Term Rewriting System R:
[X, Y, X1, X2]
active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(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(length(cons(X, Y))) -> S(length1(Y))
ACTIVE(length(cons(X, Y))) -> LENGTH1(Y)
ACTIVE(length1(X)) -> LENGTH(X)
ACTIVE(from(X)) -> FROM(active(X))
ACTIVE(from(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)
ACTIVE(s(X)) -> S(active(X))
ACTIVE(s(X)) -> ACTIVE(X)
FROM(mark(X)) -> FROM(X)
FROM(ok(X)) -> FROM(X)
CONS(mark(X1), X2) -> CONS(X1, X2)
CONS(ok(X1), ok(X2)) -> CONS(X1, X2)
S(mark(X)) -> S(X)
S(ok(X)) -> S(X)
PROPER(from(X)) -> FROM(proper(X))
PROPER(from(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> CONS(proper(X1), proper(X2))
PROPER(cons(X1, X2)) -> PROPER(X1)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(s(X)) -> S(proper(X))
PROPER(s(X)) -> PROPER(X)
PROPER(length(X)) -> LENGTH(proper(X))
PROPER(length(X)) -> PROPER(X)
PROPER(length1(X)) -> LENGTH1(proper(X))
PROPER(length1(X)) -> PROPER(X)
LENGTH(ok(X)) -> LENGTH(X)
LENGTH1(ok(X)) -> LENGTH1(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 eight 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
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pairs:

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


Rules:


active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(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)


Additionally, the following rules can be oriented:

active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  x1  
  POL(proper(x1))=  1  
  POL(mark(x1))=  0  
  POL(ok(x1))=  1 + x1  
  POL(CONS(x1, x2))=  x2  
  POL(length1(x1))=  x1  
  POL(top(x1))=  0  
  POL(active(x1))=  0  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1  
  POL(nil)=  0  
  POL(s(x1))=  x1  
  POL(length(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 9
             ↳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
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pair:

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


Rules:


active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(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)


Additionally, the following rules can be oriented:

active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  x1  
  POL(proper(x1))=  0  
  POL(mark(x1))=  1 + x1  
  POL(ok(x1))=  0  
  POL(CONS(x1, x2))=  x1  
  POL(length1(x1))=  0  
  POL(top(x1))=  0  
  POL(active(x1))=  1 + x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1  
  POL(nil)=  0  
  POL(s(x1))=  x1  
  POL(length(x1))=  0  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 9
             ↳Polo
             ...
               →DP Problem 10
                 ↳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
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pair:


Rules:


active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(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
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pairs:

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


Rules:


active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(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)


Additionally, the following rules can be oriented:

active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  x1  
  POL(FROM(x1))=  x1  
  POL(proper(x1))=  0  
  POL(mark(x1))=  1 + x1  
  POL(ok(x1))=  x1  
  POL(length1(x1))=  0  
  POL(top(x1))=  0  
  POL(active(x1))=  1 + x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1  
  POL(nil)=  0  
  POL(s(x1))=  x1  
  POL(length(x1))=  0  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 11
             ↳Polynomial Ordering
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pair:

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


Rules:


active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(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)


Additionally, the following rules can be oriented:

active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  x1  
  POL(FROM(x1))=  x1  
  POL(proper(x1))=  1  
  POL(mark(x1))=  0  
  POL(ok(x1))=  1 + x1  
  POL(length1(x1))=  x1  
  POL(top(x1))=  0  
  POL(active(x1))=  0  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1  
  POL(nil)=  0  
  POL(s(x1))=  x1  
  POL(length(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 11
             ↳Polo
             ...
               →DP Problem 12
                 ↳Dependency Graph
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pair:


Rules:


active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(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
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pairs:

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


Rules:


active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(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)


Additionally, the following rules can be oriented:

active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  x1  
  POL(proper(x1))=  0  
  POL(mark(x1))=  1 + x1  
  POL(ok(x1))=  x1  
  POL(length1(x1))=  0  
  POL(top(x1))=  0  
  POL(active(x1))=  1 + x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1  
  POL(nil)=  0  
  POL(s(x1))=  x1  
  POL(S(x1))=  x1  
  POL(length(x1))=  0  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
           →DP Problem 13
             ↳Polynomial Ordering
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pair:

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


Rules:


active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(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)


Additionally, the following rules can be oriented:

active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  x1  
  POL(proper(x1))=  1  
  POL(mark(x1))=  0  
  POL(ok(x1))=  1 + x1  
  POL(length1(x1))=  x1  
  POL(top(x1))=  0  
  POL(active(x1))=  0  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1  
  POL(nil)=  0  
  POL(s(x1))=  x1  
  POL(S(x1))=  x1  
  POL(length(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
           →DP Problem 13
             ↳Polo
             ...
               →DP Problem 14
                 ↳Dependency Graph
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pair:


Rules:


active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(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
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pair:

LENGTH1(ok(X)) -> LENGTH1(X)


Rules:


active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

LENGTH1(ok(X)) -> LENGTH1(X)


Additionally, the following rules can be oriented:

active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  x1  
  POL(proper(x1))=  1  
  POL(mark(x1))=  0  
  POL(ok(x1))=  1 + x1  
  POL(length1(x1))=  x1  
  POL(top(x1))=  0  
  POL(active(x1))=  0  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1  
  POL(LENGTH1(x1))=  x1  
  POL(nil)=  0  
  POL(s(x1))=  x1  
  POL(length(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 15
             ↳Dependency Graph
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pair:


Rules:


active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(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
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pair:

LENGTH(ok(X)) -> LENGTH(X)


Rules:


active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

LENGTH(ok(X)) -> LENGTH(X)


Additionally, the following rules can be oriented:

active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  x1  
  POL(proper(x1))=  1  
  POL(mark(x1))=  0  
  POL(ok(x1))=  1 + x1  
  POL(length1(x1))=  x1  
  POL(top(x1))=  0  
  POL(active(x1))=  0  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1  
  POL(nil)=  0  
  POL(s(x1))=  x1  
  POL(LENGTH(x1))=  x1  
  POL(length(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
             ↳Dependency Graph
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pair:


Rules:


active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(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
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pairs:

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


Rules:


active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(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)


Additionally, the following rules can be oriented:

active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  x1  
  POL(proper(x1))=  x1  
  POL(mark(x1))=  0  
  POL(ok(x1))=  0  
  POL(length1(x1))=  0  
  POL(top(x1))=  0  
  POL(active(x1))=  x1  
  POL(ACTIVE(x1))=  x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1  
  POL(nil)=  0  
  POL(s(x1))=  1 + x1  
  POL(length(x1))=  0  

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 17
             ↳Polynomial Ordering
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pairs:

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


Rules:


active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(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)


Additionally, the following rules can be oriented:

active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  1 + x1  
  POL(proper(x1))=  x1  
  POL(mark(x1))=  0  
  POL(ok(x1))=  0  
  POL(length1(x1))=  0  
  POL(top(x1))=  0  
  POL(active(x1))=  x1  
  POL(ACTIVE(x1))=  x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1  
  POL(nil)=  0  
  POL(s(x1))=  0  
  POL(length(x1))=  0  

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 17
             ↳Polo
             ...
               →DP Problem 18
                 ↳Polynomial Ordering
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pair:

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


Rules:


active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(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)


Additionally, the following rules can be oriented:

active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  0  
  POL(proper(x1))=  x1  
  POL(mark(x1))=  0  
  POL(ok(x1))=  0  
  POL(length1(x1))=  0  
  POL(top(x1))=  0  
  POL(active(x1))=  x1  
  POL(ACTIVE(x1))=  x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  1 + x1  
  POL(nil)=  0  
  POL(s(x1))=  0  
  POL(length(x1))=  0  

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 17
             ↳Polo
             ...
               →DP Problem 19
                 ↳Dependency Graph
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pair:


Rules:


active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(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
         ↳Polynomial Ordering
       →DP Problem 8
         ↳Remaining


Dependency Pairs:

PROPER(length1(X)) -> PROPER(X)
PROPER(length(X)) -> PROPER(X)
PROPER(s(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)
PROPER(from(X)) -> PROPER(X)


Rules:


active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

PROPER(length(X)) -> PROPER(X)


Additionally, the following rules can be oriented:

active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  x1  
  POL(proper(x1))=  x1  
  POL(PROPER(x1))=  1 + x1  
  POL(mark(x1))=  0  
  POL(ok(x1))=  0  
  POL(length1(x1))=  x1  
  POL(top(x1))=  0  
  POL(active(x1))=  x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1 + x2  
  POL(nil)=  0  
  POL(s(x1))=  x1  
  POL(length(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
         ↳Polo
           →DP Problem 20
             ↳Polynomial Ordering
       →DP Problem 8
         ↳Remaining


Dependency Pairs:

PROPER(length1(X)) -> PROPER(X)
PROPER(s(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)
PROPER(from(X)) -> PROPER(X)


Rules:


active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

PROPER(length1(X)) -> PROPER(X)


Additionally, the following rules can be oriented:

active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  x1  
  POL(proper(x1))=  x1  
  POL(PROPER(x1))=  x1  
  POL(mark(x1))=  0  
  POL(ok(x1))=  0  
  POL(length1(x1))=  1 + x1  
  POL(top(x1))=  0  
  POL(active(x1))=  x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1 + x2  
  POL(nil)=  0  
  POL(s(x1))=  x1  
  POL(length(x1))=  0  

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
         ↳Polo
           →DP Problem 20
             ↳Polo
             ...
               →DP Problem 21
                 ↳Polynomial Ordering
       →DP Problem 8
         ↳Remaining


Dependency Pairs:

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


Rules:


active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(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)


Additionally, the following rules can be oriented:

active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  x1  
  POL(proper(x1))=  x1  
  POL(PROPER(x1))=  x1  
  POL(mark(x1))=  0  
  POL(ok(x1))=  0  
  POL(length1(x1))=  0  
  POL(top(x1))=  0  
  POL(active(x1))=  x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1 + x2  
  POL(nil)=  0  
  POL(s(x1))=  1 + x1  
  POL(length(x1))=  0  

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
         ↳Polo
           →DP Problem 20
             ↳Polo
             ...
               →DP Problem 22
                 ↳Polynomial Ordering
       →DP Problem 8
         ↳Remaining


Dependency Pairs:

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


Rules:


active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(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)


Additionally, the following rules can be oriented:

active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  1 + x1  
  POL(proper(x1))=  x1  
  POL(PROPER(x1))=  x1  
  POL(mark(x1))=  0  
  POL(ok(x1))=  0  
  POL(length1(x1))=  0  
  POL(top(x1))=  0  
  POL(active(x1))=  x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1 + x2  
  POL(nil)=  0  
  POL(s(x1))=  0  
  POL(length(x1))=  0  

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
         ↳Polo
           →DP Problem 20
             ↳Polo
             ...
               →DP Problem 23
                 ↳Polynomial Ordering
       →DP Problem 8
         ↳Remaining


Dependency Pairs:

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


Rules:


active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(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)


Additionally, the following rules can be oriented:

active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  0  
  POL(proper(x1))=  x1  
  POL(PROPER(x1))=  x1  
  POL(mark(x1))=  0  
  POL(ok(x1))=  0  
  POL(length1(x1))=  0  
  POL(top(x1))=  0  
  POL(active(x1))=  x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  1 + x1 + x2  
  POL(nil)=  0  
  POL(s(x1))=  0  
  POL(length(x1))=  0  

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
         ↳Polo
           →DP Problem 20
             ↳Polo
             ...
               →DP Problem 24
                 ↳Dependency Graph
       →DP Problem 8
         ↳Remaining


Dependency Pair:


Rules:


active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(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
         ↳Polo
       →DP Problem 8
         ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

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


Rules:


active(from(X)) -> mark(cons(X, from(s(X))))
active(length(nil)) -> mark(0)
active(length(cons(X, Y))) -> mark(s(length1(Y)))
active(length1(X)) -> mark(length(X))
active(from(X)) -> from(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(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))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(from(X)) -> from(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(length(X)) -> length(proper(X))
proper(nil) -> ok(nil)
proper(0) -> ok(0)
proper(length1(X)) -> length1(proper(X))
length(ok(X)) -> ok(length(X))
length1(ok(X)) -> ok(length1(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))




Termination of R could not be shown.
Duration:
0:01 minutes