Term Rewriting System R:
[X, Y, X1, X2]
active(nats) -> mark(adx(zeros))
active(zeros) -> mark(cons(0, zeros))
active(incr(cons(X, Y))) -> mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) -> mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) -> mark(X)
active(tl(cons(X, Y))) -> mark(Y)
active(adx(X)) -> adx(active(X))
active(incr(X)) -> incr(active(X))
active(hd(X)) -> hd(active(X))
active(tl(X)) -> tl(active(X))
adx(mark(X)) -> mark(adx(X))
adx(ok(X)) -> ok(adx(X))
incr(mark(X)) -> mark(incr(X))
incr(ok(X)) -> ok(incr(X))
hd(mark(X)) -> mark(hd(X))
hd(ok(X)) -> ok(hd(X))
tl(mark(X)) -> mark(tl(X))
tl(ok(X)) -> ok(tl(X))
proper(nats) -> ok(nats)
proper(adx(X)) -> adx(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(incr(X)) -> incr(proper(X))
proper(s(X)) -> s(proper(X))
proper(hd(X)) -> hd(proper(X))
proper(tl(X)) -> tl(proper(X))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(ok(X)) -> ok(s(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(nats) -> ADX(zeros)
ACTIVE(zeros) -> CONS(0, zeros)
ACTIVE(incr(cons(X, Y))) -> CONS(s(X), incr(Y))
ACTIVE(incr(cons(X, Y))) -> S(X)
ACTIVE(incr(cons(X, Y))) -> INCR(Y)
ACTIVE(adx(cons(X, Y))) -> INCR(cons(X, adx(Y)))
ACTIVE(adx(cons(X, Y))) -> CONS(X, adx(Y))
ACTIVE(adx(cons(X, Y))) -> ADX(Y)
ACTIVE(adx(X)) -> ADX(active(X))
ACTIVE(adx(X)) -> ACTIVE(X)
ACTIVE(incr(X)) -> INCR(active(X))
ACTIVE(incr(X)) -> ACTIVE(X)
ACTIVE(hd(X)) -> HD(active(X))
ACTIVE(hd(X)) -> ACTIVE(X)
ACTIVE(tl(X)) -> TL(active(X))
ACTIVE(tl(X)) -> ACTIVE(X)
ADX(mark(X)) -> ADX(X)
ADX(ok(X)) -> ADX(X)
INCR(mark(X)) -> INCR(X)
INCR(ok(X)) -> INCR(X)
HD(mark(X)) -> HD(X)
HD(ok(X)) -> HD(X)
TL(mark(X)) -> TL(X)
TL(ok(X)) -> TL(X)
PROPER(adx(X)) -> ADX(proper(X))
PROPER(adx(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(incr(X)) -> INCR(proper(X))
PROPER(incr(X)) -> PROPER(X)
PROPER(s(X)) -> S(proper(X))
PROPER(s(X)) -> PROPER(X)
PROPER(hd(X)) -> HD(proper(X))
PROPER(hd(X)) -> PROPER(X)
PROPER(tl(X)) -> TL(proper(X))
PROPER(tl(X)) -> PROPER(X)
CONS(ok(X1), ok(X2)) -> CONS(X1, X2)
S(ok(X)) -> S(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 nine 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
         ↳Polo
       →DP Problem 9
         ↳Remaining


Dependency Pair:

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


Rules:


active(nats) -> mark(adx(zeros))
active(zeros) -> mark(cons(0, zeros))
active(incr(cons(X, Y))) -> mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) -> mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) -> mark(X)
active(tl(cons(X, Y))) -> mark(Y)
active(adx(X)) -> adx(active(X))
active(incr(X)) -> incr(active(X))
active(hd(X)) -> hd(active(X))
active(tl(X)) -> tl(active(X))
adx(mark(X)) -> mark(adx(X))
adx(ok(X)) -> ok(adx(X))
incr(mark(X)) -> mark(incr(X))
incr(ok(X)) -> ok(incr(X))
hd(mark(X)) -> mark(hd(X))
hd(ok(X)) -> ok(hd(X))
tl(mark(X)) -> mark(tl(X))
tl(ok(X)) -> ok(tl(X))
proper(nats) -> ok(nats)
proper(adx(X)) -> adx(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(incr(X)) -> incr(proper(X))
proper(s(X)) -> s(proper(X))
proper(hd(X)) -> hd(proper(X))
proper(tl(X)) -> tl(proper(X))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(ok(X)) -> ok(s(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(ok(x1))=  1 + x1  
  POL(CONS(x1, x2))=  x1  

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


active(nats) -> mark(adx(zeros))
active(zeros) -> mark(cons(0, zeros))
active(incr(cons(X, Y))) -> mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) -> mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) -> mark(X)
active(tl(cons(X, Y))) -> mark(Y)
active(adx(X)) -> adx(active(X))
active(incr(X)) -> incr(active(X))
active(hd(X)) -> hd(active(X))
active(tl(X)) -> tl(active(X))
adx(mark(X)) -> mark(adx(X))
adx(ok(X)) -> ok(adx(X))
incr(mark(X)) -> mark(incr(X))
incr(ok(X)) -> ok(incr(X))
hd(mark(X)) -> mark(hd(X))
hd(ok(X)) -> ok(hd(X))
tl(mark(X)) -> mark(tl(X))
tl(ok(X)) -> ok(tl(X))
proper(nats) -> ok(nats)
proper(adx(X)) -> adx(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(incr(X)) -> incr(proper(X))
proper(s(X)) -> s(proper(X))
proper(hd(X)) -> hd(proper(X))
proper(tl(X)) -> tl(proper(X))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(ok(X)) -> ok(s(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
         ↳Polo
       →DP Problem 9
         ↳Remaining


Dependency Pair:

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


Rules:


active(nats) -> mark(adx(zeros))
active(zeros) -> mark(cons(0, zeros))
active(incr(cons(X, Y))) -> mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) -> mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) -> mark(X)
active(tl(cons(X, Y))) -> mark(Y)
active(adx(X)) -> adx(active(X))
active(incr(X)) -> incr(active(X))
active(hd(X)) -> hd(active(X))
active(tl(X)) -> tl(active(X))
adx(mark(X)) -> mark(adx(X))
adx(ok(X)) -> ok(adx(X))
incr(mark(X)) -> mark(incr(X))
incr(ok(X)) -> ok(incr(X))
hd(mark(X)) -> mark(hd(X))
hd(ok(X)) -> ok(hd(X))
tl(mark(X)) -> mark(tl(X))
tl(ok(X)) -> ok(tl(X))
proper(nats) -> ok(nats)
proper(adx(X)) -> adx(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(incr(X)) -> incr(proper(X))
proper(s(X)) -> s(proper(X))
proper(hd(X)) -> hd(proper(X))
proper(tl(X)) -> tl(proper(X))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(ok(X)) -> ok(s(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(ok(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳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
         ↳Polo
       →DP Problem 8
         ↳Polo
       →DP Problem 9
         ↳Remaining


Dependency Pair:


Rules:


active(nats) -> mark(adx(zeros))
active(zeros) -> mark(cons(0, zeros))
active(incr(cons(X, Y))) -> mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) -> mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) -> mark(X)
active(tl(cons(X, Y))) -> mark(Y)
active(adx(X)) -> adx(active(X))
active(incr(X)) -> incr(active(X))
active(hd(X)) -> hd(active(X))
active(tl(X)) -> tl(active(X))
adx(mark(X)) -> mark(adx(X))
adx(ok(X)) -> ok(adx(X))
incr(mark(X)) -> mark(incr(X))
incr(ok(X)) -> ok(incr(X))
hd(mark(X)) -> mark(hd(X))
hd(ok(X)) -> ok(hd(X))
tl(mark(X)) -> mark(tl(X))
tl(ok(X)) -> ok(tl(X))
proper(nats) -> ok(nats)
proper(adx(X)) -> adx(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(incr(X)) -> incr(proper(X))
proper(s(X)) -> s(proper(X))
proper(hd(X)) -> hd(proper(X))
proper(tl(X)) -> tl(proper(X))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(ok(X)) -> ok(s(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
         ↳Polo
       →DP Problem 9
         ↳Remaining


Dependency Pairs:

INCR(ok(X)) -> INCR(X)
INCR(mark(X)) -> INCR(X)


Rules:


active(nats) -> mark(adx(zeros))
active(zeros) -> mark(cons(0, zeros))
active(incr(cons(X, Y))) -> mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) -> mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) -> mark(X)
active(tl(cons(X, Y))) -> mark(Y)
active(adx(X)) -> adx(active(X))
active(incr(X)) -> incr(active(X))
active(hd(X)) -> hd(active(X))
active(tl(X)) -> tl(active(X))
adx(mark(X)) -> mark(adx(X))
adx(ok(X)) -> ok(adx(X))
incr(mark(X)) -> mark(incr(X))
incr(ok(X)) -> ok(incr(X))
hd(mark(X)) -> mark(hd(X))
hd(ok(X)) -> ok(hd(X))
tl(mark(X)) -> mark(tl(X))
tl(ok(X)) -> ok(tl(X))
proper(nats) -> ok(nats)
proper(adx(X)) -> adx(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(incr(X)) -> incr(proper(X))
proper(s(X)) -> s(proper(X))
proper(hd(X)) -> hd(proper(X))
proper(tl(X)) -> tl(proper(X))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(ok(X)) -> ok(s(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

INCR(ok(X)) -> INCR(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(INCR(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
         ↳Polo
       →DP Problem 8
         ↳Polo
       →DP Problem 9
         ↳Remaining


Dependency Pair:

INCR(mark(X)) -> INCR(X)


Rules:


active(nats) -> mark(adx(zeros))
active(zeros) -> mark(cons(0, zeros))
active(incr(cons(X, Y))) -> mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) -> mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) -> mark(X)
active(tl(cons(X, Y))) -> mark(Y)
active(adx(X)) -> adx(active(X))
active(incr(X)) -> incr(active(X))
active(hd(X)) -> hd(active(X))
active(tl(X)) -> tl(active(X))
adx(mark(X)) -> mark(adx(X))
adx(ok(X)) -> ok(adx(X))
incr(mark(X)) -> mark(incr(X))
incr(ok(X)) -> ok(incr(X))
hd(mark(X)) -> mark(hd(X))
hd(ok(X)) -> ok(hd(X))
tl(mark(X)) -> mark(tl(X))
tl(ok(X)) -> ok(tl(X))
proper(nats) -> ok(nats)
proper(adx(X)) -> adx(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(incr(X)) -> incr(proper(X))
proper(s(X)) -> s(proper(X))
proper(hd(X)) -> hd(proper(X))
proper(tl(X)) -> tl(proper(X))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(ok(X)) -> ok(s(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

INCR(mark(X)) -> INCR(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(INCR(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
         ↳Polo
       →DP Problem 8
         ↳Polo
       →DP Problem 9
         ↳Remaining


Dependency Pair:


Rules:


active(nats) -> mark(adx(zeros))
active(zeros) -> mark(cons(0, zeros))
active(incr(cons(X, Y))) -> mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) -> mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) -> mark(X)
active(tl(cons(X, Y))) -> mark(Y)
active(adx(X)) -> adx(active(X))
active(incr(X)) -> incr(active(X))
active(hd(X)) -> hd(active(X))
active(tl(X)) -> tl(active(X))
adx(mark(X)) -> mark(adx(X))
adx(ok(X)) -> ok(adx(X))
incr(mark(X)) -> mark(incr(X))
incr(ok(X)) -> ok(incr(X))
hd(mark(X)) -> mark(hd(X))
hd(ok(X)) -> ok(hd(X))
tl(mark(X)) -> mark(tl(X))
tl(ok(X)) -> ok(tl(X))
proper(nats) -> ok(nats)
proper(adx(X)) -> adx(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(incr(X)) -> incr(proper(X))
proper(s(X)) -> s(proper(X))
proper(hd(X)) -> hd(proper(X))
proper(tl(X)) -> tl(proper(X))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(ok(X)) -> ok(s(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
         ↳Polo
       →DP Problem 9
         ↳Remaining


Dependency Pairs:

ADX(ok(X)) -> ADX(X)
ADX(mark(X)) -> ADX(X)


Rules:


active(nats) -> mark(adx(zeros))
active(zeros) -> mark(cons(0, zeros))
active(incr(cons(X, Y))) -> mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) -> mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) -> mark(X)
active(tl(cons(X, Y))) -> mark(Y)
active(adx(X)) -> adx(active(X))
active(incr(X)) -> incr(active(X))
active(hd(X)) -> hd(active(X))
active(tl(X)) -> tl(active(X))
adx(mark(X)) -> mark(adx(X))
adx(ok(X)) -> ok(adx(X))
incr(mark(X)) -> mark(incr(X))
incr(ok(X)) -> ok(incr(X))
hd(mark(X)) -> mark(hd(X))
hd(ok(X)) -> ok(hd(X))
tl(mark(X)) -> mark(tl(X))
tl(ok(X)) -> ok(tl(X))
proper(nats) -> ok(nats)
proper(adx(X)) -> adx(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(incr(X)) -> incr(proper(X))
proper(s(X)) -> s(proper(X))
proper(hd(X)) -> hd(proper(X))
proper(tl(X)) -> tl(proper(X))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(ok(X)) -> ok(s(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

ADX(ok(X)) -> ADX(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(ADX(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
         ↳Polo
       →DP Problem 8
         ↳Polo
       →DP Problem 9
         ↳Remaining


Dependency Pair:

ADX(mark(X)) -> ADX(X)


Rules:


active(nats) -> mark(adx(zeros))
active(zeros) -> mark(cons(0, zeros))
active(incr(cons(X, Y))) -> mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) -> mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) -> mark(X)
active(tl(cons(X, Y))) -> mark(Y)
active(adx(X)) -> adx(active(X))
active(incr(X)) -> incr(active(X))
active(hd(X)) -> hd(active(X))
active(tl(X)) -> tl(active(X))
adx(mark(X)) -> mark(adx(X))
adx(ok(X)) -> ok(adx(X))
incr(mark(X)) -> mark(incr(X))
incr(ok(X)) -> ok(incr(X))
hd(mark(X)) -> mark(hd(X))
hd(ok(X)) -> ok(hd(X))
tl(mark(X)) -> mark(tl(X))
tl(ok(X)) -> ok(tl(X))
proper(nats) -> ok(nats)
proper(adx(X)) -> adx(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(incr(X)) -> incr(proper(X))
proper(s(X)) -> s(proper(X))
proper(hd(X)) -> hd(proper(X))
proper(tl(X)) -> tl(proper(X))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(ok(X)) -> ok(s(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

ADX(mark(X)) -> ADX(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(ADX(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
         ↳Polo
       →DP Problem 8
         ↳Polo
       →DP Problem 9
         ↳Remaining


Dependency Pair:


Rules:


active(nats) -> mark(adx(zeros))
active(zeros) -> mark(cons(0, zeros))
active(incr(cons(X, Y))) -> mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) -> mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) -> mark(X)
active(tl(cons(X, Y))) -> mark(Y)
active(adx(X)) -> adx(active(X))
active(incr(X)) -> incr(active(X))
active(hd(X)) -> hd(active(X))
active(tl(X)) -> tl(active(X))
adx(mark(X)) -> mark(adx(X))
adx(ok(X)) -> ok(adx(X))
incr(mark(X)) -> mark(incr(X))
incr(ok(X)) -> ok(incr(X))
hd(mark(X)) -> mark(hd(X))
hd(ok(X)) -> ok(hd(X))
tl(mark(X)) -> mark(tl(X))
tl(ok(X)) -> ok(tl(X))
proper(nats) -> ok(nats)
proper(adx(X)) -> adx(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(incr(X)) -> incr(proper(X))
proper(s(X)) -> s(proper(X))
proper(hd(X)) -> hd(proper(X))
proper(tl(X)) -> tl(proper(X))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(ok(X)) -> ok(s(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
         ↳Polo
       →DP Problem 9
         ↳Remaining


Dependency Pairs:

HD(ok(X)) -> HD(X)
HD(mark(X)) -> HD(X)


Rules:


active(nats) -> mark(adx(zeros))
active(zeros) -> mark(cons(0, zeros))
active(incr(cons(X, Y))) -> mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) -> mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) -> mark(X)
active(tl(cons(X, Y))) -> mark(Y)
active(adx(X)) -> adx(active(X))
active(incr(X)) -> incr(active(X))
active(hd(X)) -> hd(active(X))
active(tl(X)) -> tl(active(X))
adx(mark(X)) -> mark(adx(X))
adx(ok(X)) -> ok(adx(X))
incr(mark(X)) -> mark(incr(X))
incr(ok(X)) -> ok(incr(X))
hd(mark(X)) -> mark(hd(X))
hd(ok(X)) -> ok(hd(X))
tl(mark(X)) -> mark(tl(X))
tl(ok(X)) -> ok(tl(X))
proper(nats) -> ok(nats)
proper(adx(X)) -> adx(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(incr(X)) -> incr(proper(X))
proper(s(X)) -> s(proper(X))
proper(hd(X)) -> hd(proper(X))
proper(tl(X)) -> tl(proper(X))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(ok(X)) -> ok(s(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

HD(ok(X)) -> HD(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(HD(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 5
         ↳Polo
           →DP Problem 16
             ↳Polynomial Ordering
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Polo
       →DP Problem 9
         ↳Remaining


Dependency Pair:

HD(mark(X)) -> HD(X)


Rules:


active(nats) -> mark(adx(zeros))
active(zeros) -> mark(cons(0, zeros))
active(incr(cons(X, Y))) -> mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) -> mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) -> mark(X)
active(tl(cons(X, Y))) -> mark(Y)
active(adx(X)) -> adx(active(X))
active(incr(X)) -> incr(active(X))
active(hd(X)) -> hd(active(X))
active(tl(X)) -> tl(active(X))
adx(mark(X)) -> mark(adx(X))
adx(ok(X)) -> ok(adx(X))
incr(mark(X)) -> mark(incr(X))
incr(ok(X)) -> ok(incr(X))
hd(mark(X)) -> mark(hd(X))
hd(ok(X)) -> ok(hd(X))
tl(mark(X)) -> mark(tl(X))
tl(ok(X)) -> ok(tl(X))
proper(nats) -> ok(nats)
proper(adx(X)) -> adx(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(incr(X)) -> incr(proper(X))
proper(s(X)) -> s(proper(X))
proper(hd(X)) -> hd(proper(X))
proper(tl(X)) -> tl(proper(X))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(ok(X)) -> ok(s(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

HD(mark(X)) -> HD(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(HD(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 5
         ↳Polo
           →DP Problem 16
             ↳Polo
             ...
               →DP Problem 17
                 ↳Dependency Graph
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Polo
       →DP Problem 9
         ↳Remaining


Dependency Pair:


Rules:


active(nats) -> mark(adx(zeros))
active(zeros) -> mark(cons(0, zeros))
active(incr(cons(X, Y))) -> mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) -> mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) -> mark(X)
active(tl(cons(X, Y))) -> mark(Y)
active(adx(X)) -> adx(active(X))
active(incr(X)) -> incr(active(X))
active(hd(X)) -> hd(active(X))
active(tl(X)) -> tl(active(X))
adx(mark(X)) -> mark(adx(X))
adx(ok(X)) -> ok(adx(X))
incr(mark(X)) -> mark(incr(X))
incr(ok(X)) -> ok(incr(X))
hd(mark(X)) -> mark(hd(X))
hd(ok(X)) -> ok(hd(X))
tl(mark(X)) -> mark(tl(X))
tl(ok(X)) -> ok(tl(X))
proper(nats) -> ok(nats)
proper(adx(X)) -> adx(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(incr(X)) -> incr(proper(X))
proper(s(X)) -> s(proper(X))
proper(hd(X)) -> hd(proper(X))
proper(tl(X)) -> tl(proper(X))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(ok(X)) -> ok(s(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
         ↳Polo
       →DP Problem 9
         ↳Remaining


Dependency Pairs:

TL(ok(X)) -> TL(X)
TL(mark(X)) -> TL(X)


Rules:


active(nats) -> mark(adx(zeros))
active(zeros) -> mark(cons(0, zeros))
active(incr(cons(X, Y))) -> mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) -> mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) -> mark(X)
active(tl(cons(X, Y))) -> mark(Y)
active(adx(X)) -> adx(active(X))
active(incr(X)) -> incr(active(X))
active(hd(X)) -> hd(active(X))
active(tl(X)) -> tl(active(X))
adx(mark(X)) -> mark(adx(X))
adx(ok(X)) -> ok(adx(X))
incr(mark(X)) -> mark(incr(X))
incr(ok(X)) -> ok(incr(X))
hd(mark(X)) -> mark(hd(X))
hd(ok(X)) -> ok(hd(X))
tl(mark(X)) -> mark(tl(X))
tl(ok(X)) -> ok(tl(X))
proper(nats) -> ok(nats)
proper(adx(X)) -> adx(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(incr(X)) -> incr(proper(X))
proper(s(X)) -> s(proper(X))
proper(hd(X)) -> hd(proper(X))
proper(tl(X)) -> tl(proper(X))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(ok(X)) -> ok(s(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

TL(ok(X)) -> TL(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(TL(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 5
         ↳Polo
       →DP Problem 6
         ↳Polo
           →DP Problem 18
             ↳Polynomial Ordering
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Polo
       →DP Problem 9
         ↳Remaining


Dependency Pair:

TL(mark(X)) -> TL(X)


Rules:


active(nats) -> mark(adx(zeros))
active(zeros) -> mark(cons(0, zeros))
active(incr(cons(X, Y))) -> mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) -> mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) -> mark(X)
active(tl(cons(X, Y))) -> mark(Y)
active(adx(X)) -> adx(active(X))
active(incr(X)) -> incr(active(X))
active(hd(X)) -> hd(active(X))
active(tl(X)) -> tl(active(X))
adx(mark(X)) -> mark(adx(X))
adx(ok(X)) -> ok(adx(X))
incr(mark(X)) -> mark(incr(X))
incr(ok(X)) -> ok(incr(X))
hd(mark(X)) -> mark(hd(X))
hd(ok(X)) -> ok(hd(X))
tl(mark(X)) -> mark(tl(X))
tl(ok(X)) -> ok(tl(X))
proper(nats) -> ok(nats)
proper(adx(X)) -> adx(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(incr(X)) -> incr(proper(X))
proper(s(X)) -> s(proper(X))
proper(hd(X)) -> hd(proper(X))
proper(tl(X)) -> tl(proper(X))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(ok(X)) -> ok(s(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

TL(mark(X)) -> TL(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(TL(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 18
             ↳Polo
             ...
               →DP Problem 19
                 ↳Dependency Graph
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Polo
       →DP Problem 9
         ↳Remaining


Dependency Pair:


Rules:


active(nats) -> mark(adx(zeros))
active(zeros) -> mark(cons(0, zeros))
active(incr(cons(X, Y))) -> mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) -> mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) -> mark(X)
active(tl(cons(X, Y))) -> mark(Y)
active(adx(X)) -> adx(active(X))
active(incr(X)) -> incr(active(X))
active(hd(X)) -> hd(active(X))
active(tl(X)) -> tl(active(X))
adx(mark(X)) -> mark(adx(X))
adx(ok(X)) -> ok(adx(X))
incr(mark(X)) -> mark(incr(X))
incr(ok(X)) -> ok(incr(X))
hd(mark(X)) -> mark(hd(X))
hd(ok(X)) -> ok(hd(X))
tl(mark(X)) -> mark(tl(X))
tl(ok(X)) -> ok(tl(X))
proper(nats) -> ok(nats)
proper(adx(X)) -> adx(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(incr(X)) -> incr(proper(X))
proper(s(X)) -> s(proper(X))
proper(hd(X)) -> hd(proper(X))
proper(tl(X)) -> tl(proper(X))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(ok(X)) -> ok(s(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
         ↳Polo
       →DP Problem 9
         ↳Remaining


Dependency Pairs:

ACTIVE(tl(X)) -> ACTIVE(X)
ACTIVE(hd(X)) -> ACTIVE(X)
ACTIVE(incr(X)) -> ACTIVE(X)
ACTIVE(adx(X)) -> ACTIVE(X)


Rules:


active(nats) -> mark(adx(zeros))
active(zeros) -> mark(cons(0, zeros))
active(incr(cons(X, Y))) -> mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) -> mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) -> mark(X)
active(tl(cons(X, Y))) -> mark(Y)
active(adx(X)) -> adx(active(X))
active(incr(X)) -> incr(active(X))
active(hd(X)) -> hd(active(X))
active(tl(X)) -> tl(active(X))
adx(mark(X)) -> mark(adx(X))
adx(ok(X)) -> ok(adx(X))
incr(mark(X)) -> mark(incr(X))
incr(ok(X)) -> ok(incr(X))
hd(mark(X)) -> mark(hd(X))
hd(ok(X)) -> ok(hd(X))
tl(mark(X)) -> mark(tl(X))
tl(ok(X)) -> ok(tl(X))
proper(nats) -> ok(nats)
proper(adx(X)) -> adx(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(incr(X)) -> incr(proper(X))
proper(s(X)) -> s(proper(X))
proper(hd(X)) -> hd(proper(X))
proper(tl(X)) -> tl(proper(X))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(ok(X)) -> ok(s(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

ACTIVE(tl(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(adx(x1))=  x1  
  POL(hd(x1))=  x1  
  POL(incr(x1))=  x1  
  POL(tl(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
         ↳Polo
       →DP Problem 9
         ↳Remaining


Dependency Pairs:

ACTIVE(hd(X)) -> ACTIVE(X)
ACTIVE(incr(X)) -> ACTIVE(X)
ACTIVE(adx(X)) -> ACTIVE(X)


Rules:


active(nats) -> mark(adx(zeros))
active(zeros) -> mark(cons(0, zeros))
active(incr(cons(X, Y))) -> mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) -> mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) -> mark(X)
active(tl(cons(X, Y))) -> mark(Y)
active(adx(X)) -> adx(active(X))
active(incr(X)) -> incr(active(X))
active(hd(X)) -> hd(active(X))
active(tl(X)) -> tl(active(X))
adx(mark(X)) -> mark(adx(X))
adx(ok(X)) -> ok(adx(X))
incr(mark(X)) -> mark(incr(X))
incr(ok(X)) -> ok(incr(X))
hd(mark(X)) -> mark(hd(X))
hd(ok(X)) -> ok(hd(X))
tl(mark(X)) -> mark(tl(X))
tl(ok(X)) -> ok(tl(X))
proper(nats) -> ok(nats)
proper(adx(X)) -> adx(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(incr(X)) -> incr(proper(X))
proper(s(X)) -> s(proper(X))
proper(hd(X)) -> hd(proper(X))
proper(tl(X)) -> tl(proper(X))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(ok(X)) -> ok(s(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

ACTIVE(hd(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(adx(x1))=  x1  
  POL(hd(x1))=  1 + x1  
  POL(incr(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
         ↳Polo
           →DP Problem 20
             ↳Polo
             ...
               →DP Problem 21
                 ↳Polynomial Ordering
       →DP Problem 8
         ↳Polo
       →DP Problem 9
         ↳Remaining


Dependency Pairs:

ACTIVE(incr(X)) -> ACTIVE(X)
ACTIVE(adx(X)) -> ACTIVE(X)


Rules:


active(nats) -> mark(adx(zeros))
active(zeros) -> mark(cons(0, zeros))
active(incr(cons(X, Y))) -> mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) -> mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) -> mark(X)
active(tl(cons(X, Y))) -> mark(Y)
active(adx(X)) -> adx(active(X))
active(incr(X)) -> incr(active(X))
active(hd(X)) -> hd(active(X))
active(tl(X)) -> tl(active(X))
adx(mark(X)) -> mark(adx(X))
adx(ok(X)) -> ok(adx(X))
incr(mark(X)) -> mark(incr(X))
incr(ok(X)) -> ok(incr(X))
hd(mark(X)) -> mark(hd(X))
hd(ok(X)) -> ok(hd(X))
tl(mark(X)) -> mark(tl(X))
tl(ok(X)) -> ok(tl(X))
proper(nats) -> ok(nats)
proper(adx(X)) -> adx(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(incr(X)) -> incr(proper(X))
proper(s(X)) -> s(proper(X))
proper(hd(X)) -> hd(proper(X))
proper(tl(X)) -> tl(proper(X))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(ok(X)) -> ok(s(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

ACTIVE(incr(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(adx(x1))=  x1  
  POL(incr(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
             ↳Polo
             ...
               →DP Problem 22
                 ↳Polynomial Ordering
       →DP Problem 8
         ↳Polo
       →DP Problem 9
         ↳Remaining


Dependency Pair:

ACTIVE(adx(X)) -> ACTIVE(X)


Rules:


active(nats) -> mark(adx(zeros))
active(zeros) -> mark(cons(0, zeros))
active(incr(cons(X, Y))) -> mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) -> mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) -> mark(X)
active(tl(cons(X, Y))) -> mark(Y)
active(adx(X)) -> adx(active(X))
active(incr(X)) -> incr(active(X))
active(hd(X)) -> hd(active(X))
active(tl(X)) -> tl(active(X))
adx(mark(X)) -> mark(adx(X))
adx(ok(X)) -> ok(adx(X))
incr(mark(X)) -> mark(incr(X))
incr(ok(X)) -> ok(incr(X))
hd(mark(X)) -> mark(hd(X))
hd(ok(X)) -> ok(hd(X))
tl(mark(X)) -> mark(tl(X))
tl(ok(X)) -> ok(tl(X))
proper(nats) -> ok(nats)
proper(adx(X)) -> adx(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(incr(X)) -> incr(proper(X))
proper(s(X)) -> s(proper(X))
proper(hd(X)) -> hd(proper(X))
proper(tl(X)) -> tl(proper(X))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(ok(X)) -> ok(s(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

ACTIVE(adx(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(adx(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
             ↳Polo
             ...
               →DP Problem 23
                 ↳Dependency Graph
       →DP Problem 8
         ↳Polo
       →DP Problem 9
         ↳Remaining


Dependency Pair:


Rules:


active(nats) -> mark(adx(zeros))
active(zeros) -> mark(cons(0, zeros))
active(incr(cons(X, Y))) -> mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) -> mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) -> mark(X)
active(tl(cons(X, Y))) -> mark(Y)
active(adx(X)) -> adx(active(X))
active(incr(X)) -> incr(active(X))
active(hd(X)) -> hd(active(X))
active(tl(X)) -> tl(active(X))
adx(mark(X)) -> mark(adx(X))
adx(ok(X)) -> ok(adx(X))
incr(mark(X)) -> mark(incr(X))
incr(ok(X)) -> ok(incr(X))
hd(mark(X)) -> mark(hd(X))
hd(ok(X)) -> ok(hd(X))
tl(mark(X)) -> mark(tl(X))
tl(ok(X)) -> ok(tl(X))
proper(nats) -> ok(nats)
proper(adx(X)) -> adx(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(incr(X)) -> incr(proper(X))
proper(s(X)) -> s(proper(X))
proper(hd(X)) -> hd(proper(X))
proper(tl(X)) -> tl(proper(X))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(ok(X)) -> ok(s(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
         ↳Polynomial Ordering
       →DP Problem 9
         ↳Remaining


Dependency Pairs:

PROPER(tl(X)) -> PROPER(X)
PROPER(hd(X)) -> PROPER(X)
PROPER(s(X)) -> PROPER(X)
PROPER(incr(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)
PROPER(adx(X)) -> PROPER(X)


Rules:


active(nats) -> mark(adx(zeros))
active(zeros) -> mark(cons(0, zeros))
active(incr(cons(X, Y))) -> mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) -> mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) -> mark(X)
active(tl(cons(X, Y))) -> mark(Y)
active(adx(X)) -> adx(active(X))
active(incr(X)) -> incr(active(X))
active(hd(X)) -> hd(active(X))
active(tl(X)) -> tl(active(X))
adx(mark(X)) -> mark(adx(X))
adx(ok(X)) -> ok(adx(X))
incr(mark(X)) -> mark(incr(X))
incr(ok(X)) -> ok(incr(X))
hd(mark(X)) -> mark(hd(X))
hd(ok(X)) -> ok(hd(X))
tl(mark(X)) -> mark(tl(X))
tl(ok(X)) -> ok(tl(X))
proper(nats) -> ok(nats)
proper(adx(X)) -> adx(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(incr(X)) -> incr(proper(X))
proper(s(X)) -> s(proper(X))
proper(hd(X)) -> hd(proper(X))
proper(tl(X)) -> tl(proper(X))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(ok(X)) -> ok(s(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(adx(x1))=  x1  
  POL(PROPER(x1))=  x1  
  POL(cons(x1, x2))=  1 + x1 + x2  
  POL(hd(x1))=  x1  
  POL(incr(x1))=  x1  
  POL(s(x1))=  x1  
  POL(tl(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
         ↳Polo
       →DP Problem 8
         ↳Polo
           →DP Problem 24
             ↳Polynomial Ordering
       →DP Problem 9
         ↳Remaining


Dependency Pairs:

PROPER(tl(X)) -> PROPER(X)
PROPER(hd(X)) -> PROPER(X)
PROPER(s(X)) -> PROPER(X)
PROPER(incr(X)) -> PROPER(X)
PROPER(adx(X)) -> PROPER(X)


Rules:


active(nats) -> mark(adx(zeros))
active(zeros) -> mark(cons(0, zeros))
active(incr(cons(X, Y))) -> mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) -> mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) -> mark(X)
active(tl(cons(X, Y))) -> mark(Y)
active(adx(X)) -> adx(active(X))
active(incr(X)) -> incr(active(X))
active(hd(X)) -> hd(active(X))
active(tl(X)) -> tl(active(X))
adx(mark(X)) -> mark(adx(X))
adx(ok(X)) -> ok(adx(X))
incr(mark(X)) -> mark(incr(X))
incr(ok(X)) -> ok(incr(X))
hd(mark(X)) -> mark(hd(X))
hd(ok(X)) -> ok(hd(X))
tl(mark(X)) -> mark(tl(X))
tl(ok(X)) -> ok(tl(X))
proper(nats) -> ok(nats)
proper(adx(X)) -> adx(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(incr(X)) -> incr(proper(X))
proper(s(X)) -> s(proper(X))
proper(hd(X)) -> hd(proper(X))
proper(tl(X)) -> tl(proper(X))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(ok(X)) -> ok(s(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

PROPER(adx(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(adx(x1))=  1 + x1  
  POL(PROPER(x1))=  x1  
  POL(hd(x1))=  x1  
  POL(incr(x1))=  x1  
  POL(s(x1))=  x1  
  POL(tl(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
         ↳Polo
       →DP Problem 8
         ↳Polo
           →DP Problem 24
             ↳Polo
             ...
               →DP Problem 25
                 ↳Polynomial Ordering
       →DP Problem 9
         ↳Remaining


Dependency Pairs:

PROPER(tl(X)) -> PROPER(X)
PROPER(hd(X)) -> PROPER(X)
PROPER(s(X)) -> PROPER(X)
PROPER(incr(X)) -> PROPER(X)


Rules:


active(nats) -> mark(adx(zeros))
active(zeros) -> mark(cons(0, zeros))
active(incr(cons(X, Y))) -> mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) -> mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) -> mark(X)
active(tl(cons(X, Y))) -> mark(Y)
active(adx(X)) -> adx(active(X))
active(incr(X)) -> incr(active(X))
active(hd(X)) -> hd(active(X))
active(tl(X)) -> tl(active(X))
adx(mark(X)) -> mark(adx(X))
adx(ok(X)) -> ok(adx(X))
incr(mark(X)) -> mark(incr(X))
incr(ok(X)) -> ok(incr(X))
hd(mark(X)) -> mark(hd(X))
hd(ok(X)) -> ok(hd(X))
tl(mark(X)) -> mark(tl(X))
tl(ok(X)) -> ok(tl(X))
proper(nats) -> ok(nats)
proper(adx(X)) -> adx(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(incr(X)) -> incr(proper(X))
proper(s(X)) -> s(proper(X))
proper(hd(X)) -> hd(proper(X))
proper(tl(X)) -> tl(proper(X))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(ok(X)) -> ok(s(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

PROPER(tl(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(hd(x1))=  x1  
  POL(incr(x1))=  x1  
  POL(s(x1))=  x1  
  POL(tl(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 8
         ↳Polo
           →DP Problem 24
             ↳Polo
             ...
               →DP Problem 26
                 ↳Polynomial Ordering
       →DP Problem 9
         ↳Remaining


Dependency Pairs:

PROPER(hd(X)) -> PROPER(X)
PROPER(s(X)) -> PROPER(X)
PROPER(incr(X)) -> PROPER(X)


Rules:


active(nats) -> mark(adx(zeros))
active(zeros) -> mark(cons(0, zeros))
active(incr(cons(X, Y))) -> mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) -> mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) -> mark(X)
active(tl(cons(X, Y))) -> mark(Y)
active(adx(X)) -> adx(active(X))
active(incr(X)) -> incr(active(X))
active(hd(X)) -> hd(active(X))
active(tl(X)) -> tl(active(X))
adx(mark(X)) -> mark(adx(X))
adx(ok(X)) -> ok(adx(X))
incr(mark(X)) -> mark(incr(X))
incr(ok(X)) -> ok(incr(X))
hd(mark(X)) -> mark(hd(X))
hd(ok(X)) -> ok(hd(X))
tl(mark(X)) -> mark(tl(X))
tl(ok(X)) -> ok(tl(X))
proper(nats) -> ok(nats)
proper(adx(X)) -> adx(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(incr(X)) -> incr(proper(X))
proper(s(X)) -> s(proper(X))
proper(hd(X)) -> hd(proper(X))
proper(tl(X)) -> tl(proper(X))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(ok(X)) -> ok(s(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

PROPER(hd(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(hd(x1))=  1 + x1  
  POL(incr(x1))=  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 6
         ↳Polo
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Polo
           →DP Problem 24
             ↳Polo
             ...
               →DP Problem 27
                 ↳Polynomial Ordering
       →DP Problem 9
         ↳Remaining


Dependency Pairs:

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


Rules:


active(nats) -> mark(adx(zeros))
active(zeros) -> mark(cons(0, zeros))
active(incr(cons(X, Y))) -> mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) -> mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) -> mark(X)
active(tl(cons(X, Y))) -> mark(Y)
active(adx(X)) -> adx(active(X))
active(incr(X)) -> incr(active(X))
active(hd(X)) -> hd(active(X))
active(tl(X)) -> tl(active(X))
adx(mark(X)) -> mark(adx(X))
adx(ok(X)) -> ok(adx(X))
incr(mark(X)) -> mark(incr(X))
incr(ok(X)) -> ok(incr(X))
hd(mark(X)) -> mark(hd(X))
hd(ok(X)) -> ok(hd(X))
tl(mark(X)) -> mark(tl(X))
tl(ok(X)) -> ok(tl(X))
proper(nats) -> ok(nats)
proper(adx(X)) -> adx(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(incr(X)) -> incr(proper(X))
proper(s(X)) -> s(proper(X))
proper(hd(X)) -> hd(proper(X))
proper(tl(X)) -> tl(proper(X))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(ok(X)) -> ok(s(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(incr(x1))=  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 6
         ↳Polo
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Polo
           →DP Problem 24
             ↳Polo
             ...
               →DP Problem 28
                 ↳Polynomial Ordering
       →DP Problem 9
         ↳Remaining


Dependency Pair:

PROPER(incr(X)) -> PROPER(X)


Rules:


active(nats) -> mark(adx(zeros))
active(zeros) -> mark(cons(0, zeros))
active(incr(cons(X, Y))) -> mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) -> mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) -> mark(X)
active(tl(cons(X, Y))) -> mark(Y)
active(adx(X)) -> adx(active(X))
active(incr(X)) -> incr(active(X))
active(hd(X)) -> hd(active(X))
active(tl(X)) -> tl(active(X))
adx(mark(X)) -> mark(adx(X))
adx(ok(X)) -> ok(adx(X))
incr(mark(X)) -> mark(incr(X))
incr(ok(X)) -> ok(incr(X))
hd(mark(X)) -> mark(hd(X))
hd(ok(X)) -> ok(hd(X))
tl(mark(X)) -> mark(tl(X))
tl(ok(X)) -> ok(tl(X))
proper(nats) -> ok(nats)
proper(adx(X)) -> adx(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(incr(X)) -> incr(proper(X))
proper(s(X)) -> s(proper(X))
proper(hd(X)) -> hd(proper(X))
proper(tl(X)) -> tl(proper(X))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(ok(X)) -> ok(s(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

PROPER(incr(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(incr(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 8
         ↳Polo
           →DP Problem 24
             ↳Polo
             ...
               →DP Problem 29
                 ↳Dependency Graph
       →DP Problem 9
         ↳Remaining


Dependency Pair:


Rules:


active(nats) -> mark(adx(zeros))
active(zeros) -> mark(cons(0, zeros))
active(incr(cons(X, Y))) -> mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) -> mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) -> mark(X)
active(tl(cons(X, Y))) -> mark(Y)
active(adx(X)) -> adx(active(X))
active(incr(X)) -> incr(active(X))
active(hd(X)) -> hd(active(X))
active(tl(X)) -> tl(active(X))
adx(mark(X)) -> mark(adx(X))
adx(ok(X)) -> ok(adx(X))
incr(mark(X)) -> mark(incr(X))
incr(ok(X)) -> ok(incr(X))
hd(mark(X)) -> mark(hd(X))
hd(ok(X)) -> ok(hd(X))
tl(mark(X)) -> mark(tl(X))
tl(ok(X)) -> ok(tl(X))
proper(nats) -> ok(nats)
proper(adx(X)) -> adx(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(incr(X)) -> incr(proper(X))
proper(s(X)) -> s(proper(X))
proper(hd(X)) -> hd(proper(X))
proper(tl(X)) -> tl(proper(X))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(ok(X)) -> ok(s(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
         ↳Polo
       →DP Problem 9
         ↳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(nats) -> mark(adx(zeros))
active(zeros) -> mark(cons(0, zeros))
active(incr(cons(X, Y))) -> mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) -> mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) -> mark(X)
active(tl(cons(X, Y))) -> mark(Y)
active(adx(X)) -> adx(active(X))
active(incr(X)) -> incr(active(X))
active(hd(X)) -> hd(active(X))
active(tl(X)) -> tl(active(X))
adx(mark(X)) -> mark(adx(X))
adx(ok(X)) -> ok(adx(X))
incr(mark(X)) -> mark(incr(X))
incr(ok(X)) -> ok(incr(X))
hd(mark(X)) -> mark(hd(X))
hd(ok(X)) -> ok(hd(X))
tl(mark(X)) -> mark(tl(X))
tl(ok(X)) -> ok(tl(X))
proper(nats) -> ok(nats)
proper(adx(X)) -> adx(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(incr(X)) -> incr(proper(X))
proper(s(X)) -> s(proper(X))
proper(hd(X)) -> hd(proper(X))
proper(tl(X)) -> tl(proper(X))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
s(ok(X)) -> ok(s(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))




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