(0) Obligation:

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

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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(add(s(X), Y)) → S(add(X, Y))
ACTIVE(add(s(X), Y)) → ADD(X, Y)
ACTIVE(first(s(X), cons(Y, Z))) → CONS(Y, first(X, Z))
ACTIVE(first(s(X), cons(Y, Z))) → FIRST(X, Z)
ACTIVE(from(X)) → CONS(X, from(s(X)))
ACTIVE(from(X)) → FROM(s(X))
ACTIVE(from(X)) → S(X)
ACTIVE(and(X1, X2)) → AND(active(X1), X2)
ACTIVE(and(X1, X2)) → ACTIVE(X1)
ACTIVE(if(X1, X2, X3)) → IF(active(X1), X2, X3)
ACTIVE(if(X1, X2, X3)) → ACTIVE(X1)
ACTIVE(add(X1, X2)) → ADD(active(X1), X2)
ACTIVE(add(X1, X2)) → ACTIVE(X1)
ACTIVE(first(X1, X2)) → FIRST(active(X1), X2)
ACTIVE(first(X1, X2)) → ACTIVE(X1)
ACTIVE(first(X1, X2)) → FIRST(X1, active(X2))
ACTIVE(first(X1, X2)) → ACTIVE(X2)
AND(mark(X1), X2) → AND(X1, X2)
IF(mark(X1), X2, X3) → IF(X1, X2, X3)
ADD(mark(X1), X2) → ADD(X1, X2)
FIRST(mark(X1), X2) → FIRST(X1, X2)
FIRST(X1, mark(X2)) → FIRST(X1, X2)
PROPER(and(X1, X2)) → AND(proper(X1), proper(X2))
PROPER(and(X1, X2)) → PROPER(X1)
PROPER(and(X1, X2)) → PROPER(X2)
PROPER(if(X1, X2, X3)) → IF(proper(X1), proper(X2), proper(X3))
PROPER(if(X1, X2, X3)) → PROPER(X1)
PROPER(if(X1, X2, X3)) → PROPER(X2)
PROPER(if(X1, X2, X3)) → PROPER(X3)
PROPER(add(X1, X2)) → ADD(proper(X1), proper(X2))
PROPER(add(X1, X2)) → PROPER(X1)
PROPER(add(X1, X2)) → PROPER(X2)
PROPER(s(X)) → S(proper(X))
PROPER(s(X)) → PROPER(X)
PROPER(first(X1, X2)) → FIRST(proper(X1), proper(X2))
PROPER(first(X1, X2)) → PROPER(X1)
PROPER(first(X1, X2)) → PROPER(X2)
PROPER(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)
AND(ok(X1), ok(X2)) → AND(X1, X2)
IF(ok(X1), ok(X2), ok(X3)) → IF(X1, X2, X3)
ADD(ok(X1), ok(X2)) → ADD(X1, X2)
S(ok(X)) → S(X)
FIRST(ok(X1), ok(X2)) → FIRST(X1, X2)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
FROM(ok(X)) → FROM(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(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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 10 SCCs with 21 less nodes.

(4) Complex Obligation (AND)

(5) 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(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.


FROM(ok(X)) → FROM(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FROM(x1)  =  FROM(x1)
ok(x1)  =  ok(x1)
active(x1)  =  active(x1)
and(x1, x2)  =  x1
true  =  true
mark(x1)  =  mark
false  =  false
if(x1, x2, x3)  =  x1
add(x1, x2)  =  x1
0  =  0
s(x1)  =  x1
first(x1, x2)  =  first(x1, x2)
nil  =  nil
cons(x1, x2)  =  x1
from(x1)  =  from(x1)
proper(x1)  =  proper(x1)
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
false > mark > proper1 > true
false > mark > proper1 > 0
false > mark > proper1 > first2 > ok1
false > mark > proper1 > first2 > nil
false > mark > proper1 > from1 > ok1
top > active1 > mark > proper1 > true
top > active1 > mark > proper1 > 0
top > active1 > mark > proper1 > first2 > ok1
top > active1 > mark > proper1 > first2 > nil
top > active1 > mark > proper1 > from1 > ok1

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(7) Obligation:

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

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) 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(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(11) 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)  =  x1
ok(x1)  =  ok(x1)
active(x1)  =  active(x1)
and(x1, x2)  =  and(x1, x2)
true  =  true
mark(x1)  =  mark
false  =  false
if(x1, x2, x3)  =  x1
add(x1, x2)  =  x1
0  =  0
s(x1)  =  x1
first(x1, x2)  =  first(x1, x2)
nil  =  nil
cons(x1, x2)  =  x1
from(x1)  =  from(x1)
proper(x1)  =  proper(x1)
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
active1 > from1 > mark > and2 > ok1
active1 > from1 > mark > first2 > ok1
active1 > from1 > mark > top > ok1
false > mark > and2 > ok1
false > mark > first2 > ok1
false > mark > top > ok1
nil > ok1
proper1 > true > mark > and2 > ok1
proper1 > true > mark > first2 > ok1
proper1 > true > mark > top > ok1
proper1 > 0 > ok1
proper1 > from1 > mark > and2 > ok1
proper1 > from1 > mark > first2 > ok1
proper1 > from1 > mark > top > ok1

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(12) Obligation:

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

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) 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(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


S(ok(X)) → S(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
S(x1)  =  S(x1)
ok(x1)  =  ok(x1)
active(x1)  =  active(x1)
and(x1, x2)  =  x1
true  =  true
mark(x1)  =  mark
false  =  false
if(x1, x2, x3)  =  x1
add(x1, x2)  =  x1
0  =  0
s(x1)  =  x1
first(x1, x2)  =  first(x1, x2)
nil  =  nil
cons(x1, x2)  =  x1
from(x1)  =  from(x1)
proper(x1)  =  proper(x1)
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
false > mark > proper1 > true
false > mark > proper1 > 0
false > mark > proper1 > first2 > ok1
false > mark > proper1 > first2 > nil
false > mark > proper1 > from1 > ok1
top > active1 > mark > proper1 > true
top > active1 > mark > proper1 > 0
top > active1 > mark > proper1 > first2 > ok1
top > active1 > mark > proper1 > first2 > nil
top > active1 > mark > proper1 > from1 > ok1

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(17) Obligation:

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

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE

(20) Obligation:

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

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

The TRS R consists of the following rules:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(21) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


FIRST(X1, mark(X2)) → FIRST(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FIRST(x1, x2)  =  FIRST(x2)
mark(x1)  =  mark(x1)
ok(x1)  =  x1
active(x1)  =  x1
and(x1, x2)  =  and(x1, x2)
true  =  true
false  =  false
if(x1, x2, x3)  =  if(x1, x2, x3)
add(x1, x2)  =  add(x1, x2)
0  =  0
s(x1)  =  s(x1)
first(x1, x2)  =  first(x1, x2)
nil  =  nil
cons(x1, x2)  =  cons
from(x1)  =  from
proper(x1)  =  x1
top(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:
FIRST1 > nil
and2 > mark1 > nil
and2 > false > nil
true > mark1 > nil
if3 > mark1 > nil
add2 > s1 > first2 > mark1 > nil
add2 > s1 > first2 > cons > nil
0 > nil
from > s1 > first2 > mark1 > nil
from > s1 > first2 > cons > nil

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(22) Obligation:

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

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

The TRS R consists of the following rules:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


FIRST(ok(X1), ok(X2)) → FIRST(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FIRST(x1, x2)  =  FIRST(x2)
mark(x1)  =  mark
ok(x1)  =  ok(x1)
active(x1)  =  active(x1)
and(x1, x2)  =  and(x1)
true  =  true
false  =  false
if(x1, x2, x3)  =  if(x1)
add(x1, x2)  =  x1
0  =  0
s(x1)  =  s(x1)
first(x1, x2)  =  first(x2)
nil  =  nil
cons(x1, x2)  =  cons(x1)
from(x1)  =  x1
proper(x1)  =  proper(x1)
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
FIRST1 > nil
active1 > and1 > mark > top > nil
active1 > and1 > ok1 > nil
active1 > and1 > false > nil
active1 > if1 > mark > top > nil
active1 > if1 > ok1 > nil
active1 > s1 > ok1 > nil
active1 > cons1 > first1 > mark > top > nil
active1 > cons1 > first1 > ok1 > nil
proper1 > and1 > mark > top > nil
proper1 > and1 > ok1 > nil
proper1 > and1 > false > nil
proper1 > true > mark > top > nil
proper1 > true > ok1 > nil
proper1 > if1 > mark > top > nil
proper1 > if1 > ok1 > nil
proper1 > 0 > nil
proper1 > s1 > ok1 > nil
proper1 > cons1 > first1 > mark > top > nil
proper1 > cons1 > first1 > ok1 > nil

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(24) Obligation:

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

FIRST(mark(X1), X2) → FIRST(X1, X2)

The TRS R consists of the following rules:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(25) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


FIRST(mark(X1), X2) → FIRST(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FIRST(x1, x2)  =  x1
mark(x1)  =  mark(x1)
active(x1)  =  x1
and(x1, x2)  =  and(x1, x2)
true  =  true
false  =  false
if(x1, x2, x3)  =  if(x1, x2, x3)
add(x1, x2)  =  add(x1, x2)
0  =  0
s(x1)  =  s
first(x1, x2)  =  first(x1, x2)
nil  =  nil
cons(x1, x2)  =  cons
from(x1)  =  from
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
and2 > mark1 > nil
and2 > false > nil
true > nil
if3 > mark1 > nil
0 > nil
s > add2 > mark1 > nil
s > first2 > cons > mark1 > nil
from > cons > mark1 > nil
top > nil

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(26) Obligation:

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

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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) PisEmptyProof (EQUIVALENT transformation)

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

(28) TRUE

(29) Obligation:

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

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

The TRS R consists of the following rules:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(30) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ADD(ok(X1), ok(X2)) → ADD(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ADD(x1, x2)  =  x2
ok(x1)  =  ok(x1)
mark(x1)  =  x1
active(x1)  =  x1
and(x1, x2)  =  and(x2)
true  =  true
false  =  false
if(x1, x2, x3)  =  if(x2, x3)
add(x1, x2)  =  add(x2)
0  =  0
s(x1)  =  x1
first(x1, x2)  =  first(x2)
nil  =  nil
cons(x1, x2)  =  x1
from(x1)  =  x1
proper(x1)  =  proper(x1)
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
true > ok1 > false
proper1 > and1 > ok1 > false
proper1 > if2 > ok1 > false
proper1 > add1 > ok1 > false
proper1 > 0 > ok1 > false
proper1 > 0 > nil > false
proper1 > first1 > ok1 > false
proper1 > first1 > nil > false
top > false

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(31) Obligation:

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

ADD(mark(X1), X2) → ADD(X1, X2)

The TRS R consists of the following rules:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(32) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ADD(mark(X1), X2) → ADD(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ADD(x1, x2)  =  ADD(x1)
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
and(x1, x2)  =  and(x1, x2)
true  =  true
false  =  false
if(x1, x2, x3)  =  if(x1, x2, x3)
add(x1, x2)  =  add(x1, x2)
0  =  0
s(x1)  =  s
first(x1, x2)  =  first(x1, x2)
nil  =  nil
cons(x1, x2)  =  cons(x1)
from(x1)  =  x1
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
ADD1 > nil
active1 > and2 > mark1 > nil
active1 > and2 > false > nil
active1 > if3 > mark1 > nil
active1 > add2 > mark1 > nil
active1 > s > mark1 > nil
active1 > s > cons1 > nil
active1 > first2 > mark1 > nil
active1 > first2 > cons1 > nil
true > nil
0 > nil
top > nil

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(33) Obligation:

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

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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) PisEmptyProof (EQUIVALENT transformation)

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

(35) TRUE

(36) Obligation:

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

IF(ok(X1), ok(X2), ok(X3)) → IF(X1, X2, X3)
IF(mark(X1), X2, X3) → IF(X1, X2, X3)

The TRS R consists of the following rules:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(37) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


IF(mark(X1), X2, X3) → IF(X1, X2, X3)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
IF(x1, x2, x3)  =  IF(x1, x3)
ok(x1)  =  x1
mark(x1)  =  mark(x1)
active(x1)  =  x1
and(x1, x2)  =  and(x1, x2)
true  =  true
false  =  false
if(x1, x2, x3)  =  if(x1, x2, x3)
add(x1, x2)  =  add(x1, x2)
0  =  0
s(x1)  =  s(x1)
first(x1, x2)  =  first(x1, x2)
nil  =  nil
cons(x1, x2)  =  cons
from(x1)  =  from
proper(x1)  =  proper(x1)
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
IF2 > nil
proper1 > and2 > mark1 > nil
proper1 > and2 > false > nil
proper1 > true > nil
proper1 > if3 > mark1 > nil
proper1 > add2 > mark1 > nil
proper1 > add2 > s1 > nil
proper1 > 0 > nil
proper1 > first2 > mark1 > nil
proper1 > first2 > cons > nil
proper1 > from > mark1 > nil
proper1 > from > s1 > nil
proper1 > from > cons > nil
top > nil

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(38) Obligation:

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

IF(ok(X1), ok(X2), ok(X3)) → IF(X1, X2, X3)

The TRS R consists of the following rules:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.


IF(ok(X1), ok(X2), ok(X3)) → IF(X1, X2, X3)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
IF(x1, x2, x3)  =  x3
ok(x1)  =  ok(x1)
active(x1)  =  active(x1)
and(x1, x2)  =  x1
true  =  true
mark(x1)  =  mark
false  =  false
if(x1, x2, x3)  =  x1
add(x1, x2)  =  x2
0  =  0
s(x1)  =  x1
first(x1, x2)  =  x1
nil  =  nil
cons(x1, x2)  =  x2
from(x1)  =  x1
proper(x1)  =  proper
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
top > proper > ok1 > active1 > mark
top > proper > true > mark
top > proper > false > mark
top > proper > 0 > mark
top > proper > nil > mark

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(40) Obligation:

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

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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) PisEmptyProof (EQUIVALENT transformation)

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

(42) TRUE

(43) 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(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(44) 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)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
AND(x1, x2)  =  x2
ok(x1)  =  ok(x1)
mark(x1)  =  x1
active(x1)  =  x1
and(x1, x2)  =  and(x2)
true  =  true
false  =  false
if(x1, x2, x3)  =  if(x2, x3)
add(x1, x2)  =  add(x2)
0  =  0
s(x1)  =  x1
first(x1, x2)  =  first(x2)
nil  =  nil
cons(x1, x2)  =  x1
from(x1)  =  x1
proper(x1)  =  proper(x1)
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
true > ok1 > false
proper1 > and1 > ok1 > false
proper1 > if2 > ok1 > false
proper1 > add1 > ok1 > false
proper1 > 0 > ok1 > false
proper1 > 0 > nil > false
proper1 > first1 > ok1 > false
proper1 > first1 > nil > false
top > false

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(45) Obligation:

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

AND(mark(X1), X2) → AND(X1, X2)

The TRS R consists of the following rules:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(46) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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)
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
and(x1, x2)  =  and(x1, x2)
true  =  true
false  =  false
if(x1, x2, x3)  =  if(x1, x2, x3)
add(x1, x2)  =  add(x1, x2)
0  =  0
s(x1)  =  s
first(x1, x2)  =  first(x1, x2)
nil  =  nil
cons(x1, x2)  =  cons(x1)
from(x1)  =  x1
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
AND1 > nil
active1 > and2 > mark1 > nil
active1 > and2 > false > nil
active1 > if3 > mark1 > nil
active1 > add2 > mark1 > nil
active1 > s > mark1 > nil
active1 > s > cons1 > nil
active1 > first2 > mark1 > nil
active1 > first2 > cons1 > nil
true > nil
0 > nil
top > nil

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(47) Obligation:

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

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(48) PisEmptyProof (EQUIVALENT transformation)

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

(49) TRUE

(50) Obligation:

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

PROPER(and(X1, X2)) → PROPER(X2)
PROPER(and(X1, X2)) → PROPER(X1)
PROPER(if(X1, X2, X3)) → PROPER(X1)
PROPER(if(X1, X2, X3)) → PROPER(X2)
PROPER(if(X1, X2, X3)) → PROPER(X3)
PROPER(add(X1, X2)) → PROPER(X1)
PROPER(add(X1, X2)) → PROPER(X2)
PROPER(s(X)) → PROPER(X)
PROPER(first(X1, X2)) → PROPER(X1)
PROPER(first(X1, X2)) → PROPER(X2)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(from(X)) → PROPER(X)

The TRS R consists of the following rules:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(51) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(and(X1, X2)) → PROPER(X2)
PROPER(and(X1, X2)) → PROPER(X1)
PROPER(if(X1, X2, X3)) → PROPER(X1)
PROPER(if(X1, X2, X3)) → PROPER(X2)
PROPER(if(X1, X2, X3)) → PROPER(X3)
PROPER(add(X1, X2)) → PROPER(X1)
PROPER(add(X1, X2)) → PROPER(X2)
PROPER(first(X1, X2)) → PROPER(X1)
PROPER(first(X1, X2)) → PROPER(X2)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  x1
and(x1, x2)  =  and(x1, x2)
if(x1, x2, x3)  =  if(x1, x2, x3)
add(x1, x2)  =  add(x1, x2)
s(x1)  =  x1
first(x1, x2)  =  first(x1, x2)
cons(x1, x2)  =  cons(x1, x2)
from(x1)  =  x1
active(x1)  =  x1
true  =  true
mark(x1)  =  mark
false  =  false
0  =  0
nil  =  nil
proper(x1)  =  x1
ok(x1)  =  ok
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
add2 > ok > and2 > mark
add2 > ok > if3 > mark
add2 > ok > cons2 > mark
add2 > ok > top > mark
first2 > nil > ok > and2 > mark
first2 > nil > ok > if3 > mark
first2 > nil > ok > cons2 > mark
first2 > nil > ok > top > mark
true > ok > and2 > mark
true > ok > if3 > mark
true > ok > cons2 > mark
true > ok > top > mark
false > ok > and2 > mark
false > ok > if3 > mark
false > ok > cons2 > mark
false > ok > top > mark
0 > nil > ok > and2 > mark
0 > nil > ok > if3 > mark
0 > nil > ok > cons2 > mark
0 > nil > ok > top > mark

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(52) Obligation:

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

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

The TRS R consists of the following rules:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(53) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(s(X)) → PROPER(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  PROPER(x1)
s(x1)  =  s(x1)
from(x1)  =  x1
active(x1)  =  active(x1)
and(x1, x2)  =  x1
true  =  true
mark(x1)  =  mark
false  =  false
if(x1, x2, x3)  =  x1
add(x1, x2)  =  x1
0  =  0
first(x1, x2)  =  first(x2)
nil  =  nil
cons(x1, x2)  =  x2
proper(x1)  =  proper(x1)
ok(x1)  =  ok(x1)
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
PROPER1 > nil
proper1 > s1 > first1 > mark > nil
proper1 > s1 > first1 > ok1 > nil
proper1 > true > mark > nil
proper1 > true > ok1 > nil
proper1 > false > mark > nil
proper1 > false > ok1 > nil
proper1 > 0 > nil
top > active1 > s1 > first1 > mark > nil
top > active1 > s1 > first1 > ok1 > nil
top > active1 > false > mark > nil
top > active1 > false > ok1 > nil

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(54) Obligation:

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

PROPER(from(X)) → PROPER(X)

The TRS R consists of the following rules:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(55) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(from(X)) → PROPER(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  PROPER(x1)
from(x1)  =  from(x1)
active(x1)  =  x1
and(x1, x2)  =  and(x2)
true  =  true
mark(x1)  =  x1
false  =  false
if(x1, x2, x3)  =  if(x2, x3)
add(x1, x2)  =  add(x1, x2)
0  =  0
s(x1)  =  s(x1)
first(x1, x2)  =  first(x2)
nil  =  nil
cons(x1, x2)  =  x1
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top(x1)

Lexicographic Path Order [LPO].
Precedence:
PROPER1 > nil
from1 > nil
and1 > false > nil
true > nil
if2 > nil
add2 > s1 > first1 > nil
0 > nil
top1 > nil

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(56) Obligation:

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

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(57) PisEmptyProof (EQUIVALENT transformation)

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

(58) TRUE

(59) Obligation:

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

ACTIVE(if(X1, X2, X3)) → ACTIVE(X1)
ACTIVE(and(X1, X2)) → ACTIVE(X1)
ACTIVE(add(X1, X2)) → ACTIVE(X1)
ACTIVE(first(X1, X2)) → ACTIVE(X1)
ACTIVE(first(X1, X2)) → ACTIVE(X2)

The TRS R consists of the following rules:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(60) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(if(X1, X2, X3)) → ACTIVE(X1)
ACTIVE(first(X1, X2)) → ACTIVE(X1)
ACTIVE(first(X1, X2)) → ACTIVE(X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  x1
if(x1, x2, x3)  =  if(x1, x2, x3)
and(x1, x2)  =  x1
add(x1, x2)  =  x1
first(x1, x2)  =  first(x1, x2)
active(x1)  =  x1
true  =  true
mark(x1)  =  mark
false  =  false
0  =  0
s(x1)  =  s
nil  =  nil
cons(x1, x2)  =  cons
from(x1)  =  from
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
false > mark
proper1 > if3 > mark
proper1 > first2 > nil
proper1 > first2 > cons > mark
proper1 > true > mark
proper1 > 0 > mark
proper1 > s > cons > mark
proper1 > from > cons > mark

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(61) Obligation:

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

ACTIVE(and(X1, X2)) → ACTIVE(X1)
ACTIVE(add(X1, X2)) → ACTIVE(X1)

The TRS R consists of the following rules:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(62) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(and(X1, X2)) → ACTIVE(X1)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  x1
and(x1, x2)  =  and(x1, x2)
add(x1, x2)  =  x1
active(x1)  =  active(x1)
true  =  true
mark(x1)  =  mark
false  =  false
if(x1, x2, x3)  =  if(x1, x2, x3)
0  =  0
s(x1)  =  s
first(x1, x2)  =  first(x2)
nil  =  nil
cons(x1, x2)  =  cons
from(x1)  =  from
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
active1 > false
active1 > if3 > mark > and2
active1 > s > first1 > mark > and2
active1 > s > first1 > nil
active1 > s > first1 > cons
active1 > from > mark > and2
active1 > from > cons

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(63) Obligation:

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

ACTIVE(add(X1, X2)) → ACTIVE(X1)

The TRS R consists of the following rules:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(64) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(add(X1, X2)) → ACTIVE(X1)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  x1
add(x1, x2)  =  add(x1, x2)
active(x1)  =  x1
and(x1, x2)  =  and(x1, x2)
true  =  true
mark(x1)  =  mark(x1)
false  =  false
if(x1, x2, x3)  =  if(x1, x2, x3)
0  =  0
s(x1)  =  s
first(x1, x2)  =  first(x1, x2)
nil  =  nil
cons(x1, x2)  =  cons
from(x1)  =  from
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top(x1)

Lexicographic Path Order [LPO].
Precedence:
and2 > false > mark1 > nil
true > mark1 > nil
if3 > mark1 > nil
0 > nil
s > add2 > mark1 > nil
s > first2 > mark1 > nil
s > first2 > cons > nil
from > mark1 > nil
from > cons > nil
top1 > nil

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(65) Obligation:

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

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(66) PisEmptyProof (EQUIVALENT transformation)

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

(67) TRUE

(68) 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(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(69) 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
and(x1, x2)  =  and(x1, x2)
true  =  true
false  =  false
if(x1, x2, x3)  =  if(x1, x2, x3)
add(x1, x2)  =  add(x1, x2)
0  =  0
s(x1)  =  s
first(x1, x2)  =  first(x1, x2)
nil  =  nil
cons(x1, x2)  =  x1
from(x1)  =  from(x1)
top(x1)  =  top(x1)

Lexicographic Path Order [LPO].
Precedence:
and2 > mark1 > top1
and2 > false
if3 > mark1 > top1
s > add2 > mark1 > top1
s > first2 > mark1 > top1
s > first2 > nil
from1 > mark1 > top1

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(70) 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(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(71) 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)  =  TOP(x1)
ok(x1)  =  ok(x1)
active(x1)  =  x1
and(x1, x2)  =  x2
true  =  true
mark(x1)  =  mark
false  =  false
if(x1, x2, x3)  =  x1
add(x1, x2)  =  x2
0  =  0
s(x1)  =  s(x1)
first(x1, x2)  =  first(x2)
nil  =  nil
cons(x1, x2)  =  x2
from(x1)  =  x1
proper(x1)  =  proper(x1)
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
TOP1 > mark
proper1 > true > ok1 > mark
proper1 > false > mark
proper1 > 0 > mark
proper1 > s1 > first1 > ok1 > mark
proper1 > s1 > first1 > nil > mark
top > mark

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(72) Obligation:

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

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(73) PisEmptyProof (EQUIVALENT transformation)

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

(74) TRUE