Term Rewriting System R:
[X, X1, X2]
active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(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))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(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(f(0)) -> CONS(0, f(s(0)))
ACTIVE(f(0)) -> F(s(0))
ACTIVE(f(0)) -> S(0)
ACTIVE(f(s(0))) -> F(p(s(0)))
ACTIVE(f(s(0))) -> P(s(0))
ACTIVE(f(X)) -> F(active(X))
ACTIVE(f(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)
ACTIVE(p(X)) -> P(active(X))
ACTIVE(p(X)) -> ACTIVE(X)
F(mark(X)) -> F(X)
F(ok(X)) -> F(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)
P(mark(X)) -> P(X)
P(ok(X)) -> P(X)
PROPER(f(X)) -> F(proper(X))
PROPER(f(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(p(X)) -> P(proper(X))
PROPER(p(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
         ↳Remaining


Dependency Pairs:

F(ok(X)) -> F(X)
F(mark(X)) -> F(X)


Rules:


active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(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))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

F(ok(X)) -> F(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(mark(x1))=  x1  
  POL(ok(x1))=  1 + x1  
  POL(F(x1))=  x1  

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
         ↳Remaining


Dependency Pair:

F(mark(X)) -> F(X)


Rules:


active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(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))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

F(mark(X)) -> F(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(mark(x1))=  1 + x1  
  POL(F(x1))=  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
         ↳Remaining


Dependency Pair:


Rules:


active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(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))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(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
         ↳Remaining


Dependency Pairs:

P(ok(X)) -> P(X)
P(mark(X)) -> P(X)


Rules:


active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(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))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

P(ok(X)) -> P(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(P(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
         ↳Remaining


Dependency Pair:

P(mark(X)) -> P(X)


Rules:


active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(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))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

P(mark(X)) -> P(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(P(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
         ↳Remaining


Dependency Pair:


Rules:


active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(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))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(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
         ↳Remaining


Dependency Pairs:

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


Rules:


active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(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))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(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 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
         ↳Remaining


Dependency Pair:

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


Rules:


active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(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))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(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 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
         ↳Remaining


Dependency Pair:


Rules:


active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(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))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(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
         ↳Remaining


Dependency Pairs:

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


Rules:


active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(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))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(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 4
         ↳Polo
           →DP Problem 14
             ↳Polynomial Ordering
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Remaining


Dependency Pair:

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


Rules:


active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(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))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(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 4
         ↳Polo
           →DP Problem 14
             ↳Polo
             ...
               →DP Problem 15
                 ↳Dependency Graph
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Remaining


Dependency Pair:


Rules:


active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(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))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(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
         ↳Remaining


Dependency Pairs:

ACTIVE(p(X)) -> ACTIVE(X)
ACTIVE(s(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)
ACTIVE(f(X)) -> ACTIVE(X)


Rules:


active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(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))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

ACTIVE(f(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(ACTIVE(x1))=  x1  
  POL(cons(x1, x2))=  x1  
  POL(s(x1))=  x1  
  POL(f(x1))=  1 + x1  
  POL(p(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
         ↳Remaining


Dependency Pairs:

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


Rules:


active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(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))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

ACTIVE(p(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(ACTIVE(x1))=  x1  
  POL(cons(x1, x2))=  x1  
  POL(s(x1))=  x1  
  POL(p(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
         ↳Remaining


Dependency Pairs:

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


Rules:


active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(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))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(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(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 18
                 ↳Polynomial Ordering
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Remaining


Dependency Pair:

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


Rules:


active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(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))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(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
         ↳Remaining


Dependency Pair:


Rules:


active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(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))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(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
         ↳Remaining


Dependency Pairs:

PROPER(p(X)) -> PROPER(X)
PROPER(s(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)
PROPER(f(X)) -> PROPER(X)


Rules:


active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(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))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

PROPER(f(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(PROPER(x1))=  x1  
  POL(cons(x1, x2))=  x1 + x2  
  POL(s(x1))=  x1  
  POL(f(x1))=  1 + x1  
  POL(p(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
         ↳Remaining


Dependency Pairs:

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


Rules:


active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(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))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

PROPER(p(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(PROPER(x1))=  x1  
  POL(cons(x1, x2))=  x1 + x2  
  POL(s(x1))=  x1  
  POL(p(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
         ↳Remaining


Dependency Pairs:

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


Rules:


active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(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))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(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(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 22
                 ↳Polynomial Ordering
       →DP Problem 7
         ↳Remaining


Dependency Pairs:

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


Rules:


active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(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))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(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
         ↳Remaining


Dependency Pair:


Rules:


active(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(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))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(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
         ↳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(f(0)) -> mark(cons(0, f(s(0))))
active(f(s(0))) -> mark(f(p(s(0))))
active(p(s(0))) -> mark(0)
active(f(X)) -> f(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(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))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
proper(f(X)) -> f(proper(X))
proper(0) -> ok(0)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))




Termination of R could not be shown.
Duration:
0:00 minutes