Term Rewriting System R:
[X, Y, Z, X1, X2]
active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

ACTIVE(from(X)) -> CONS(X, from(s(X)))
ACTIVE(from(X)) -> FROM(s(X))
ACTIVE(from(X)) -> S(X)
ACTIVE(2nd(X)) -> 2ND(active(X))
ACTIVE(2nd(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)
ACTIVE(from(X)) -> FROM(active(X))
ACTIVE(from(X)) -> ACTIVE(X)
ACTIVE(s(X)) -> S(active(X))
ACTIVE(s(X)) -> ACTIVE(X)
2ND(mark(X)) -> 2ND(X)
2ND(ok(X)) -> 2ND(X)
CONS(mark(X1), X2) -> CONS(X1, X2)
CONS(ok(X1), ok(X2)) -> CONS(X1, X2)
FROM(mark(X)) -> FROM(X)
FROM(ok(X)) -> FROM(X)
S(mark(X)) -> S(X)
S(ok(X)) -> S(X)
PROPER(2nd(X)) -> 2ND(proper(X))
PROPER(2nd(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> CONS(proper(X1), proper(X2))
PROPER(cons(X1, X2)) -> PROPER(X1)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(from(X)) -> FROM(proper(X))
PROPER(from(X)) -> PROPER(X)
PROPER(s(X)) -> S(proper(X))
PROPER(s(X)) -> PROPER(X)
TOP(mark(X)) -> TOP(proper(X))
TOP(mark(X)) -> PROPER(X)
TOP(ok(X)) -> TOP(active(X))
TOP(ok(X)) -> ACTIVE(X)

Furthermore, R contains seven SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Nar


Dependency Pairs:

CONS(ok(X1), ok(X2)) -> CONS(X1, X2)
CONS(mark(X1), X2) -> CONS(X1, X2)


Rules:


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





The following dependency pair can be strictly oriented:

CONS(ok(X1), ok(X2)) -> CONS(X1, X2)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(mark(x1))=  0  
  POL(ok(x1))=  1 + x1  
  POL(CONS(x1, x2))=  x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 8
             ↳Polynomial Ordering
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Nar


Dependency Pair:

CONS(mark(X1), X2) -> CONS(X1, X2)


Rules:


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





The following dependency pair can be strictly oriented:

CONS(mark(X1), X2) -> CONS(X1, X2)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(mark(x1))=  1 + x1  
  POL(CONS(x1, x2))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 8
             ↳Polo
             ...
               →DP Problem 9
                 ↳Dependency Graph
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Nar


Dependency Pair:


Rules:


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





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polynomial Ordering
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Nar


Dependency Pairs:

FROM(ok(X)) -> FROM(X)
FROM(mark(X)) -> FROM(X)


Rules:


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





The following dependency pair can be strictly oriented:

FROM(ok(X)) -> FROM(X)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(FROM(x1))=  x1  
  POL(mark(x1))=  x1  
  POL(ok(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 10
             ↳Polynomial Ordering
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Nar


Dependency Pair:

FROM(mark(X)) -> FROM(X)


Rules:


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





The following dependency pair can be strictly oriented:

FROM(mark(X)) -> FROM(X)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(FROM(x1))=  x1  
  POL(mark(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 10
             ↳Polo
             ...
               →DP Problem 11
                 ↳Dependency Graph
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Nar


Dependency Pair:


Rules:


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





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polynomial Ordering
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Nar


Dependency Pairs:

S(ok(X)) -> S(X)
S(mark(X)) -> S(X)


Rules:


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





The following dependency pair can be strictly oriented:

S(ok(X)) -> S(X)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(S(x1))=  x1  
  POL(mark(x1))=  x1  
  POL(ok(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
           →DP Problem 12
             ↳Polynomial Ordering
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Nar


Dependency Pair:

S(mark(X)) -> S(X)


Rules:


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





The following dependency pair can be strictly oriented:

S(mark(X)) -> S(X)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(S(x1))=  x1  
  POL(mark(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
           →DP Problem 12
             ↳Polo
             ...
               →DP Problem 13
                 ↳Dependency Graph
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Nar


Dependency Pair:


Rules:


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





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polynomial Ordering
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Nar


Dependency Pairs:

2ND(ok(X)) -> 2ND(X)
2ND(mark(X)) -> 2ND(X)


Rules:


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





The following dependency pair can be strictly oriented:

2ND(ok(X)) -> 2ND(X)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(2ND(x1))=  x1  
  POL(mark(x1))=  x1  
  POL(ok(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
           →DP Problem 14
             ↳Polynomial Ordering
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Nar


Dependency Pair:

2ND(mark(X)) -> 2ND(X)


Rules:


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





The following dependency pair can be strictly oriented:

2ND(mark(X)) -> 2ND(X)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(2ND(x1))=  x1  
  POL(mark(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
           →DP Problem 14
             ↳Polo
             ...
               →DP Problem 15
                 ↳Dependency Graph
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Nar


Dependency Pair:


Rules:


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





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polynomial Ordering
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Nar


Dependency Pairs:

ACTIVE(s(X)) -> ACTIVE(X)
ACTIVE(from(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)
ACTIVE(2nd(X)) -> ACTIVE(X)


Rules:


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





The following dependency pair can be strictly oriented:

ACTIVE(2nd(X)) -> ACTIVE(X)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  x1  
  POL(ACTIVE(x1))=  x1  
  POL(2nd(x1))=  1 + x1  
  POL(cons(x1, x2))=  x1  
  POL(s(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
           →DP Problem 16
             ↳Polynomial Ordering
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Nar


Dependency Pairs:

ACTIVE(s(X)) -> ACTIVE(X)
ACTIVE(from(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)


Rules:


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





The following dependency pair can be strictly oriented:

ACTIVE(s(X)) -> ACTIVE(X)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  x1  
  POL(ACTIVE(x1))=  x1  
  POL(cons(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
           →DP Problem 16
             ↳Polo
             ...
               →DP Problem 17
                 ↳Polynomial Ordering
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Nar


Dependency Pairs:

ACTIVE(from(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)


Rules:


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





The following dependency pair can be strictly oriented:

ACTIVE(from(X)) -> ACTIVE(X)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  1 + x1  
  POL(ACTIVE(x1))=  x1  
  POL(cons(x1, x2))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
           →DP Problem 16
             ↳Polo
             ...
               →DP Problem 18
                 ↳Polynomial Ordering
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Nar


Dependency Pair:

ACTIVE(cons(X1, X2)) -> ACTIVE(X1)


Rules:


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





The following dependency pair can be strictly oriented:

ACTIVE(cons(X1, X2)) -> ACTIVE(X1)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(ACTIVE(x1))=  x1  
  POL(cons(x1, x2))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
           →DP Problem 16
             ↳Polo
             ...
               →DP Problem 19
                 ↳Dependency Graph
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Nar


Dependency Pair:


Rules:


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





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polynomial Ordering
       →DP Problem 7
         ↳Nar


Dependency Pairs:

PROPER(s(X)) -> PROPER(X)
PROPER(from(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)
PROPER(2nd(X)) -> PROPER(X)


Rules:


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





The following dependency pair can be strictly oriented:

PROPER(2nd(X)) -> PROPER(X)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  x1  
  POL(2nd(x1))=  1 + x1  
  POL(PROPER(x1))=  x1  
  POL(cons(x1, x2))=  x1 + x2  
  POL(s(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
           →DP Problem 20
             ↳Polynomial Ordering
       →DP Problem 7
         ↳Nar


Dependency Pairs:

PROPER(s(X)) -> PROPER(X)
PROPER(from(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)


Rules:


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





The following dependency pair can be strictly oriented:

PROPER(s(X)) -> PROPER(X)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  x1  
  POL(PROPER(x1))=  x1  
  POL(cons(x1, x2))=  x1 + x2  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
           →DP Problem 20
             ↳Polo
             ...
               →DP Problem 21
                 ↳Polynomial Ordering
       →DP Problem 7
         ↳Nar


Dependency Pairs:

PROPER(from(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)


Rules:


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





The following dependency pair can be strictly oriented:

PROPER(from(X)) -> PROPER(X)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  1 + x1  
  POL(PROPER(x1))=  x1  
  POL(cons(x1, x2))=  x1 + x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
           →DP Problem 20
             ↳Polo
             ...
               →DP Problem 22
                 ↳Polynomial Ordering
       →DP Problem 7
         ↳Nar


Dependency Pairs:

PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)


Rules:


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





The following dependency pairs can be strictly oriented:

PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(PROPER(x1))=  x1  
  POL(cons(x1, x2))=  1 + x1 + x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
           →DP Problem 20
             ↳Polo
             ...
               →DP Problem 23
                 ↳Dependency Graph
       →DP Problem 7
         ↳Nar


Dependency Pair:


Rules:


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





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Narrowing Transformation


Dependency Pairs:

TOP(ok(X)) -> TOP(active(X))
TOP(mark(X)) -> TOP(proper(X))


Rules:


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





On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TOP(mark(X)) -> TOP(proper(X))
four new Dependency Pairs are created:

TOP(mark(2nd(X''))) -> TOP(2nd(proper(X'')))
TOP(mark(cons(X1', X2'))) -> TOP(cons(proper(X1'), proper(X2')))
TOP(mark(from(X''))) -> TOP(from(proper(X'')))
TOP(mark(s(X''))) -> TOP(s(proper(X'')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Nar
           →DP Problem 24
             ↳Narrowing Transformation


Dependency Pairs:

TOP(mark(s(X''))) -> TOP(s(proper(X'')))
TOP(mark(from(X''))) -> TOP(from(proper(X'')))
TOP(mark(cons(X1', X2'))) -> TOP(cons(proper(X1'), proper(X2')))
TOP(mark(2nd(X''))) -> TOP(2nd(proper(X'')))
TOP(ok(X)) -> TOP(active(X))


Rules:


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





On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TOP(ok(X)) -> TOP(active(X))
six new Dependency Pairs are created:

TOP(ok(2nd(cons(X'', cons(Y', Z'))))) -> TOP(mark(Y'))
TOP(ok(from(X''))) -> TOP(mark(cons(X'', from(s(X'')))))
TOP(ok(2nd(X''))) -> TOP(2nd(active(X'')))
TOP(ok(cons(X1', X2'))) -> TOP(cons(active(X1'), X2'))
TOP(ok(from(X''))) -> TOP(from(active(X'')))
TOP(ok(s(X''))) -> TOP(s(active(X'')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Nar
           →DP Problem 24
             ↳Nar
             ...
               →DP Problem 25
                 ↳Narrowing Transformation


Dependency Pairs:

TOP(ok(s(X''))) -> TOP(s(active(X'')))
TOP(ok(from(X''))) -> TOP(from(active(X'')))
TOP(ok(cons(X1', X2'))) -> TOP(cons(active(X1'), X2'))
TOP(ok(2nd(X''))) -> TOP(2nd(active(X'')))
TOP(ok(from(X''))) -> TOP(mark(cons(X'', from(s(X'')))))
TOP(ok(2nd(cons(X'', cons(Y', Z'))))) -> TOP(mark(Y'))
TOP(mark(from(X''))) -> TOP(from(proper(X'')))
TOP(mark(cons(X1', X2'))) -> TOP(cons(proper(X1'), proper(X2')))
TOP(mark(2nd(X''))) -> TOP(2nd(proper(X'')))
TOP(mark(s(X''))) -> TOP(s(proper(X'')))


Rules:


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





On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TOP(mark(2nd(X''))) -> TOP(2nd(proper(X'')))
four new Dependency Pairs are created:

TOP(mark(2nd(2nd(X')))) -> TOP(2nd(2nd(proper(X'))))
TOP(mark(2nd(cons(X1', X2')))) -> TOP(2nd(cons(proper(X1'), proper(X2'))))
TOP(mark(2nd(from(X')))) -> TOP(2nd(from(proper(X'))))
TOP(mark(2nd(s(X')))) -> TOP(2nd(s(proper(X'))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Nar
           →DP Problem 24
             ↳Nar
             ...
               →DP Problem 26
                 ↳Narrowing Transformation


Dependency Pairs:

TOP(mark(2nd(s(X')))) -> TOP(2nd(s(proper(X'))))
TOP(mark(2nd(from(X')))) -> TOP(2nd(from(proper(X'))))
TOP(ok(from(X''))) -> TOP(from(active(X'')))
TOP(ok(cons(X1', X2'))) -> TOP(cons(active(X1'), X2'))
TOP(ok(2nd(X''))) -> TOP(2nd(active(X'')))
TOP(mark(2nd(cons(X1', X2')))) -> TOP(2nd(cons(proper(X1'), proper(X2'))))
TOP(ok(from(X''))) -> TOP(mark(cons(X'', from(s(X'')))))
TOP(mark(2nd(2nd(X')))) -> TOP(2nd(2nd(proper(X'))))
TOP(ok(2nd(cons(X'', cons(Y', Z'))))) -> TOP(mark(Y'))
TOP(mark(s(X''))) -> TOP(s(proper(X'')))
TOP(mark(from(X''))) -> TOP(from(proper(X'')))
TOP(mark(cons(X1', X2'))) -> TOP(cons(proper(X1'), proper(X2')))
TOP(ok(s(X''))) -> TOP(s(active(X'')))


Rules:


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





On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TOP(mark(cons(X1', X2'))) -> TOP(cons(proper(X1'), proper(X2')))
eight new Dependency Pairs are created:

TOP(mark(cons(2nd(X'), X2'))) -> TOP(cons(2nd(proper(X')), proper(X2')))
TOP(mark(cons(cons(X1'', X2''), X2'))) -> TOP(cons(cons(proper(X1''), proper(X2'')), proper(X2')))
TOP(mark(cons(from(X'), X2'))) -> TOP(cons(from(proper(X')), proper(X2')))
TOP(mark(cons(s(X'), X2'))) -> TOP(cons(s(proper(X')), proper(X2')))
TOP(mark(cons(X1', 2nd(X')))) -> TOP(cons(proper(X1'), 2nd(proper(X'))))
TOP(mark(cons(X1', cons(X1'', X2'')))) -> TOP(cons(proper(X1'), cons(proper(X1''), proper(X2''))))
TOP(mark(cons(X1', from(X')))) -> TOP(cons(proper(X1'), from(proper(X'))))
TOP(mark(cons(X1', s(X')))) -> TOP(cons(proper(X1'), s(proper(X'))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Nar
           →DP Problem 24
             ↳Nar
             ...
               →DP Problem 27
                 ↳Narrowing Transformation


Dependency Pairs:

TOP(mark(cons(X1', s(X')))) -> TOP(cons(proper(X1'), s(proper(X'))))
TOP(mark(cons(X1', from(X')))) -> TOP(cons(proper(X1'), from(proper(X'))))
TOP(mark(cons(X1', cons(X1'', X2'')))) -> TOP(cons(proper(X1'), cons(proper(X1''), proper(X2''))))
TOP(mark(cons(X1', 2nd(X')))) -> TOP(cons(proper(X1'), 2nd(proper(X'))))
TOP(mark(cons(s(X'), X2'))) -> TOP(cons(s(proper(X')), proper(X2')))
TOP(mark(cons(from(X'), X2'))) -> TOP(cons(from(proper(X')), proper(X2')))
TOP(mark(cons(cons(X1'', X2''), X2'))) -> TOP(cons(cons(proper(X1''), proper(X2'')), proper(X2')))
TOP(mark(cons(2nd(X'), X2'))) -> TOP(cons(2nd(proper(X')), proper(X2')))
TOP(mark(2nd(from(X')))) -> TOP(2nd(from(proper(X'))))
TOP(ok(s(X''))) -> TOP(s(active(X'')))
TOP(ok(from(X''))) -> TOP(from(active(X'')))
TOP(ok(cons(X1', X2'))) -> TOP(cons(active(X1'), X2'))
TOP(ok(2nd(X''))) -> TOP(2nd(active(X'')))
TOP(mark(2nd(cons(X1', X2')))) -> TOP(2nd(cons(proper(X1'), proper(X2'))))
TOP(ok(from(X''))) -> TOP(mark(cons(X'', from(s(X'')))))
TOP(mark(2nd(2nd(X')))) -> TOP(2nd(2nd(proper(X'))))
TOP(ok(2nd(cons(X'', cons(Y', Z'))))) -> TOP(mark(Y'))
TOP(mark(s(X''))) -> TOP(s(proper(X'')))
TOP(mark(from(X''))) -> TOP(from(proper(X'')))
TOP(mark(2nd(s(X')))) -> TOP(2nd(s(proper(X'))))


Rules:


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





On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TOP(mark(from(X''))) -> TOP(from(proper(X'')))
four new Dependency Pairs are created:

TOP(mark(from(2nd(X')))) -> TOP(from(2nd(proper(X'))))
TOP(mark(from(cons(X1', X2')))) -> TOP(from(cons(proper(X1'), proper(X2'))))
TOP(mark(from(from(X')))) -> TOP(from(from(proper(X'))))
TOP(mark(from(s(X')))) -> TOP(from(s(proper(X'))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Nar
           →DP Problem 24
             ↳Nar
             ...
               →DP Problem 28
                 ↳Narrowing Transformation


Dependency Pairs:

TOP(mark(from(s(X')))) -> TOP(from(s(proper(X'))))
TOP(mark(from(from(X')))) -> TOP(from(from(proper(X'))))
TOP(mark(from(cons(X1', X2')))) -> TOP(from(cons(proper(X1'), proper(X2'))))
TOP(mark(from(2nd(X')))) -> TOP(from(2nd(proper(X'))))
TOP(mark(cons(X1', from(X')))) -> TOP(cons(proper(X1'), from(proper(X'))))
TOP(mark(cons(X1', cons(X1'', X2'')))) -> TOP(cons(proper(X1'), cons(proper(X1''), proper(X2''))))
TOP(mark(cons(X1', 2nd(X')))) -> TOP(cons(proper(X1'), 2nd(proper(X'))))
TOP(mark(cons(s(X'), X2'))) -> TOP(cons(s(proper(X')), proper(X2')))
TOP(mark(cons(from(X'), X2'))) -> TOP(cons(from(proper(X')), proper(X2')))
TOP(mark(cons(cons(X1'', X2''), X2'))) -> TOP(cons(cons(proper(X1''), proper(X2'')), proper(X2')))
TOP(mark(cons(2nd(X'), X2'))) -> TOP(cons(2nd(proper(X')), proper(X2')))
TOP(mark(2nd(s(X')))) -> TOP(2nd(s(proper(X'))))
TOP(mark(2nd(from(X')))) -> TOP(2nd(from(proper(X'))))
TOP(ok(s(X''))) -> TOP(s(active(X'')))
TOP(ok(from(X''))) -> TOP(from(active(X'')))
TOP(ok(cons(X1', X2'))) -> TOP(cons(active(X1'), X2'))
TOP(ok(2nd(X''))) -> TOP(2nd(active(X'')))
TOP(mark(2nd(cons(X1', X2')))) -> TOP(2nd(cons(proper(X1'), proper(X2'))))
TOP(ok(from(X''))) -> TOP(mark(cons(X'', from(s(X'')))))
TOP(mark(2nd(2nd(X')))) -> TOP(2nd(2nd(proper(X'))))
TOP(ok(2nd(cons(X'', cons(Y', Z'))))) -> TOP(mark(Y'))
TOP(mark(s(X''))) -> TOP(s(proper(X'')))
TOP(mark(cons(X1', s(X')))) -> TOP(cons(proper(X1'), s(proper(X'))))


Rules:


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





On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TOP(mark(s(X''))) -> TOP(s(proper(X'')))
four new Dependency Pairs are created:

TOP(mark(s(2nd(X')))) -> TOP(s(2nd(proper(X'))))
TOP(mark(s(cons(X1', X2')))) -> TOP(s(cons(proper(X1'), proper(X2'))))
TOP(mark(s(from(X')))) -> TOP(s(from(proper(X'))))
TOP(mark(s(s(X')))) -> TOP(s(s(proper(X'))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Nar
           →DP Problem 24
             ↳Nar
             ...
               →DP Problem 29
                 ↳Narrowing Transformation


Dependency Pairs:

TOP(mark(s(s(X')))) -> TOP(s(s(proper(X'))))
TOP(mark(s(from(X')))) -> TOP(s(from(proper(X'))))
TOP(mark(s(cons(X1', X2')))) -> TOP(s(cons(proper(X1'), proper(X2'))))
TOP(mark(s(2nd(X')))) -> TOP(s(2nd(proper(X'))))
TOP(mark(from(from(X')))) -> TOP(from(from(proper(X'))))
TOP(mark(from(cons(X1', X2')))) -> TOP(from(cons(proper(X1'), proper(X2'))))
TOP(mark(from(2nd(X')))) -> TOP(from(2nd(proper(X'))))
TOP(mark(cons(X1', s(X')))) -> TOP(cons(proper(X1'), s(proper(X'))))
TOP(mark(cons(X1', from(X')))) -> TOP(cons(proper(X1'), from(proper(X'))))
TOP(mark(cons(X1', cons(X1'', X2'')))) -> TOP(cons(proper(X1'), cons(proper(X1''), proper(X2''))))
TOP(mark(cons(X1', 2nd(X')))) -> TOP(cons(proper(X1'), 2nd(proper(X'))))
TOP(mark(cons(s(X'), X2'))) -> TOP(cons(s(proper(X')), proper(X2')))
TOP(mark(cons(from(X'), X2'))) -> TOP(cons(from(proper(X')), proper(X2')))
TOP(mark(cons(cons(X1'', X2''), X2'))) -> TOP(cons(cons(proper(X1''), proper(X2'')), proper(X2')))
TOP(mark(cons(2nd(X'), X2'))) -> TOP(cons(2nd(proper(X')), proper(X2')))
TOP(mark(2nd(s(X')))) -> TOP(2nd(s(proper(X'))))
TOP(mark(2nd(from(X')))) -> TOP(2nd(from(proper(X'))))
TOP(ok(s(X''))) -> TOP(s(active(X'')))
TOP(ok(from(X''))) -> TOP(from(active(X'')))
TOP(ok(cons(X1', X2'))) -> TOP(cons(active(X1'), X2'))
TOP(ok(2nd(X''))) -> TOP(2nd(active(X'')))
TOP(mark(2nd(cons(X1', X2')))) -> TOP(2nd(cons(proper(X1'), proper(X2'))))
TOP(ok(from(X''))) -> TOP(mark(cons(X'', from(s(X'')))))
TOP(mark(2nd(2nd(X')))) -> TOP(2nd(2nd(proper(X'))))
TOP(ok(2nd(cons(X'', cons(Y', Z'))))) -> TOP(mark(Y'))
TOP(mark(from(s(X')))) -> TOP(from(s(proper(X'))))


Rules:


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





On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TOP(ok(2nd(X''))) -> TOP(2nd(active(X'')))
six new Dependency Pairs are created:

TOP(ok(2nd(2nd(cons(X', cons(Y', Z')))))) -> TOP(2nd(mark(Y')))
TOP(ok(2nd(from(X')))) -> TOP(2nd(mark(cons(X', from(s(X'))))))
TOP(ok(2nd(2nd(X')))) -> TOP(2nd(2nd(active(X'))))
TOP(ok(2nd(cons(X1', X2')))) -> TOP(2nd(cons(active(X1'), X2')))
TOP(ok(2nd(from(X')))) -> TOP(2nd(from(active(X'))))
TOP(ok(2nd(s(X')))) -> TOP(2nd(s(active(X'))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Nar
           →DP Problem 24
             ↳Nar
             ...
               →DP Problem 30
                 ↳Narrowing Transformation


Dependency Pairs:

TOP(ok(2nd(s(X')))) -> TOP(2nd(s(active(X'))))
TOP(ok(2nd(from(X')))) -> TOP(2nd(from(active(X'))))
TOP(ok(2nd(cons(X1', X2')))) -> TOP(2nd(cons(active(X1'), X2')))
TOP(ok(2nd(2nd(X')))) -> TOP(2nd(2nd(active(X'))))
TOP(ok(2nd(from(X')))) -> TOP(2nd(mark(cons(X', from(s(X'))))))
TOP(ok(2nd(2nd(cons(X', cons(Y', Z')))))) -> TOP(2nd(mark(Y')))
TOP(mark(s(from(X')))) -> TOP(s(from(proper(X'))))
TOP(mark(s(cons(X1', X2')))) -> TOP(s(cons(proper(X1'), proper(X2'))))
TOP(mark(s(2nd(X')))) -> TOP(s(2nd(proper(X'))))
TOP(mark(from(s(X')))) -> TOP(from(s(proper(X'))))
TOP(mark(from(from(X')))) -> TOP(from(from(proper(X'))))
TOP(mark(from(cons(X1', X2')))) -> TOP(from(cons(proper(X1'), proper(X2'))))
TOP(mark(from(2nd(X')))) -> TOP(from(2nd(proper(X'))))
TOP(mark(cons(X1', s(X')))) -> TOP(cons(proper(X1'), s(proper(X'))))
TOP(mark(cons(X1', from(X')))) -> TOP(cons(proper(X1'), from(proper(X'))))
TOP(mark(cons(X1', cons(X1'', X2'')))) -> TOP(cons(proper(X1'), cons(proper(X1''), proper(X2''))))
TOP(mark(cons(X1', 2nd(X')))) -> TOP(cons(proper(X1'), 2nd(proper(X'))))
TOP(mark(cons(s(X'), X2'))) -> TOP(cons(s(proper(X')), proper(X2')))
TOP(mark(cons(from(X'), X2'))) -> TOP(cons(from(proper(X')), proper(X2')))
TOP(mark(cons(cons(X1'', X2''), X2'))) -> TOP(cons(cons(proper(X1''), proper(X2'')), proper(X2')))
TOP(mark(cons(2nd(X'), X2'))) -> TOP(cons(2nd(proper(X')), proper(X2')))
TOP(mark(2nd(s(X')))) -> TOP(2nd(s(proper(X'))))
TOP(mark(2nd(from(X')))) -> TOP(2nd(from(proper(X'))))
TOP(ok(s(X''))) -> TOP(s(active(X'')))
TOP(ok(from(X''))) -> TOP(from(active(X'')))
TOP(ok(cons(X1', X2'))) -> TOP(cons(active(X1'), X2'))
TOP(mark(2nd(cons(X1', X2')))) -> TOP(2nd(cons(proper(X1'), proper(X2'))))
TOP(ok(from(X''))) -> TOP(mark(cons(X'', from(s(X'')))))
TOP(mark(2nd(2nd(X')))) -> TOP(2nd(2nd(proper(X'))))
TOP(ok(2nd(cons(X'', cons(Y', Z'))))) -> TOP(mark(Y'))
TOP(mark(s(s(X')))) -> TOP(s(s(proper(X'))))


Rules:


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





On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TOP(ok(cons(X1', X2'))) -> TOP(cons(active(X1'), X2'))
six new Dependency Pairs are created:

TOP(ok(cons(2nd(cons(X', cons(Y', Z'))), X2'))) -> TOP(cons(mark(Y'), X2'))
TOP(ok(cons(from(X'), X2'))) -> TOP(cons(mark(cons(X', from(s(X')))), X2'))
TOP(ok(cons(2nd(X'), X2'))) -> TOP(cons(2nd(active(X')), X2'))
TOP(ok(cons(cons(X1'', X2''), X2'))) -> TOP(cons(cons(active(X1''), X2''), X2'))
TOP(ok(cons(from(X'), X2'))) -> TOP(cons(from(active(X')), X2'))
TOP(ok(cons(s(X'), X2'))) -> TOP(cons(s(active(X')), X2'))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Nar
           →DP Problem 24
             ↳Nar
             ...
               →DP Problem 31
                 ↳Narrowing Transformation


Dependency Pairs:

TOP(ok(cons(s(X'), X2'))) -> TOP(cons(s(active(X')), X2'))
TOP(ok(cons(from(X'), X2'))) -> TOP(cons(from(active(X')), X2'))
TOP(ok(cons(cons(X1'', X2''), X2'))) -> TOP(cons(cons(active(X1''), X2''), X2'))
TOP(ok(cons(2nd(X'), X2'))) -> TOP(cons(2nd(active(X')), X2'))
TOP(ok(cons(from(X'), X2'))) -> TOP(cons(mark(cons(X', from(s(X')))), X2'))
TOP(ok(cons(2nd(cons(X', cons(Y', Z'))), X2'))) -> TOP(cons(mark(Y'), X2'))
TOP(ok(2nd(from(X')))) -> TOP(2nd(from(active(X'))))
TOP(ok(2nd(cons(X1', X2')))) -> TOP(2nd(cons(active(X1'), X2')))
TOP(ok(2nd(2nd(X')))) -> TOP(2nd(2nd(active(X'))))
TOP(ok(2nd(from(X')))) -> TOP(2nd(mark(cons(X', from(s(X'))))))
TOP(ok(2nd(2nd(cons(X', cons(Y', Z')))))) -> TOP(2nd(mark(Y')))
TOP(mark(s(s(X')))) -> TOP(s(s(proper(X'))))
TOP(mark(s(from(X')))) -> TOP(s(from(proper(X'))))
TOP(mark(s(cons(X1', X2')))) -> TOP(s(cons(proper(X1'), proper(X2'))))
TOP(mark(s(2nd(X')))) -> TOP(s(2nd(proper(X'))))
TOP(mark(from(s(X')))) -> TOP(from(s(proper(X'))))
TOP(mark(from(from(X')))) -> TOP(from(from(proper(X'))))
TOP(mark(from(cons(X1', X2')))) -> TOP(from(cons(proper(X1'), proper(X2'))))
TOP(mark(from(2nd(X')))) -> TOP(from(2nd(proper(X'))))
TOP(mark(cons(X1', s(X')))) -> TOP(cons(proper(X1'), s(proper(X'))))
TOP(mark(cons(X1', from(X')))) -> TOP(cons(proper(X1'), from(proper(X'))))
TOP(mark(cons(X1', cons(X1'', X2'')))) -> TOP(cons(proper(X1'), cons(proper(X1''), proper(X2''))))
TOP(mark(cons(X1', 2nd(X')))) -> TOP(cons(proper(X1'), 2nd(proper(X'))))
TOP(mark(cons(s(X'), X2'))) -> TOP(cons(s(proper(X')), proper(X2')))
TOP(mark(cons(from(X'), X2'))) -> TOP(cons(from(proper(X')), proper(X2')))
TOP(mark(cons(cons(X1'', X2''), X2'))) -> TOP(cons(cons(proper(X1''), proper(X2'')), proper(X2')))
TOP(mark(cons(2nd(X'), X2'))) -> TOP(cons(2nd(proper(X')), proper(X2')))
TOP(mark(2nd(s(X')))) -> TOP(2nd(s(proper(X'))))
TOP(mark(2nd(from(X')))) -> TOP(2nd(from(proper(X'))))
TOP(ok(s(X''))) -> TOP(s(active(X'')))
TOP(ok(from(X''))) -> TOP(from(active(X'')))
TOP(mark(2nd(cons(X1', X2')))) -> TOP(2nd(cons(proper(X1'), proper(X2'))))
TOP(ok(from(X''))) -> TOP(mark(cons(X'', from(s(X'')))))
TOP(mark(2nd(2nd(X')))) -> TOP(2nd(2nd(proper(X'))))
TOP(ok(2nd(cons(X'', cons(Y', Z'))))) -> TOP(mark(Y'))
TOP(ok(2nd(s(X')))) -> TOP(2nd(s(active(X'))))


Rules:


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





On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TOP(ok(from(X''))) -> TOP(from(active(X'')))
six new Dependency Pairs are created:

TOP(ok(from(2nd(cons(X', cons(Y', Z')))))) -> TOP(from(mark(Y')))
TOP(ok(from(from(X')))) -> TOP(from(mark(cons(X', from(s(X'))))))
TOP(ok(from(2nd(X')))) -> TOP(from(2nd(active(X'))))
TOP(ok(from(cons(X1', X2')))) -> TOP(from(cons(active(X1'), X2')))
TOP(ok(from(from(X')))) -> TOP(from(from(active(X'))))
TOP(ok(from(s(X')))) -> TOP(from(s(active(X'))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Nar
           →DP Problem 24
             ↳Nar
             ...
               →DP Problem 32
                 ↳Narrowing Transformation


Dependency Pairs:

TOP(ok(from(s(X')))) -> TOP(from(s(active(X'))))
TOP(ok(from(from(X')))) -> TOP(from(from(active(X'))))
TOP(ok(from(cons(X1', X2')))) -> TOP(from(cons(active(X1'), X2')))
TOP(ok(from(2nd(X')))) -> TOP(from(2nd(active(X'))))
TOP(ok(from(from(X')))) -> TOP(from(mark(cons(X', from(s(X'))))))
TOP(ok(from(2nd(cons(X', cons(Y', Z')))))) -> TOP(from(mark(Y')))
TOP(ok(cons(from(X'), X2'))) -> TOP(cons(from(active(X')), X2'))
TOP(ok(cons(cons(X1'', X2''), X2'))) -> TOP(cons(cons(active(X1''), X2''), X2'))
TOP(ok(cons(2nd(X'), X2'))) -> TOP(cons(2nd(active(X')), X2'))
TOP(ok(cons(from(X'), X2'))) -> TOP(cons(mark(cons(X', from(s(X')))), X2'))
TOP(ok(cons(2nd(cons(X', cons(Y', Z'))), X2'))) -> TOP(cons(mark(Y'), X2'))
TOP(ok(2nd(s(X')))) -> TOP(2nd(s(active(X'))))
TOP(ok(2nd(from(X')))) -> TOP(2nd(from(active(X'))))
TOP(ok(2nd(cons(X1', X2')))) -> TOP(2nd(cons(active(X1'), X2')))
TOP(ok(2nd(2nd(X')))) -> TOP(2nd(2nd(active(X'))))
TOP(ok(2nd(from(X')))) -> TOP(2nd(mark(cons(X', from(s(X'))))))
TOP(ok(2nd(2nd(cons(X', cons(Y', Z')))))) -> TOP(2nd(mark(Y')))
TOP(mark(s(s(X')))) -> TOP(s(s(proper(X'))))
TOP(mark(s(from(X')))) -> TOP(s(from(proper(X'))))
TOP(mark(s(cons(X1', X2')))) -> TOP(s(cons(proper(X1'), proper(X2'))))
TOP(mark(s(2nd(X')))) -> TOP(s(2nd(proper(X'))))
TOP(mark(from(s(X')))) -> TOP(from(s(proper(X'))))
TOP(mark(from(from(X')))) -> TOP(from(from(proper(X'))))
TOP(mark(from(cons(X1', X2')))) -> TOP(from(cons(proper(X1'), proper(X2'))))
TOP(mark(from(2nd(X')))) -> TOP(from(2nd(proper(X'))))
TOP(mark(cons(X1', s(X')))) -> TOP(cons(proper(X1'), s(proper(X'))))
TOP(mark(cons(X1', from(X')))) -> TOP(cons(proper(X1'), from(proper(X'))))
TOP(mark(cons(X1', cons(X1'', X2'')))) -> TOP(cons(proper(X1'), cons(proper(X1''), proper(X2''))))
TOP(mark(cons(X1', 2nd(X')))) -> TOP(cons(proper(X1'), 2nd(proper(X'))))
TOP(mark(cons(s(X'), X2'))) -> TOP(cons(s(proper(X')), proper(X2')))
TOP(mark(cons(from(X'), X2'))) -> TOP(cons(from(proper(X')), proper(X2')))
TOP(mark(cons(cons(X1'', X2''), X2'))) -> TOP(cons(cons(proper(X1''), proper(X2'')), proper(X2')))
TOP(mark(cons(2nd(X'), X2'))) -> TOP(cons(2nd(proper(X')), proper(X2')))
TOP(mark(2nd(s(X')))) -> TOP(2nd(s(proper(X'))))
TOP(mark(2nd(from(X')))) -> TOP(2nd(from(proper(X'))))
TOP(ok(s(X''))) -> TOP(s(active(X'')))
TOP(mark(2nd(cons(X1', X2')))) -> TOP(2nd(cons(proper(X1'), proper(X2'))))
TOP(ok(from(X''))) -> TOP(mark(cons(X'', from(s(X'')))))
TOP(mark(2nd(2nd(X')))) -> TOP(2nd(2nd(proper(X'))))
TOP(ok(2nd(cons(X'', cons(Y', Z'))))) -> TOP(mark(Y'))
TOP(ok(cons(s(X'), X2'))) -> TOP(cons(s(active(X')), X2'))


Rules:


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





On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TOP(ok(s(X''))) -> TOP(s(active(X'')))
six new Dependency Pairs are created:

TOP(ok(s(2nd(cons(X', cons(Y', Z')))))) -> TOP(s(mark(Y')))
TOP(ok(s(from(X')))) -> TOP(s(mark(cons(X', from(s(X'))))))
TOP(ok(s(2nd(X')))) -> TOP(s(2nd(active(X'))))
TOP(ok(s(cons(X1', X2')))) -> TOP(s(cons(active(X1'), X2')))
TOP(ok(s(from(X')))) -> TOP(s(from(active(X'))))
TOP(ok(s(s(X')))) -> TOP(s(s(active(X'))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Nar
           →DP Problem 24
             ↳Nar
             ...
               →DP Problem 33
                 ↳Polynomial Ordering


Dependency Pairs:

TOP(ok(s(s(X')))) -> TOP(s(s(active(X'))))
TOP(ok(s(from(X')))) -> TOP(s(from(active(X'))))
TOP(ok(s(cons(X1', X2')))) -> TOP(s(cons(active(X1'), X2')))
TOP(ok(s(2nd(X')))) -> TOP(s(2nd(active(X'))))
TOP(ok(s(from(X')))) -> TOP(s(mark(cons(X', from(s(X'))))))
TOP(ok(s(2nd(cons(X', cons(Y', Z')))))) -> TOP(s(mark(Y')))
TOP(ok(from(from(X')))) -> TOP(from(from(active(X'))))
TOP(ok(from(cons(X1', X2')))) -> TOP(from(cons(active(X1'), X2')))
TOP(ok(from(2nd(X')))) -> TOP(from(2nd(active(X'))))
TOP(ok(from(from(X')))) -> TOP(from(mark(cons(X', from(s(X'))))))
TOP(ok(from(2nd(cons(X', cons(Y', Z')))))) -> TOP(from(mark(Y')))
TOP(ok(cons(s(X'), X2'))) -> TOP(cons(s(active(X')), X2'))
TOP(ok(cons(from(X'), X2'))) -> TOP(cons(from(active(X')), X2'))
TOP(ok(cons(cons(X1'', X2''), X2'))) -> TOP(cons(cons(active(X1''), X2''), X2'))
TOP(ok(cons(2nd(X'), X2'))) -> TOP(cons(2nd(active(X')), X2'))
TOP(ok(cons(from(X'), X2'))) -> TOP(cons(mark(cons(X', from(s(X')))), X2'))
TOP(ok(cons(2nd(cons(X', cons(Y', Z'))), X2'))) -> TOP(cons(mark(Y'), X2'))
TOP(ok(2nd(s(X')))) -> TOP(2nd(s(active(X'))))
TOP(ok(2nd(from(X')))) -> TOP(2nd(from(active(X'))))
TOP(ok(2nd(cons(X1', X2')))) -> TOP(2nd(cons(active(X1'), X2')))
TOP(ok(2nd(2nd(X')))) -> TOP(2nd(2nd(active(X'))))
TOP(ok(2nd(from(X')))) -> TOP(2nd(mark(cons(X', from(s(X'))))))
TOP(ok(2nd(2nd(cons(X', cons(Y', Z')))))) -> TOP(2nd(mark(Y')))
TOP(mark(s(s(X')))) -> TOP(s(s(proper(X'))))
TOP(mark(s(from(X')))) -> TOP(s(from(proper(X'))))
TOP(mark(s(cons(X1', X2')))) -> TOP(s(cons(proper(X1'), proper(X2'))))
TOP(mark(s(2nd(X')))) -> TOP(s(2nd(proper(X'))))
TOP(mark(from(s(X')))) -> TOP(from(s(proper(X'))))
TOP(mark(from(from(X')))) -> TOP(from(from(proper(X'))))
TOP(mark(from(cons(X1', X2')))) -> TOP(from(cons(proper(X1'), proper(X2'))))
TOP(mark(from(2nd(X')))) -> TOP(from(2nd(proper(X'))))
TOP(mark(cons(X1', s(X')))) -> TOP(cons(proper(X1'), s(proper(X'))))
TOP(mark(cons(X1', from(X')))) -> TOP(cons(proper(X1'), from(proper(X'))))
TOP(mark(cons(X1', cons(X1'', X2'')))) -> TOP(cons(proper(X1'), cons(proper(X1''), proper(X2''))))
TOP(mark(cons(X1', 2nd(X')))) -> TOP(cons(proper(X1'), 2nd(proper(X'))))
TOP(mark(cons(s(X'), X2'))) -> TOP(cons(s(proper(X')), proper(X2')))
TOP(mark(cons(from(X'), X2'))) -> TOP(cons(from(proper(X')), proper(X2')))
TOP(mark(cons(cons(X1'', X2''), X2'))) -> TOP(cons(cons(proper(X1''), proper(X2'')), proper(X2')))
TOP(mark(cons(2nd(X'), X2'))) -> TOP(cons(2nd(proper(X')), proper(X2')))
TOP(mark(2nd(s(X')))) -> TOP(2nd(s(proper(X'))))
TOP(mark(2nd(from(X')))) -> TOP(2nd(from(proper(X'))))
TOP(mark(2nd(cons(X1', X2')))) -> TOP(2nd(cons(proper(X1'), proper(X2'))))
TOP(ok(from(X''))) -> TOP(mark(cons(X'', from(s(X'')))))
TOP(mark(2nd(2nd(X')))) -> TOP(2nd(2nd(proper(X'))))
TOP(ok(2nd(cons(X'', cons(Y', Z'))))) -> TOP(mark(Y'))
TOP(ok(from(s(X')))) -> TOP(from(s(active(X'))))


Rules:


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





The following dependency pairs can be strictly oriented:

TOP(ok(s(s(X')))) -> TOP(s(s(active(X'))))
TOP(ok(s(from(X')))) -> TOP(s(from(active(X'))))
TOP(ok(s(cons(X1', X2')))) -> TOP(s(cons(active(X1'), X2')))
TOP(ok(s(2nd(X')))) -> TOP(s(2nd(active(X'))))
TOP(ok(s(from(X')))) -> TOP(s(mark(cons(X', from(s(X'))))))
TOP(ok(s(2nd(cons(X', cons(Y', Z')))))) -> TOP(s(mark(Y')))
TOP(ok(from(from(X')))) -> TOP(from(from(active(X'))))
TOP(ok(from(cons(X1', X2')))) -> TOP(from(cons(active(X1'), X2')))
TOP(ok(from(2nd(X')))) -> TOP(from(2nd(active(X'))))
TOP(ok(from(from(X')))) -> TOP(from(mark(cons(X', from(s(X'))))))
TOP(ok(from(2nd(cons(X', cons(Y', Z')))))) -> TOP(from(mark(Y')))
TOP(ok(cons(s(X'), X2'))) -> TOP(cons(s(active(X')), X2'))
TOP(ok(cons(from(X'), X2'))) -> TOP(cons(from(active(X')), X2'))
TOP(ok(cons(cons(X1'', X2''), X2'))) -> TOP(cons(cons(active(X1''), X2''), X2'))
TOP(ok(cons(2nd(X'), X2'))) -> TOP(cons(2nd(active(X')), X2'))
TOP(ok(cons(from(X'), X2'))) -> TOP(cons(mark(cons(X', from(s(X')))), X2'))
TOP(ok(cons(2nd(cons(X', cons(Y', Z'))), X2'))) -> TOP(cons(mark(Y'), X2'))
TOP(ok(2nd(s(X')))) -> TOP(2nd(s(active(X'))))
TOP(ok(2nd(from(X')))) -> TOP(2nd(from(active(X'))))
TOP(ok(2nd(cons(X1', X2')))) -> TOP(2nd(cons(active(X1'), X2')))
TOP(ok(2nd(2nd(X')))) -> TOP(2nd(2nd(active(X'))))
TOP(ok(2nd(from(X')))) -> TOP(2nd(mark(cons(X', from(s(X'))))))
TOP(ok(2nd(2nd(cons(X', cons(Y', Z')))))) -> TOP(2nd(mark(Y')))
TOP(ok(from(X''))) -> TOP(mark(cons(X'', from(s(X'')))))
TOP(ok(2nd(cons(X'', cons(Y', Z'))))) -> TOP(mark(Y'))
TOP(ok(from(s(X')))) -> TOP(from(s(active(X'))))


Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  x1  
  POL(active(x1))=  0  
  POL(proper(x1))=  0  
  POL(2nd(x1))=  x1  
  POL(cons(x1, x2))=  x1  
  POL(s(x1))=  x1  
  POL(mark(x1))=  0  
  POL(ok(x1))=  1  
  POL(TOP(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Nar
           →DP Problem 24
             ↳Nar
             ...
               →DP Problem 34
                 ↳Polynomial Ordering


Dependency Pairs:

TOP(mark(s(s(X')))) -> TOP(s(s(proper(X'))))
TOP(mark(s(from(X')))) -> TOP(s(from(proper(X'))))
TOP(mark(s(cons(X1', X2')))) -> TOP(s(cons(proper(X1'), proper(X2'))))
TOP(mark(s(2nd(X')))) -> TOP(s(2nd(proper(X'))))
TOP(mark(from(s(X')))) -> TOP(from(s(proper(X'))))
TOP(mark(from(from(X')))) -> TOP(from(from(proper(X'))))
TOP(mark(from(cons(X1', X2')))) -> TOP(from(cons(proper(X1'), proper(X2'))))
TOP(mark(from(2nd(X')))) -> TOP(from(2nd(proper(X'))))
TOP(mark(cons(X1', s(X')))) -> TOP(cons(proper(X1'), s(proper(X'))))
TOP(mark(cons(X1', from(X')))) -> TOP(cons(proper(X1'), from(proper(X'))))
TOP(mark(cons(X1', cons(X1'', X2'')))) -> TOP(cons(proper(X1'), cons(proper(X1''), proper(X2''))))
TOP(mark(cons(X1', 2nd(X')))) -> TOP(cons(proper(X1'), 2nd(proper(X'))))
TOP(mark(cons(s(X'), X2'))) -> TOP(cons(s(proper(X')), proper(X2')))
TOP(mark(cons(from(X'), X2'))) -> TOP(cons(from(proper(X')), proper(X2')))
TOP(mark(cons(cons(X1'', X2''), X2'))) -> TOP(cons(cons(proper(X1''), proper(X2'')), proper(X2')))
TOP(mark(cons(2nd(X'), X2'))) -> TOP(cons(2nd(proper(X')), proper(X2')))
TOP(mark(2nd(s(X')))) -> TOP(2nd(s(proper(X'))))
TOP(mark(2nd(from(X')))) -> TOP(2nd(from(proper(X'))))
TOP(mark(2nd(cons(X1', X2')))) -> TOP(2nd(cons(proper(X1'), proper(X2'))))
TOP(mark(2nd(2nd(X')))) -> TOP(2nd(2nd(proper(X'))))


Rules:


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





The following dependency pairs can be strictly oriented:

TOP(mark(s(s(X')))) -> TOP(s(s(proper(X'))))
TOP(mark(s(from(X')))) -> TOP(s(from(proper(X'))))
TOP(mark(s(cons(X1', X2')))) -> TOP(s(cons(proper(X1'), proper(X2'))))
TOP(mark(s(2nd(X')))) -> TOP(s(2nd(proper(X'))))
TOP(mark(from(s(X')))) -> TOP(from(s(proper(X'))))
TOP(mark(from(from(X')))) -> TOP(from(from(proper(X'))))
TOP(mark(from(cons(X1', X2')))) -> TOP(from(cons(proper(X1'), proper(X2'))))
TOP(mark(from(2nd(X')))) -> TOP(from(2nd(proper(X'))))
TOP(mark(cons(X1', s(X')))) -> TOP(cons(proper(X1'), s(proper(X'))))
TOP(mark(cons(X1', from(X')))) -> TOP(cons(proper(X1'), from(proper(X'))))
TOP(mark(cons(X1', cons(X1'', X2'')))) -> TOP(cons(proper(X1'), cons(proper(X1''), proper(X2''))))
TOP(mark(cons(X1', 2nd(X')))) -> TOP(cons(proper(X1'), 2nd(proper(X'))))
TOP(mark(cons(s(X'), X2'))) -> TOP(cons(s(proper(X')), proper(X2')))
TOP(mark(cons(from(X'), X2'))) -> TOP(cons(from(proper(X')), proper(X2')))
TOP(mark(cons(cons(X1'', X2''), X2'))) -> TOP(cons(cons(proper(X1''), proper(X2'')), proper(X2')))
TOP(mark(cons(2nd(X'), X2'))) -> TOP(cons(2nd(proper(X')), proper(X2')))
TOP(mark(2nd(s(X')))) -> TOP(2nd(s(proper(X'))))
TOP(mark(2nd(from(X')))) -> TOP(2nd(from(proper(X'))))
TOP(mark(2nd(cons(X1', X2')))) -> TOP(2nd(cons(proper(X1'), proper(X2'))))
TOP(mark(2nd(2nd(X')))) -> TOP(2nd(2nd(proper(X'))))


Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  x1  
  POL(proper(x1))=  0  
  POL(2nd(x1))=  x1  
  POL(cons(x1, x2))=  x1  
  POL(s(x1))=  x1  
  POL(mark(x1))=  1  
  POL(ok(x1))=  0  
  POL(TOP(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Nar
           →DP Problem 24
             ↳Nar
             ...
               →DP Problem 35
                 ↳Dependency Graph


Dependency Pair:


Rules:


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





Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:40 minutes