(X1, X2)) → __(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(isNePal(X)) → isNePal(proper(X))
__(ok(X1), ok(X2)) → ok(__(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNePal(ok(X)) → ok(isNePal(X))
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:

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

The TRS R consists of the following rules:

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
active(__(X1, X2)) → __(active(X1), X2)
active(__(X1, X2)) → __(X1, active(X2))
active(and(X1, X2)) → and(active(X1), X2)
active(isNePal(X)) → isNePal(active(X))
__(mark(X1), X2) → mark(__(X1, X2))
__(X1, mark(X2)) → mark(__(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
isNePal(mark(X)) → mark(isNePal(X))
proper(__(X1, X2)) → __(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(isNePal(X)) → isNePal(proper(X))
__(ok(X1), ok(X2)) → ok(__(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNePal(ok(X)) → ok(isNePal(X))
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.


AND(ok(X1), ok(X2)) → AND(X1, X2)
AND(mark(X1), X2) → AND(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
AND(x1, x2)  =  AND(x1)
ok(x1)  =  ok(x1)
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
__(x1, x2)  =  __(x1, x2)
nil  =  nil
and(x1, x2)  =  and(x1, x2)
tt  =  tt
isNePal(x1)  =  x1
proper(x1)  =  proper(x1)
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
proper1 > [active1, 2, and2, top] > ok1 > [AND1, mark1]
proper1 > [active1, 2, and2, top] > tt
proper1 > nil > ok1 > [AND1, mark1]


The following usable rules [FROCOS05] were oriented:

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
active(__(X1, X2)) → __(active(X1), X2)
active(__(X1, X2)) → __(X1, active(X2))
active(and(X1, X2)) → and(active(X1), X2)
active(isNePal(X)) → isNePal(active(X))
__(mark(X1), X2) → mark(__(X1, X2))
__(X1, mark(X2)) → mark(__(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
isNePal(mark(X)) → mark(isNePal(X))
proper(__(X1, X2)) → __(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(isNePal(X)) → isNePal(proper(X))
__(ok(X1), ok(X2)) → ok(__(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNePal(ok(X)) → ok(isNePal(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(14) Obligation:

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

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
active(__(X1, X2)) → __(active(X1), X2)
active(__(X1, X2)) → __(X1, active(X2))
active(and(X1, X2)) → and(active(X1), X2)
active(isNePal(X)) → isNePal(active(X))
__(mark(X1), X2) → mark(__(X1, X2))
__(X1, mark(X2)) → mark(__(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
isNePal(mark(X)) → mark(isNePal(X))
proper(__(X1, X2)) → __(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(isNePal(X)) → isNePal(proper(X))
__(ok(X1), ok(X2)) → ok(__(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNePal(ok(X)) → ok(isNePal(X))
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) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE

(17) Obligation:

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

__1(X1, mark(X2)) → __1(X1, X2)
__1(mark(X1), X2) → __1(X1, X2)
__1(ok(X1), ok(X2)) → __1(X1, X2)

The TRS R consists of the following rules:

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
active(__(X1, X2)) → __(active(X1), X2)
active(__(X1, X2)) → __(X1, active(X2))
active(and(X1, X2)) → and(active(X1), X2)
active(isNePal(X)) → isNePal(active(X))
__(mark(X1), X2) → mark(__(X1, X2))
__(X1, mark(X2)) → mark(__(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
isNePal(mark(X)) → mark(isNePal(X))
proper(__(X1, X2)) → __(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(isNePal(X)) → isNePal(proper(X))
__(ok(X1), ok(X2)) → ok(__(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNePal(ok(X)) → ok(isNePal(X))
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.

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


__1(X1, mark(X2)) → __1(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
__1(x1, x2)  =  x2
mark(x1)  =  mark(x1)
ok(x1)  =  x1
active(x1)  =  active(x1)
__(x1, x2)  =  __(x1, x2)
nil  =  nil
and(x1, x2)  =  and(x1, x2)
tt  =  tt
isNePal(x1)  =  isNePal(x1)
proper(x1)  =  x1
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
[active1, top] > _2 > mark1
[active1, top] > and2 > mark1
[active1, top] > tt > mark1
[active1, top] > isNePal1 > mark1
nil > mark1


The following usable rules [FROCOS05] were oriented:

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
active(__(X1, X2)) → __(active(X1), X2)
active(__(X1, X2)) → __(X1, active(X2))
active(and(X1, X2)) → and(active(X1), X2)
active(isNePal(X)) → isNePal(active(X))
__(mark(X1), X2) → mark(__(X1, X2))
__(X1, mark(X2)) → mark(__(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
isNePal(mark(X)) → mark(isNePal(X))
proper(__(X1, X2)) → __(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(isNePal(X)) → isNePal(proper(X))
__(ok(X1), ok(X2)) → ok(__(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNePal(ok(X)) → ok(isNePal(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(19) Obligation:

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

__1(mark(X1), X2) → __1(X1, X2)
__1(ok(X1), ok(X2)) → __1(X1, X2)

The TRS R consists of the following rules:

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
active(__(X1, X2)) → __(active(X1), X2)
active(__(X1, X2)) → __(X1, active(X2))
active(and(X1, X2)) → and(active(X1), X2)
active(isNePal(X)) → isNePal(active(X))
__(mark(X1), X2) → mark(__(X1, X2))
__(X1, mark(X2)) → mark(__(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
isNePal(mark(X)) → mark(isNePal(X))
proper(__(X1, X2)) → __(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(isNePal(X)) → isNePal(proper(X))
__(ok(X1), ok(X2)) → ok(__(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNePal(ok(X)) → ok(isNePal(X))
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.


__1(mark(X1), X2) → __1(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
__1(x1, x2)  =  x1
mark(x1)  =  mark(x1)
ok(x1)  =  x1
active(x1)  =  x1
__(x1, x2)  =  __(x1, x2)
nil  =  nil
and(x1, x2)  =  and(x1, x2)
tt  =  tt
isNePal(x1)  =  x1
proper(x1)  =  x1
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
_2 > mark1 > top
_2 > tt > top
nil > mark1 > top
and2 > mark1 > top


The following usable rules [FROCOS05] were oriented:

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
active(__(X1, X2)) → __(active(X1), X2)
active(__(X1, X2)) → __(X1, active(X2))
active(and(X1, X2)) → and(active(X1), X2)
active(isNePal(X)) → isNePal(active(X))
__(mark(X1), X2) → mark(__(X1, X2))
__(X1, mark(X2)) → mark(__(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
isNePal(mark(X)) → mark(isNePal(X))
proper(__(X1, X2)) → __(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(isNePal(X)) → isNePal(proper(X))
__(ok(X1), ok(X2)) → ok(__(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNePal(ok(X)) → ok(isNePal(X))
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:

__1(ok(X1), ok(X2)) → __1(X1, X2)

The TRS R consists of the following rules:

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
active(__(X1, X2)) → __(active(X1), X2)
active(__(X1, X2)) → __(X1, active(X2))
active(and(X1, X2)) → and(active(X1), X2)
active(isNePal(X)) → isNePal(active(X))
__(mark(X1), X2) → mark(__(X1, X2))
__(X1, mark(X2)) → mark(__(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
isNePal(mark(X)) → mark(isNePal(X))
proper(__(X1, X2)) → __(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(isNePal(X)) → isNePal(proper(X))
__(ok(X1), ok(X2)) → ok(__(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNePal(ok(X)) → ok(isNePal(X))
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.


__1(ok(X1), ok(X2)) → __1(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
__1(x1, x2)  =  __1(x1)
ok(x1)  =  ok(x1)
active(x1)  =  active(x1)
__(x1, x2)  =  x1
mark(x1)  =  mark
nil  =  nil
and(x1, x2)  =  and(x1, x2)
tt  =  tt
isNePal(x1)  =  x1
proper(x1)  =  proper(x1)
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
_^11 > [ok1, mark]
nil > [ok1, mark]
tt > [ok1, mark]
top > [active1, and2, proper1] > [ok1, mark]


The following usable rules [FROCOS05] were oriented:

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
active(__(X1, X2)) → __(active(X1), X2)
active(__(X1, X2)) → __(X1, active(X2))
active(and(X1, X2)) → and(active(X1), X2)
active(isNePal(X)) → isNePal(active(X))
__(mark(X1), X2) → mark(__(X1, X2))
__(X1, mark(X2)) → mark(__(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
isNePal(mark(X)) → mark(isNePal(X))
proper(__(X1, X2)) → __(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(isNePal(X)) → isNePal(proper(X))
__(ok(X1), ok(X2)) → ok(__(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNePal(ok(X)) → ok(isNePal(X))
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(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
active(__(X1, X2)) → __(active(X1), X2)
active(__(X1, X2)) → __(X1, active(X2))
active(and(X1, X2)) → and(active(X1), X2)
active(isNePal(X)) → isNePal(active(X))
__(mark(X1), X2) → mark(__(X1, X2))
__(X1, mark(X2)) → mark(__(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
isNePal(mark(X)) → mark(isNePal(X))
proper(__(X1, X2)) → __(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(isNePal(X)) → isNePal(proper(X))
__(ok(X1), ok(X2)) → ok(__(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNePal(ok(X)) → ok(isNePal(X))
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:

PROPER(__(X1, X2)) → PROPER(X2)
PROPER(__(X1, X2)) → PROPER(X1)
PROPER(and(X1, X2)) → PROPER(X1)
PROPER(and(X1, X2)) → PROPER(X2)
PROPER(isNePal(X)) → PROPER(X)

The TRS R consists of the following rules:

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
active(__(X1, X2)) → __(active(X1), X2)
active(__(X1, X2)) → __(X1, active(X2))
active(and(X1, X2)) → and(active(X1), X2)
active(isNePal(X)) → isNePal(active(X))
__(mark(X1), X2) → mark(__(X1, X2))
__(X1, mark(X2)) → mark(__(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
isNePal(mark(X)) → mark(isNePal(X))
proper(__(X1, X2)) → __(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(isNePal(X)) → isNePal(proper(X))
__(ok(X1), ok(X2)) → ok(__(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNePal(ok(X)) → ok(isNePal(X))
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.


PROPER(__(X1, X2)) → PROPER(X2)
PROPER(__(X1, X2)) → PROPER(X1)
PROPER(and(X1, X2)) → PROPER(X1)
PROPER(and(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)
__(x1, x2)  =  __(x1, x2)
and(x1, x2)  =  and(x1, x2)
isNePal(x1)  =  x1
active(x1)  =  x1
mark(x1)  =  x1
nil  =  nil
tt  =  tt
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
PROPER1 > [2, and2, tt]
proper1 > [nil, ok, top] > [2, and2, tt]


The following usable rules [FROCOS05] were oriented:

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
active(__(X1, X2)) → __(active(X1), X2)
active(__(X1, X2)) → __(X1, active(X2))
active(and(X1, X2)) → and(active(X1), X2)
active(isNePal(X)) → isNePal(active(X))
__(mark(X1), X2) → mark(__(X1, X2))
__(X1, mark(X2)) → mark(__(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
isNePal(mark(X)) → mark(isNePal(X))
proper(__(X1, X2)) → __(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(isNePal(X)) → isNePal(proper(X))
__(ok(X1), ok(X2)) → ok(__(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNePal(ok(X)) → ok(isNePal(X))
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:

PROPER(isNePal(X)) → PROPER(X)

The TRS R consists of the following rules:

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
active(__(X1, X2)) → __(active(X1), X2)
active(__(X1, X2)) → __(X1, active(X2))
active(and(X1, X2)) → and(active(X1), X2)
active(isNePal(X)) → isNePal(active(X))
__(mark(X1), X2) → mark(__(X1, X2))
__(X1, mark(X2)) → mark(__(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
isNePal(mark(X)) → mark(isNePal(X))
proper(__(X1, X2)) → __(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(isNePal(X)) → isNePal(proper(X))
__(ok(X1), ok(X2)) → ok(__(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNePal(ok(X)) → ok(isNePal(X))
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.


PROPER(isNePal(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
isNePal(x1)  =  isNePal(x1)
active(x1)  =  active(x1)
__(x1, x2)  =  __(x1, x2)
mark(x1)  =  mark(x1)
nil  =  nil
and(x1, x2)  =  and(x1, x2)
tt  =  tt
proper(x1)  =  x1
ok(x1)  =  ok
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
[isNePal1, active1, 2] > and2 > [mark1, tt] > [ok, top]
nil > [ok, top]


The following usable rules [FROCOS05] were oriented:

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
active(__(X1, X2)) → __(active(X1), X2)
active(__(X1, X2)) → __(X1, active(X2))
active(and(X1, X2)) → and(active(X1), X2)
active(isNePal(X)) → isNePal(active(X))
__(mark(X1), X2) → mark(__(X1, X2))
__(X1, mark(X2)) → mark(__(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
isNePal(mark(X)) → mark(isNePal(X))
proper(__(X1, X2)) → __(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(isNePal(X)) → isNePal(proper(X))
__(ok(X1), ok(X2)) → ok(__(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNePal(ok(X)) → ok(isNePal(X))
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(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
active(__(X1, X2)) → __(active(X1), X2)
active(__(X1, X2)) → __(X1, active(X2))
active(and(X1, X2)) → and(active(X1), X2)
active(isNePal(X)) → isNePal(active(X))
__(mark(X1), X2) → mark(__(X1, X2))
__(X1, mark(X2)) → mark(__(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
isNePal(mark(X)) → mark(isNePal(X))
proper(__(X1, X2)) → __(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(isNePal(X)) → isNePal(proper(X))
__(ok(X1), ok(X2)) → ok(__(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNePal(ok(X)) → ok(isNePal(X))
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:

ACTIVE(__(X1, X2)) → ACTIVE(X2)
ACTIVE(__(X1, X2)) → ACTIVE(X1)
ACTIVE(and(X1, X2)) → ACTIVE(X1)
ACTIVE(isNePal(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
active(__(X1, X2)) → __(active(X1), X2)
active(__(X1, X2)) → __(X1, active(X2))
active(and(X1, X2)) → and(active(X1), X2)
active(isNePal(X)) → isNePal(active(X))
__(mark(X1), X2) → mark(__(X1, X2))
__(X1, mark(X2)) → mark(__(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
isNePal(mark(X)) → mark(isNePal(X))
proper(__(X1, X2)) → __(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(isNePal(X)) → isNePal(proper(X))
__(ok(X1), ok(X2)) → ok(__(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNePal(ok(X)) → ok(isNePal(X))
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.


ACTIVE(__(X1, X2)) → ACTIVE(X2)
ACTIVE(__(X1, X2)) → ACTIVE(X1)
ACTIVE(and(X1, X2)) → ACTIVE(X1)
ACTIVE(isNePal(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)
__(x1, x2)  =  __(x1, x2)
and(x1, x2)  =  and(x1, x2)
isNePal(x1)  =  isNePal(x1)
active(x1)  =  x1
mark(x1)  =  mark
nil  =  nil
tt  =  tt
proper(x1)  =  x1
ok(x1)  =  ok
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
ACTIVE1 > [and2, mark, tt, ok, top]
_2 > [and2, mark, tt, ok, top]
isNePal1 > [and2, mark, tt, ok, top]
nil > [and2, mark, tt, ok, top]


The following usable rules [FROCOS05] were oriented:

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
active(__(X1, X2)) → __(active(X1), X2)
active(__(X1, X2)) → __(X1, active(X2))
active(and(X1, X2)) → and(active(X1), X2)
active(isNePal(X)) → isNePal(active(X))
__(mark(X1), X2) → mark(__(X1, X2))
__(X1, mark(X2)) → mark(__(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
isNePal(mark(X)) → mark(isNePal(X))
proper(__(X1, X2)) → __(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(isNePal(X)) → isNePal(proper(X))
__(ok(X1), ok(X2)) → ok(__(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNePal(ok(X)) → ok(isNePal(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(35) Obligation:

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

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
active(__(X1, X2)) → __(active(X1), X2)
active(__(X1, X2)) → __(X1, active(X2))
active(and(X1, X2)) → and(active(X1), X2)
active(isNePal(X)) → isNePal(active(X))
__(mark(X1), X2) → mark(__(X1, X2))
__(X1, mark(X2)) → mark(__(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
isNePal(mark(X)) → mark(isNePal(X))
proper(__(X1, X2)) → __(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(isNePal(X)) → isNePal(proper(X))
__(ok(X1), ok(X2)) → ok(__(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNePal(ok(X)) → ok(isNePal(X))
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) PisEmptyProof (EQUIVALENT transformation)

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

(37) TRUE

(38) 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(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
active(__(X1, X2)) → __(active(X1), X2)
active(__(X1, X2)) → __(X1, active(X2))
active(and(X1, X2)) → and(active(X1), X2)
active(isNePal(X)) → isNePal(active(X))
__(mark(X1), X2) → mark(__(X1, X2))
__(X1, mark(X2)) → mark(__(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
isNePal(mark(X)) → mark(isNePal(X))
proper(__(X1, X2)) → __(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(isNePal(X)) → isNePal(proper(X))
__(ok(X1), ok(X2)) → ok(__(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNePal(ok(X)) → ok(isNePal(X))
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.

(39) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


TOP(mark(X)) → TOP(proper(X))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TOP(x1)  =  TOP(x1)
ok(x1)  =  x1
active(x1)  =  x1
mark(x1)  =  mark(x1)
proper(x1)  =  x1
__(x1, x2)  =  __(x1, x2)
nil  =  nil
and(x1, x2)  =  and(x1, x2)
tt  =  tt
isNePal(x1)  =  isNePal(x1)
top(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:
_2 > tt > [mark1, isNePal1] > TOP1
nil > [mark1, isNePal1] > TOP1
and2 > [mark1, isNePal1] > TOP1


The following usable rules [FROCOS05] were oriented:

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
active(__(X1, X2)) → __(active(X1), X2)
active(__(X1, X2)) → __(X1, active(X2))
active(and(X1, X2)) → and(active(X1), X2)
active(isNePal(X)) → isNePal(active(X))
__(mark(X1), X2) → mark(__(X1, X2))
__(X1, mark(X2)) → mark(__(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
isNePal(mark(X)) → mark(isNePal(X))
proper(__(X1, X2)) → __(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(isNePal(X)) → isNePal(proper(X))
__(ok(X1), ok(X2)) → ok(__(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNePal(ok(X)) → ok(isNePal(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(40) Obligation:

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

TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
active(__(X1, X2)) → __(active(X1), X2)
active(__(X1, X2)) → __(X1, active(X2))
active(and(X1, X2)) → and(active(X1), X2)
active(isNePal(X)) → isNePal(active(X))
__(mark(X1), X2) → mark(__(X1, X2))
__(X1, mark(X2)) → mark(__(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
isNePal(mark(X)) → mark(isNePal(X))
proper(__(X1, X2)) → __(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(isNePal(X)) → isNePal(proper(X))
__(ok(X1), ok(X2)) → ok(__(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNePal(ok(X)) → ok(isNePal(X))
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.


TOP(ok(X)) → TOP(active(X))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TOP(x1)  =  x1
ok(x1)  =  ok(x1)
active(x1)  =  x1
__(x1, x2)  =  x2
mark(x1)  =  mark
nil  =  nil
and(x1, x2)  =  x2
tt  =  tt
isNePal(x1)  =  x1
proper(x1)  =  proper
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
proper > nil > ok1 > mark
proper > tt > ok1 > mark
top > mark


The following usable rules [FROCOS05] were oriented:

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
active(__(X1, X2)) → __(active(X1), X2)
active(__(X1, X2)) → __(X1, active(X2))
active(and(X1, X2)) → and(active(X1), X2)
active(isNePal(X)) → isNePal(active(X))
__(mark(X1), X2) → mark(__(X1, X2))
__(X1, mark(X2)) → mark(__(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
isNePal(mark(X)) → mark(isNePal(X))
proper(__(X1, X2)) → __(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(isNePal(X)) → isNePal(proper(X))
__(ok(X1), ok(X2)) → ok(__(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNePal(ok(X)) → ok(isNePal(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(42) Obligation:

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

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
active(__(X1, X2)) → __(active(X1), X2)
active(__(X1, X2)) → __(X1, active(X2))
active(and(X1, X2)) → and(active(X1), X2)
active(isNePal(X)) → isNePal(active(X))
__(mark(X1), X2) → mark(__(X1, X2))
__(X1, mark(X2)) → mark(__(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
isNePal(mark(X)) → mark(isNePal(X))
proper(__(X1, X2)) → __(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(isNePal(X)) → isNePal(proper(X))
__(ok(X1), ok(X2)) → ok(__(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNePal(ok(X)) → ok(isNePal(X))
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) PisEmptyProof (EQUIVALENT transformation)

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

(44) TRUE