Term Rewriting System R:
[Y, X, X1, X2, X3]
active(minus(0, Y)) -> mark(0)
active(minus(s(X), s(Y))) -> mark(minus(X, Y))
active(geq(X, 0)) -> mark(true)
active(geq(0, s(Y))) -> mark(false)
active(geq(s(X), s(Y))) -> mark(geq(X, Y))
active(div(0, s(Y))) -> mark(0)
active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(s(X)) -> s(active(X))
active(div(X1, X2)) -> div(active(X1), X2)
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(minus(X1, X2)) -> minus(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(geq(X1, X2)) -> geq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(div(X1, X2)) -> div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) -> ok(minus(X1, X2))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
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(minus(s(X), s(Y))) -> MINUS(X, Y)
ACTIVE(geq(s(X), s(Y))) -> GEQ(X, Y)
ACTIVE(div(s(X), s(Y))) -> IF(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
ACTIVE(div(s(X), s(Y))) -> GEQ(X, Y)
ACTIVE(div(s(X), s(Y))) -> S(div(minus(X, Y), s(Y)))
ACTIVE(div(s(X), s(Y))) -> DIV(minus(X, Y), s(Y))
ACTIVE(div(s(X), s(Y))) -> MINUS(X, Y)
ACTIVE(s(X)) -> S(active(X))
ACTIVE(s(X)) -> ACTIVE(X)
ACTIVE(div(X1, X2)) -> DIV(active(X1), X2)
ACTIVE(div(X1, X2)) -> ACTIVE(X1)
ACTIVE(if(X1, X2, X3)) -> IF(active(X1), X2, X3)
ACTIVE(if(X1, X2, X3)) -> ACTIVE(X1)
S(mark(X)) -> S(X)
S(ok(X)) -> S(X)
DIV(mark(X1), X2) -> DIV(X1, X2)
DIV(ok(X1), ok(X2)) -> DIV(X1, X2)
IF(mark(X1), X2, X3) -> IF(X1, X2, X3)
IF(ok(X1), ok(X2), ok(X3)) -> IF(X1, X2, X3)
PROPER(minus(X1, X2)) -> MINUS(proper(X1), proper(X2))
PROPER(minus(X1, X2)) -> PROPER(X1)
PROPER(minus(X1, X2)) -> PROPER(X2)
PROPER(s(X)) -> S(proper(X))
PROPER(s(X)) -> PROPER(X)
PROPER(geq(X1, X2)) -> GEQ(proper(X1), proper(X2))
PROPER(geq(X1, X2)) -> PROPER(X1)
PROPER(geq(X1, X2)) -> PROPER(X2)
PROPER(div(X1, X2)) -> DIV(proper(X1), proper(X2))
PROPER(div(X1, X2)) -> PROPER(X1)
PROPER(div(X1, X2)) -> PROPER(X2)
PROPER(if(X1, X2, X3)) -> IF(proper(X1), proper(X2), proper(X3))
PROPER(if(X1, X2, X3)) -> PROPER(X1)
PROPER(if(X1, X2, X3)) -> PROPER(X2)
PROPER(if(X1, X2, X3)) -> PROPER(X3)
MINUS(ok(X1), ok(X2)) -> MINUS(X1, X2)
GEQ(ok(X1), ok(X2)) -> GEQ(X1, X2)
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 Pair:

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


Rules:


active(minus(0, Y)) -> mark(0)
active(minus(s(X), s(Y))) -> mark(minus(X, Y))
active(geq(X, 0)) -> mark(true)
active(geq(0, s(Y))) -> mark(false)
active(geq(s(X), s(Y))) -> mark(geq(X, Y))
active(div(0, s(Y))) -> mark(0)
active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(s(X)) -> s(active(X))
active(div(X1, X2)) -> div(active(X1), X2)
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(minus(X1, X2)) -> minus(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(geq(X1, X2)) -> geq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(div(X1, X2)) -> div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) -> ok(minus(X1, X2))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

MINUS(ok(X1), ok(X2)) -> MINUS(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(MINUS(x1, x2))=  x1  
  POL(ok(x1))=  1 + x1  

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


active(minus(0, Y)) -> mark(0)
active(minus(s(X), s(Y))) -> mark(minus(X, Y))
active(geq(X, 0)) -> mark(true)
active(geq(0, s(Y))) -> mark(false)
active(geq(s(X), s(Y))) -> mark(geq(X, Y))
active(div(0, s(Y))) -> mark(0)
active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(s(X)) -> s(active(X))
active(div(X1, X2)) -> div(active(X1), X2)
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(minus(X1, X2)) -> minus(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(geq(X1, X2)) -> geq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(div(X1, X2)) -> div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) -> ok(minus(X1, X2))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
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 Pair:

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


Rules:


active(minus(0, Y)) -> mark(0)
active(minus(s(X), s(Y))) -> mark(minus(X, Y))
active(geq(X, 0)) -> mark(true)
active(geq(0, s(Y))) -> mark(false)
active(geq(s(X), s(Y))) -> mark(geq(X, Y))
active(div(0, s(Y))) -> mark(0)
active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(s(X)) -> s(active(X))
active(div(X1, X2)) -> div(active(X1), X2)
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(minus(X1, X2)) -> minus(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(geq(X1, X2)) -> geq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(div(X1, X2)) -> div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) -> ok(minus(X1, X2))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

GEQ(ok(X1), ok(X2)) -> GEQ(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(GEQ(x1, x2))=  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
             ↳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(minus(0, Y)) -> mark(0)
active(minus(s(X), s(Y))) -> mark(minus(X, Y))
active(geq(X, 0)) -> mark(true)
active(geq(0, s(Y))) -> mark(false)
active(geq(s(X), s(Y))) -> mark(geq(X, Y))
active(div(0, s(Y))) -> mark(0)
active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(s(X)) -> s(active(X))
active(div(X1, X2)) -> div(active(X1), X2)
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(minus(X1, X2)) -> minus(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(geq(X1, X2)) -> geq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(div(X1, X2)) -> div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) -> ok(minus(X1, X2))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
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:

IF(ok(X1), ok(X2), ok(X3)) -> IF(X1, X2, X3)
IF(mark(X1), X2, X3) -> IF(X1, X2, X3)


Rules:


active(minus(0, Y)) -> mark(0)
active(minus(s(X), s(Y))) -> mark(minus(X, Y))
active(geq(X, 0)) -> mark(true)
active(geq(0, s(Y))) -> mark(false)
active(geq(s(X), s(Y))) -> mark(geq(X, Y))
active(div(0, s(Y))) -> mark(0)
active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(s(X)) -> s(active(X))
active(div(X1, X2)) -> div(active(X1), X2)
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(minus(X1, X2)) -> minus(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(geq(X1, X2)) -> geq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(div(X1, X2)) -> div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) -> ok(minus(X1, X2))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

IF(ok(X1), ok(X2), ok(X3)) -> IF(X1, X2, X3)


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(IF(x1, x2, x3))=  x2  

resulting in one new DP problem. 



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


Dependency Pair:

IF(mark(X1), X2, X3) -> IF(X1, X2, X3)


Rules:


active(minus(0, Y)) -> mark(0)
active(minus(s(X), s(Y))) -> mark(minus(X, Y))
active(geq(X, 0)) -> mark(true)
active(geq(0, s(Y))) -> mark(false)
active(geq(s(X), s(Y))) -> mark(geq(X, Y))
active(div(0, s(Y))) -> mark(0)
active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(s(X)) -> s(active(X))
active(div(X1, X2)) -> div(active(X1), X2)
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(minus(X1, X2)) -> minus(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(geq(X1, X2)) -> geq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(div(X1, X2)) -> div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) -> ok(minus(X1, X2))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

IF(mark(X1), X2, X3) -> IF(X1, X2, X3)


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(IF(x1, x2, x3))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
           →DP Problem 11
             ↳Polo
             ...
               →DP Problem 12
                 ↳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(minus(0, Y)) -> mark(0)
active(minus(s(X), s(Y))) -> mark(minus(X, Y))
active(geq(X, 0)) -> mark(true)
active(geq(0, s(Y))) -> mark(false)
active(geq(s(X), s(Y))) -> mark(geq(X, Y))
active(div(0, s(Y))) -> mark(0)
active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(s(X)) -> s(active(X))
active(div(X1, X2)) -> div(active(X1), X2)
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(minus(X1, X2)) -> minus(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(geq(X1, X2)) -> geq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(div(X1, X2)) -> div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) -> ok(minus(X1, X2))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
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 Pairs:

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


Rules:


active(minus(0, Y)) -> mark(0)
active(minus(s(X), s(Y))) -> mark(minus(X, Y))
active(geq(X, 0)) -> mark(true)
active(geq(0, s(Y))) -> mark(false)
active(geq(s(X), s(Y))) -> mark(geq(X, Y))
active(div(0, s(Y))) -> mark(0)
active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(s(X)) -> s(active(X))
active(div(X1, X2)) -> div(active(X1), X2)
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(minus(X1, X2)) -> minus(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(geq(X1, X2)) -> geq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(div(X1, X2)) -> div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) -> ok(minus(X1, X2))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
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 13
             ↳Polynomial Ordering
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pair:

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


Rules:


active(minus(0, Y)) -> mark(0)
active(minus(s(X), s(Y))) -> mark(minus(X, Y))
active(geq(X, 0)) -> mark(true)
active(geq(0, s(Y))) -> mark(false)
active(geq(s(X), s(Y))) -> mark(geq(X, Y))
active(div(0, s(Y))) -> mark(0)
active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(s(X)) -> s(active(X))
active(div(X1, X2)) -> div(active(X1), X2)
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(minus(X1, X2)) -> minus(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(geq(X1, X2)) -> geq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(div(X1, X2)) -> div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) -> ok(minus(X1, X2))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
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 13
             ↳Polo
             ...
               →DP Problem 14
                 ↳Dependency Graph
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pair:


Rules:


active(minus(0, Y)) -> mark(0)
active(minus(s(X), s(Y))) -> mark(minus(X, Y))
active(geq(X, 0)) -> mark(true)
active(geq(0, s(Y))) -> mark(false)
active(geq(s(X), s(Y))) -> mark(geq(X, Y))
active(div(0, s(Y))) -> mark(0)
active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(s(X)) -> s(active(X))
active(div(X1, X2)) -> div(active(X1), X2)
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(minus(X1, X2)) -> minus(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(geq(X1, X2)) -> geq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(div(X1, X2)) -> div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) -> ok(minus(X1, X2))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
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 Pairs:

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


Rules:


active(minus(0, Y)) -> mark(0)
active(minus(s(X), s(Y))) -> mark(minus(X, Y))
active(geq(X, 0)) -> mark(true)
active(geq(0, s(Y))) -> mark(false)
active(geq(s(X), s(Y))) -> mark(geq(X, Y))
active(div(0, s(Y))) -> mark(0)
active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(s(X)) -> s(active(X))
active(div(X1, X2)) -> div(active(X1), X2)
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(minus(X1, X2)) -> minus(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(geq(X1, X2)) -> geq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(div(X1, X2)) -> div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) -> ok(minus(X1, X2))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

DIV(ok(X1), ok(X2)) -> DIV(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(DIV(x1, x2))=  x2  
  POL(mark(x1))=  0  
  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 15
             ↳Polynomial Ordering
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pair:

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


Rules:


active(minus(0, Y)) -> mark(0)
active(minus(s(X), s(Y))) -> mark(minus(X, Y))
active(geq(X, 0)) -> mark(true)
active(geq(0, s(Y))) -> mark(false)
active(geq(s(X), s(Y))) -> mark(geq(X, Y))
active(div(0, s(Y))) -> mark(0)
active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(s(X)) -> s(active(X))
active(div(X1, X2)) -> div(active(X1), X2)
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(minus(X1, X2)) -> minus(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(geq(X1, X2)) -> geq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(div(X1, X2)) -> div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) -> ok(minus(X1, X2))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

DIV(mark(X1), X2) -> DIV(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(DIV(x1, x2))=  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 15
             ↳Polo
             ...
               →DP Problem 16
                 ↳Dependency Graph
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pair:


Rules:


active(minus(0, Y)) -> mark(0)
active(minus(s(X), s(Y))) -> mark(minus(X, Y))
active(geq(X, 0)) -> mark(true)
active(geq(0, s(Y))) -> mark(false)
active(geq(s(X), s(Y))) -> mark(geq(X, Y))
active(div(0, s(Y))) -> mark(0)
active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(s(X)) -> s(active(X))
active(div(X1, X2)) -> div(active(X1), X2)
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(minus(X1, X2)) -> minus(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(geq(X1, X2)) -> geq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(div(X1, X2)) -> div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) -> ok(minus(X1, X2))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
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(if(X1, X2, X3)) -> ACTIVE(X1)
ACTIVE(div(X1, X2)) -> ACTIVE(X1)
ACTIVE(s(X)) -> ACTIVE(X)


Rules:


active(minus(0, Y)) -> mark(0)
active(minus(s(X), s(Y))) -> mark(minus(X, Y))
active(geq(X, 0)) -> mark(true)
active(geq(0, s(Y))) -> mark(false)
active(geq(s(X), s(Y))) -> mark(geq(X, Y))
active(div(0, s(Y))) -> mark(0)
active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(s(X)) -> s(active(X))
active(div(X1, X2)) -> div(active(X1), X2)
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(minus(X1, X2)) -> minus(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(geq(X1, X2)) -> geq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(div(X1, X2)) -> div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) -> ok(minus(X1, X2))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

ACTIVE(if(X1, X2, X3)) -> 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(if(x1, x2, x3))=  1 + x1  
  POL(s(x1))=  x1  
  POL(div(x1, x2))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
           →DP Problem 17
             ↳Polynomial Ordering
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pairs:

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


Rules:


active(minus(0, Y)) -> mark(0)
active(minus(s(X), s(Y))) -> mark(minus(X, Y))
active(geq(X, 0)) -> mark(true)
active(geq(0, s(Y))) -> mark(false)
active(geq(s(X), s(Y))) -> mark(geq(X, Y))
active(div(0, s(Y))) -> mark(0)
active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(s(X)) -> s(active(X))
active(div(X1, X2)) -> div(active(X1), X2)
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(minus(X1, X2)) -> minus(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(geq(X1, X2)) -> geq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(div(X1, X2)) -> div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) -> ok(minus(X1, X2))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

ACTIVE(div(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(s(x1))=  x1  
  POL(div(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 6
         ↳Polo
           →DP Problem 17
             ↳Polo
             ...
               →DP Problem 18
                 ↳Polynomial Ordering
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pair:

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


Rules:


active(minus(0, Y)) -> mark(0)
active(minus(s(X), s(Y))) -> mark(minus(X, Y))
active(geq(X, 0)) -> mark(true)
active(geq(0, s(Y))) -> mark(false)
active(geq(s(X), s(Y))) -> mark(geq(X, Y))
active(div(0, s(Y))) -> mark(0)
active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(s(X)) -> s(active(X))
active(div(X1, X2)) -> div(active(X1), X2)
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(minus(X1, X2)) -> minus(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(geq(X1, X2)) -> geq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(div(X1, X2)) -> div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) -> ok(minus(X1, X2))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
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(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 17
             ↳Polo
             ...
               →DP Problem 19
                 ↳Dependency Graph
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pair:


Rules:


active(minus(0, Y)) -> mark(0)
active(minus(s(X), s(Y))) -> mark(minus(X, Y))
active(geq(X, 0)) -> mark(true)
active(geq(0, s(Y))) -> mark(false)
active(geq(s(X), s(Y))) -> mark(geq(X, Y))
active(div(0, s(Y))) -> mark(0)
active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(s(X)) -> s(active(X))
active(div(X1, X2)) -> div(active(X1), X2)
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(minus(X1, X2)) -> minus(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(geq(X1, X2)) -> geq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(div(X1, X2)) -> div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) -> ok(minus(X1, X2))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
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(if(X1, X2, X3)) -> PROPER(X3)
PROPER(if(X1, X2, X3)) -> PROPER(X2)
PROPER(if(X1, X2, X3)) -> PROPER(X1)
PROPER(div(X1, X2)) -> PROPER(X2)
PROPER(div(X1, X2)) -> PROPER(X1)
PROPER(geq(X1, X2)) -> PROPER(X2)
PROPER(geq(X1, X2)) -> PROPER(X1)
PROPER(s(X)) -> PROPER(X)
PROPER(minus(X1, X2)) -> PROPER(X2)
PROPER(minus(X1, X2)) -> PROPER(X1)


Rules:


active(minus(0, Y)) -> mark(0)
active(minus(s(X), s(Y))) -> mark(minus(X, Y))
active(geq(X, 0)) -> mark(true)
active(geq(0, s(Y))) -> mark(false)
active(geq(s(X), s(Y))) -> mark(geq(X, Y))
active(div(0, s(Y))) -> mark(0)
active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(s(X)) -> s(active(X))
active(div(X1, X2)) -> div(active(X1), X2)
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(minus(X1, X2)) -> minus(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(geq(X1, X2)) -> geq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(div(X1, X2)) -> div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) -> ok(minus(X1, X2))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pairs can be strictly oriented:

PROPER(if(X1, X2, X3)) -> PROPER(X3)
PROPER(if(X1, X2, X3)) -> PROPER(X2)
PROPER(if(X1, X2, X3)) -> 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(if(x1, x2, x3))=  1 + x1 + x2 + x3  
  POL(geq(x1, x2))=  x1 + x2  
  POL(PROPER(x1))=  x1  
  POL(minus(x1, x2))=  x1 + x2  
  POL(s(x1))=  x1  
  POL(div(x1, x2))=  x1 + x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Polo
           →DP Problem 20
             ↳Polynomial Ordering
       →DP Problem 8
         ↳Remaining


Dependency Pairs:

PROPER(div(X1, X2)) -> PROPER(X2)
PROPER(div(X1, X2)) -> PROPER(X1)
PROPER(geq(X1, X2)) -> PROPER(X2)
PROPER(geq(X1, X2)) -> PROPER(X1)
PROPER(s(X)) -> PROPER(X)
PROPER(minus(X1, X2)) -> PROPER(X2)
PROPER(minus(X1, X2)) -> PROPER(X1)


Rules:


active(minus(0, Y)) -> mark(0)
active(minus(s(X), s(Y))) -> mark(minus(X, Y))
active(geq(X, 0)) -> mark(true)
active(geq(0, s(Y))) -> mark(false)
active(geq(s(X), s(Y))) -> mark(geq(X, Y))
active(div(0, s(Y))) -> mark(0)
active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(s(X)) -> s(active(X))
active(div(X1, X2)) -> div(active(X1), X2)
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(minus(X1, X2)) -> minus(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(geq(X1, X2)) -> geq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(div(X1, X2)) -> div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) -> ok(minus(X1, X2))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pairs can be strictly oriented:

PROPER(div(X1, X2)) -> PROPER(X2)
PROPER(div(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(geq(x1, x2))=  x1 + x2  
  POL(PROPER(x1))=  x1  
  POL(minus(x1, x2))=  x1 + x2  
  POL(s(x1))=  x1  
  POL(div(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 7
         ↳Polo
           →DP Problem 20
             ↳Polo
             ...
               →DP Problem 21
                 ↳Polynomial Ordering
       →DP Problem 8
         ↳Remaining


Dependency Pairs:

PROPER(geq(X1, X2)) -> PROPER(X2)
PROPER(geq(X1, X2)) -> PROPER(X1)
PROPER(s(X)) -> PROPER(X)
PROPER(minus(X1, X2)) -> PROPER(X2)
PROPER(minus(X1, X2)) -> PROPER(X1)


Rules:


active(minus(0, Y)) -> mark(0)
active(minus(s(X), s(Y))) -> mark(minus(X, Y))
active(geq(X, 0)) -> mark(true)
active(geq(0, s(Y))) -> mark(false)
active(geq(s(X), s(Y))) -> mark(geq(X, Y))
active(div(0, s(Y))) -> mark(0)
active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(s(X)) -> s(active(X))
active(div(X1, X2)) -> div(active(X1), X2)
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(minus(X1, X2)) -> minus(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(geq(X1, X2)) -> geq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(div(X1, X2)) -> div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) -> ok(minus(X1, X2))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pairs can be strictly oriented:

PROPER(geq(X1, X2)) -> PROPER(X2)
PROPER(geq(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(geq(x1, x2))=  1 + x1 + x2  
  POL(PROPER(x1))=  x1  
  POL(minus(x1, x2))=  x1 + x2  
  POL(s(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Polo
           →DP Problem 20
             ↳Polo
             ...
               →DP Problem 22
                 ↳Polynomial Ordering
       →DP Problem 8
         ↳Remaining


Dependency Pairs:

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


Rules:


active(minus(0, Y)) -> mark(0)
active(minus(s(X), s(Y))) -> mark(minus(X, Y))
active(geq(X, 0)) -> mark(true)
active(geq(0, s(Y))) -> mark(false)
active(geq(s(X), s(Y))) -> mark(geq(X, Y))
active(div(0, s(Y))) -> mark(0)
active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(s(X)) -> s(active(X))
active(div(X1, X2)) -> div(active(X1), X2)
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(minus(X1, X2)) -> minus(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(geq(X1, X2)) -> geq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(div(X1, X2)) -> div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) -> ok(minus(X1, X2))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
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(minus(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 7
         ↳Polo
           →DP Problem 20
             ↳Polo
             ...
               →DP Problem 23
                 ↳Polynomial Ordering
       →DP Problem 8
         ↳Remaining


Dependency Pairs:

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


Rules:


active(minus(0, Y)) -> mark(0)
active(minus(s(X), s(Y))) -> mark(minus(X, Y))
active(geq(X, 0)) -> mark(true)
active(geq(0, s(Y))) -> mark(false)
active(geq(s(X), s(Y))) -> mark(geq(X, Y))
active(div(0, s(Y))) -> mark(0)
active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(s(X)) -> s(active(X))
active(div(X1, X2)) -> div(active(X1), X2)
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(minus(X1, X2)) -> minus(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(geq(X1, X2)) -> geq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(div(X1, X2)) -> div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) -> ok(minus(X1, X2))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pairs can be strictly oriented:

PROPER(minus(X1, X2)) -> PROPER(X2)
PROPER(minus(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(minus(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 7
         ↳Polo
           →DP Problem 20
             ↳Polo
             ...
               →DP Problem 24
                 ↳Dependency Graph
       →DP Problem 8
         ↳Remaining


Dependency Pair:


Rules:


active(minus(0, Y)) -> mark(0)
active(minus(s(X), s(Y))) -> mark(minus(X, Y))
active(geq(X, 0)) -> mark(true)
active(geq(0, s(Y))) -> mark(false)
active(geq(s(X), s(Y))) -> mark(geq(X, Y))
active(div(0, s(Y))) -> mark(0)
active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(s(X)) -> s(active(X))
active(div(X1, X2)) -> div(active(X1), X2)
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(minus(X1, X2)) -> minus(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(geq(X1, X2)) -> geq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(div(X1, X2)) -> div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) -> ok(minus(X1, X2))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
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(minus(0, Y)) -> mark(0)
active(minus(s(X), s(Y))) -> mark(minus(X, Y))
active(geq(X, 0)) -> mark(true)
active(geq(0, s(Y))) -> mark(false)
active(geq(s(X), s(Y))) -> mark(geq(X, Y))
active(div(0, s(Y))) -> mark(0)
active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(s(X)) -> s(active(X))
active(div(X1, X2)) -> div(active(X1), X2)
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(minus(X1, X2)) -> minus(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(geq(X1, X2)) -> geq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(div(X1, X2)) -> div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) -> ok(minus(X1, X2))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))




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