Term Rewriting System R:
[X, Y, X1, X2, X3]
active(p(0)) -> mark(0)
active(p(s(X))) -> mark(X)
active(leq(0, Y)) -> mark(true)
active(leq(s(X), 0)) -> mark(false)
active(leq(s(X), s(Y))) -> mark(leq(X, Y))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(diff(X, Y)) -> mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) -> p(active(X))
active(s(X)) -> s(active(X))
active(leq(X1, X2)) -> leq(active(X1), X2)
active(leq(X1, X2)) -> leq(X1, active(X2))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(diff(X1, X2)) -> diff(active(X1), X2)
active(diff(X1, X2)) -> diff(X1, active(X2))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
leq(mark(X1), X2) -> mark(leq(X1, X2))
leq(X1, mark(X2)) -> mark(leq(X1, X2))
leq(ok(X1), ok(X2)) -> ok(leq(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
diff(mark(X1), X2) -> mark(diff(X1, X2))
diff(X1, mark(X2)) -> mark(diff(X1, X2))
diff(ok(X1), ok(X2)) -> ok(diff(X1, X2))
proper(p(X)) -> p(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(leq(X1, X2)) -> leq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) -> diff(proper(X1), proper(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(leq(s(X), s(Y))) -> LEQ(X, Y)
ACTIVE(diff(X, Y)) -> IF(leq(X, Y), 0, s(diff(p(X), Y)))
ACTIVE(diff(X, Y)) -> LEQ(X, Y)
ACTIVE(diff(X, Y)) -> S(diff(p(X), Y))
ACTIVE(diff(X, Y)) -> DIFF(p(X), Y)
ACTIVE(diff(X, Y)) -> P(X)
ACTIVE(p(X)) -> P(active(X))
ACTIVE(p(X)) -> ACTIVE(X)
ACTIVE(s(X)) -> S(active(X))
ACTIVE(s(X)) -> ACTIVE(X)
ACTIVE(leq(X1, X2)) -> LEQ(active(X1), X2)
ACTIVE(leq(X1, X2)) -> ACTIVE(X1)
ACTIVE(leq(X1, X2)) -> LEQ(X1, active(X2))
ACTIVE(leq(X1, X2)) -> ACTIVE(X2)
ACTIVE(if(X1, X2, X3)) -> IF(active(X1), X2, X3)
ACTIVE(if(X1, X2, X3)) -> ACTIVE(X1)
ACTIVE(diff(X1, X2)) -> DIFF(active(X1), X2)
ACTIVE(diff(X1, X2)) -> ACTIVE(X1)
ACTIVE(diff(X1, X2)) -> DIFF(X1, active(X2))
ACTIVE(diff(X1, X2)) -> ACTIVE(X2)
P(mark(X)) -> P(X)
P(ok(X)) -> P(X)
S(mark(X)) -> S(X)
S(ok(X)) -> S(X)
LEQ(mark(X1), X2) -> LEQ(X1, X2)
LEQ(X1, mark(X2)) -> LEQ(X1, X2)
LEQ(ok(X1), ok(X2)) -> LEQ(X1, X2)
IF(mark(X1), X2, X3) -> IF(X1, X2, X3)
IF(ok(X1), ok(X2), ok(X3)) -> IF(X1, X2, X3)
DIFF(mark(X1), X2) -> DIFF(X1, X2)
DIFF(X1, mark(X2)) -> DIFF(X1, X2)
DIFF(ok(X1), ok(X2)) -> DIFF(X1, X2)
PROPER(p(X)) -> P(proper(X))
PROPER(p(X)) -> PROPER(X)
PROPER(s(X)) -> S(proper(X))
PROPER(s(X)) -> PROPER(X)
PROPER(leq(X1, X2)) -> LEQ(proper(X1), proper(X2))
PROPER(leq(X1, X2)) -> PROPER(X1)
PROPER(leq(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)
PROPER(diff(X1, X2)) -> DIFF(proper(X1), proper(X2))
PROPER(diff(X1, X2)) -> PROPER(X1)
PROPER(diff(X1, X2)) -> PROPER(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 Pairs:

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


Rules:


active(p(0)) -> mark(0)
active(p(s(X))) -> mark(X)
active(leq(0, Y)) -> mark(true)
active(leq(s(X), 0)) -> mark(false)
active(leq(s(X), s(Y))) -> mark(leq(X, Y))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(diff(X, Y)) -> mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) -> p(active(X))
active(s(X)) -> s(active(X))
active(leq(X1, X2)) -> leq(active(X1), X2)
active(leq(X1, X2)) -> leq(X1, active(X2))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(diff(X1, X2)) -> diff(active(X1), X2)
active(diff(X1, X2)) -> diff(X1, active(X2))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
leq(mark(X1), X2) -> mark(leq(X1, X2))
leq(X1, mark(X2)) -> mark(leq(X1, X2))
leq(ok(X1), ok(X2)) -> ok(leq(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
diff(mark(X1), X2) -> mark(diff(X1, X2))
diff(X1, mark(X2)) -> mark(diff(X1, X2))
diff(ok(X1), ok(X2)) -> ok(diff(X1, X2))
proper(p(X)) -> p(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(leq(X1, X2)) -> leq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) -> diff(proper(X1), proper(X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

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

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 9
             ↳Polynomial Ordering
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pairs:

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


Rules:


active(p(0)) -> mark(0)
active(p(s(X))) -> mark(X)
active(leq(0, Y)) -> mark(true)
active(leq(s(X), 0)) -> mark(false)
active(leq(s(X), s(Y))) -> mark(leq(X, Y))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(diff(X, Y)) -> mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) -> p(active(X))
active(s(X)) -> s(active(X))
active(leq(X1, X2)) -> leq(active(X1), X2)
active(leq(X1, X2)) -> leq(X1, active(X2))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(diff(X1, X2)) -> diff(active(X1), X2)
active(diff(X1, X2)) -> diff(X1, active(X2))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
leq(mark(X1), X2) -> mark(leq(X1, X2))
leq(X1, mark(X2)) -> mark(leq(X1, X2))
leq(ok(X1), ok(X2)) -> ok(leq(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
diff(mark(X1), X2) -> mark(diff(X1, X2))
diff(X1, mark(X2)) -> mark(diff(X1, X2))
diff(ok(X1), ok(X2)) -> ok(diff(X1, X2))
proper(p(X)) -> p(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(leq(X1, X2)) -> leq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) -> diff(proper(X1), proper(X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

LEQ(mark(X1), X2) -> LEQ(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(LEQ(x1, x2))=  x1  
  POL(mark(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 9
             ↳Polo
             ...
               →DP Problem 10
                 ↳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:

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


Rules:


active(p(0)) -> mark(0)
active(p(s(X))) -> mark(X)
active(leq(0, Y)) -> mark(true)
active(leq(s(X), 0)) -> mark(false)
active(leq(s(X), s(Y))) -> mark(leq(X, Y))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(diff(X, Y)) -> mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) -> p(active(X))
active(s(X)) -> s(active(X))
active(leq(X1, X2)) -> leq(active(X1), X2)
active(leq(X1, X2)) -> leq(X1, active(X2))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(diff(X1, X2)) -> diff(active(X1), X2)
active(diff(X1, X2)) -> diff(X1, active(X2))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
leq(mark(X1), X2) -> mark(leq(X1, X2))
leq(X1, mark(X2)) -> mark(leq(X1, X2))
leq(ok(X1), ok(X2)) -> ok(leq(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
diff(mark(X1), X2) -> mark(diff(X1, X2))
diff(X1, mark(X2)) -> mark(diff(X1, X2))
diff(ok(X1), ok(X2)) -> ok(diff(X1, X2))
proper(p(X)) -> p(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(leq(X1, X2)) -> leq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) -> diff(proper(X1), proper(X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

LEQ(X1, mark(X2)) -> LEQ(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(LEQ(x1, x2))=  x2  
  POL(mark(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 9
             ↳Polo
             ...
               →DP Problem 11
                 ↳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(p(0)) -> mark(0)
active(p(s(X))) -> mark(X)
active(leq(0, Y)) -> mark(true)
active(leq(s(X), 0)) -> mark(false)
active(leq(s(X), s(Y))) -> mark(leq(X, Y))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(diff(X, Y)) -> mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) -> p(active(X))
active(s(X)) -> s(active(X))
active(leq(X1, X2)) -> leq(active(X1), X2)
active(leq(X1, X2)) -> leq(X1, active(X2))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(diff(X1, X2)) -> diff(active(X1), X2)
active(diff(X1, X2)) -> diff(X1, active(X2))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
leq(mark(X1), X2) -> mark(leq(X1, X2))
leq(X1, mark(X2)) -> mark(leq(X1, X2))
leq(ok(X1), ok(X2)) -> ok(leq(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
diff(mark(X1), X2) -> mark(diff(X1, X2))
diff(X1, mark(X2)) -> mark(diff(X1, X2))
diff(ok(X1), ok(X2)) -> ok(diff(X1, X2))
proper(p(X)) -> p(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(leq(X1, X2)) -> leq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) -> diff(proper(X1), proper(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 Pairs:

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


Rules:


active(p(0)) -> mark(0)
active(p(s(X))) -> mark(X)
active(leq(0, Y)) -> mark(true)
active(leq(s(X), 0)) -> mark(false)
active(leq(s(X), s(Y))) -> mark(leq(X, Y))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(diff(X, Y)) -> mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) -> p(active(X))
active(s(X)) -> s(active(X))
active(leq(X1, X2)) -> leq(active(X1), X2)
active(leq(X1, X2)) -> leq(X1, active(X2))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(diff(X1, X2)) -> diff(active(X1), X2)
active(diff(X1, X2)) -> diff(X1, active(X2))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
leq(mark(X1), X2) -> mark(leq(X1, X2))
leq(X1, mark(X2)) -> mark(leq(X1, X2))
leq(ok(X1), ok(X2)) -> ok(leq(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
diff(mark(X1), X2) -> mark(diff(X1, X2))
diff(X1, mark(X2)) -> mark(diff(X1, X2))
diff(ok(X1), ok(X2)) -> ok(diff(X1, X2))
proper(p(X)) -> p(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(leq(X1, X2)) -> leq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) -> diff(proper(X1), proper(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 12
             ↳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:

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


Rules:


active(p(0)) -> mark(0)
active(p(s(X))) -> mark(X)
active(leq(0, Y)) -> mark(true)
active(leq(s(X), 0)) -> mark(false)
active(leq(s(X), s(Y))) -> mark(leq(X, Y))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(diff(X, Y)) -> mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) -> p(active(X))
active(s(X)) -> s(active(X))
active(leq(X1, X2)) -> leq(active(X1), X2)
active(leq(X1, X2)) -> leq(X1, active(X2))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(diff(X1, X2)) -> diff(active(X1), X2)
active(diff(X1, X2)) -> diff(X1, active(X2))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
leq(mark(X1), X2) -> mark(leq(X1, X2))
leq(X1, mark(X2)) -> mark(leq(X1, X2))
leq(ok(X1), ok(X2)) -> ok(leq(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
diff(mark(X1), X2) -> mark(diff(X1, X2))
diff(X1, mark(X2)) -> mark(diff(X1, X2))
diff(ok(X1), ok(X2)) -> ok(diff(X1, X2))
proper(p(X)) -> p(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(leq(X1, X2)) -> leq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) -> diff(proper(X1), proper(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 12
             ↳Polo
             ...
               →DP Problem 13
                 ↳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(p(0)) -> mark(0)
active(p(s(X))) -> mark(X)
active(leq(0, Y)) -> mark(true)
active(leq(s(X), 0)) -> mark(false)
active(leq(s(X), s(Y))) -> mark(leq(X, Y))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(diff(X, Y)) -> mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) -> p(active(X))
active(s(X)) -> s(active(X))
active(leq(X1, X2)) -> leq(active(X1), X2)
active(leq(X1, X2)) -> leq(X1, active(X2))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(diff(X1, X2)) -> diff(active(X1), X2)
active(diff(X1, X2)) -> diff(X1, active(X2))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
leq(mark(X1), X2) -> mark(leq(X1, X2))
leq(X1, mark(X2)) -> mark(leq(X1, X2))
leq(ok(X1), ok(X2)) -> ok(leq(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
diff(mark(X1), X2) -> mark(diff(X1, X2))
diff(X1, mark(X2)) -> mark(diff(X1, X2))
diff(ok(X1), ok(X2)) -> ok(diff(X1, X2))
proper(p(X)) -> p(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(leq(X1, X2)) -> leq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) -> diff(proper(X1), proper(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:

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


Rules:


active(p(0)) -> mark(0)
active(p(s(X))) -> mark(X)
active(leq(0, Y)) -> mark(true)
active(leq(s(X), 0)) -> mark(false)
active(leq(s(X), s(Y))) -> mark(leq(X, Y))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(diff(X, Y)) -> mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) -> p(active(X))
active(s(X)) -> s(active(X))
active(leq(X1, X2)) -> leq(active(X1), X2)
active(leq(X1, X2)) -> leq(X1, active(X2))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(diff(X1, X2)) -> diff(active(X1), X2)
active(diff(X1, X2)) -> diff(X1, active(X2))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
leq(mark(X1), X2) -> mark(leq(X1, X2))
leq(X1, mark(X2)) -> mark(leq(X1, X2))
leq(ok(X1), ok(X2)) -> ok(leq(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
diff(mark(X1), X2) -> mark(diff(X1, X2))
diff(X1, mark(X2)) -> mark(diff(X1, X2))
diff(ok(X1), ok(X2)) -> ok(diff(X1, X2))
proper(p(X)) -> p(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(leq(X1, X2)) -> leq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) -> diff(proper(X1), proper(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 14
             ↳Polynomial Ordering
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pair:

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


Rules:


active(p(0)) -> mark(0)
active(p(s(X))) -> mark(X)
active(leq(0, Y)) -> mark(true)
active(leq(s(X), 0)) -> mark(false)
active(leq(s(X), s(Y))) -> mark(leq(X, Y))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(diff(X, Y)) -> mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) -> p(active(X))
active(s(X)) -> s(active(X))
active(leq(X1, X2)) -> leq(active(X1), X2)
active(leq(X1, X2)) -> leq(X1, active(X2))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(diff(X1, X2)) -> diff(active(X1), X2)
active(diff(X1, X2)) -> diff(X1, active(X2))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
leq(mark(X1), X2) -> mark(leq(X1, X2))
leq(X1, mark(X2)) -> mark(leq(X1, X2))
leq(ok(X1), ok(X2)) -> ok(leq(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
diff(mark(X1), X2) -> mark(diff(X1, X2))
diff(X1, mark(X2)) -> mark(diff(X1, X2))
diff(ok(X1), ok(X2)) -> ok(diff(X1, X2))
proper(p(X)) -> p(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(leq(X1, X2)) -> leq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) -> diff(proper(X1), proper(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 14
             ↳Polo
             ...
               →DP Problem 15
                 ↳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(p(0)) -> mark(0)
active(p(s(X))) -> mark(X)
active(leq(0, Y)) -> mark(true)
active(leq(s(X), 0)) -> mark(false)
active(leq(s(X), s(Y))) -> mark(leq(X, Y))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(diff(X, Y)) -> mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) -> p(active(X))
active(s(X)) -> s(active(X))
active(leq(X1, X2)) -> leq(active(X1), X2)
active(leq(X1, X2)) -> leq(X1, active(X2))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(diff(X1, X2)) -> diff(active(X1), X2)
active(diff(X1, X2)) -> diff(X1, active(X2))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
leq(mark(X1), X2) -> mark(leq(X1, X2))
leq(X1, mark(X2)) -> mark(leq(X1, X2))
leq(ok(X1), ok(X2)) -> ok(leq(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
diff(mark(X1), X2) -> mark(diff(X1, X2))
diff(X1, mark(X2)) -> mark(diff(X1, X2))
diff(ok(X1), ok(X2)) -> ok(diff(X1, X2))
proper(p(X)) -> p(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(leq(X1, X2)) -> leq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) -> diff(proper(X1), proper(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:

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


Rules:


active(p(0)) -> mark(0)
active(p(s(X))) -> mark(X)
active(leq(0, Y)) -> mark(true)
active(leq(s(X), 0)) -> mark(false)
active(leq(s(X), s(Y))) -> mark(leq(X, Y))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(diff(X, Y)) -> mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) -> p(active(X))
active(s(X)) -> s(active(X))
active(leq(X1, X2)) -> leq(active(X1), X2)
active(leq(X1, X2)) -> leq(X1, active(X2))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(diff(X1, X2)) -> diff(active(X1), X2)
active(diff(X1, X2)) -> diff(X1, active(X2))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
leq(mark(X1), X2) -> mark(leq(X1, X2))
leq(X1, mark(X2)) -> mark(leq(X1, X2))
leq(ok(X1), ok(X2)) -> ok(leq(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
diff(mark(X1), X2) -> mark(diff(X1, X2))
diff(X1, mark(X2)) -> mark(diff(X1, X2))
diff(ok(X1), ok(X2)) -> ok(diff(X1, X2))
proper(p(X)) -> p(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(leq(X1, X2)) -> leq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) -> diff(proper(X1), proper(X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

DIFF(ok(X1), ok(X2)) -> DIFF(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(DIFF(x1, x2))=  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 16
             ↳Polynomial Ordering
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pairs:

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


Rules:


active(p(0)) -> mark(0)
active(p(s(X))) -> mark(X)
active(leq(0, Y)) -> mark(true)
active(leq(s(X), 0)) -> mark(false)
active(leq(s(X), s(Y))) -> mark(leq(X, Y))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(diff(X, Y)) -> mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) -> p(active(X))
active(s(X)) -> s(active(X))
active(leq(X1, X2)) -> leq(active(X1), X2)
active(leq(X1, X2)) -> leq(X1, active(X2))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(diff(X1, X2)) -> diff(active(X1), X2)
active(diff(X1, X2)) -> diff(X1, active(X2))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
leq(mark(X1), X2) -> mark(leq(X1, X2))
leq(X1, mark(X2)) -> mark(leq(X1, X2))
leq(ok(X1), ok(X2)) -> ok(leq(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
diff(mark(X1), X2) -> mark(diff(X1, X2))
diff(X1, mark(X2)) -> mark(diff(X1, X2))
diff(ok(X1), ok(X2)) -> ok(diff(X1, X2))
proper(p(X)) -> p(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(leq(X1, X2)) -> leq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) -> diff(proper(X1), proper(X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

DIFF(mark(X1), X2) -> DIFF(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(DIFF(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 16
             ↳Polo
             ...
               →DP Problem 17
                 ↳Polynomial Ordering
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pair:

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


Rules:


active(p(0)) -> mark(0)
active(p(s(X))) -> mark(X)
active(leq(0, Y)) -> mark(true)
active(leq(s(X), 0)) -> mark(false)
active(leq(s(X), s(Y))) -> mark(leq(X, Y))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(diff(X, Y)) -> mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) -> p(active(X))
active(s(X)) -> s(active(X))
active(leq(X1, X2)) -> leq(active(X1), X2)
active(leq(X1, X2)) -> leq(X1, active(X2))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(diff(X1, X2)) -> diff(active(X1), X2)
active(diff(X1, X2)) -> diff(X1, active(X2))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
leq(mark(X1), X2) -> mark(leq(X1, X2))
leq(X1, mark(X2)) -> mark(leq(X1, X2))
leq(ok(X1), ok(X2)) -> ok(leq(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
diff(mark(X1), X2) -> mark(diff(X1, X2))
diff(X1, mark(X2)) -> mark(diff(X1, X2))
diff(ok(X1), ok(X2)) -> ok(diff(X1, X2))
proper(p(X)) -> p(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(leq(X1, X2)) -> leq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) -> diff(proper(X1), proper(X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

DIFF(X1, mark(X2)) -> DIFF(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(DIFF(x1, x2))=  x2  
  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 16
             ↳Polo
             ...
               →DP Problem 18
                 ↳Dependency Graph
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pair:


Rules:


active(p(0)) -> mark(0)
active(p(s(X))) -> mark(X)
active(leq(0, Y)) -> mark(true)
active(leq(s(X), 0)) -> mark(false)
active(leq(s(X), s(Y))) -> mark(leq(X, Y))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(diff(X, Y)) -> mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) -> p(active(X))
active(s(X)) -> s(active(X))
active(leq(X1, X2)) -> leq(active(X1), X2)
active(leq(X1, X2)) -> leq(X1, active(X2))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(diff(X1, X2)) -> diff(active(X1), X2)
active(diff(X1, X2)) -> diff(X1, active(X2))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
leq(mark(X1), X2) -> mark(leq(X1, X2))
leq(X1, mark(X2)) -> mark(leq(X1, X2))
leq(ok(X1), ok(X2)) -> ok(leq(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
diff(mark(X1), X2) -> mark(diff(X1, X2))
diff(X1, mark(X2)) -> mark(diff(X1, X2))
diff(ok(X1), ok(X2)) -> ok(diff(X1, X2))
proper(p(X)) -> p(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(leq(X1, X2)) -> leq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) -> diff(proper(X1), proper(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:

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


Rules:


active(p(0)) -> mark(0)
active(p(s(X))) -> mark(X)
active(leq(0, Y)) -> mark(true)
active(leq(s(X), 0)) -> mark(false)
active(leq(s(X), s(Y))) -> mark(leq(X, Y))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(diff(X, Y)) -> mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) -> p(active(X))
active(s(X)) -> s(active(X))
active(leq(X1, X2)) -> leq(active(X1), X2)
active(leq(X1, X2)) -> leq(X1, active(X2))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(diff(X1, X2)) -> diff(active(X1), X2)
active(diff(X1, X2)) -> diff(X1, active(X2))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
leq(mark(X1), X2) -> mark(leq(X1, X2))
leq(X1, mark(X2)) -> mark(leq(X1, X2))
leq(ok(X1), ok(X2)) -> ok(leq(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
diff(mark(X1), X2) -> mark(diff(X1, X2))
diff(X1, mark(X2)) -> mark(diff(X1, X2))
diff(ok(X1), ok(X2)) -> ok(diff(X1, X2))
proper(p(X)) -> p(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(leq(X1, X2)) -> leq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) -> diff(proper(X1), proper(X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

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


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

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

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
           →DP Problem 19
             ↳Polynomial Ordering
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pair:

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


Rules:


active(p(0)) -> mark(0)
active(p(s(X))) -> mark(X)
active(leq(0, Y)) -> mark(true)
active(leq(s(X), 0)) -> mark(false)
active(leq(s(X), s(Y))) -> mark(leq(X, Y))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(diff(X, Y)) -> mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) -> p(active(X))
active(s(X)) -> s(active(X))
active(leq(X1, X2)) -> leq(active(X1), X2)
active(leq(X1, X2)) -> leq(X1, active(X2))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(diff(X1, X2)) -> diff(active(X1), X2)
active(diff(X1, X2)) -> diff(X1, active(X2))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
leq(mark(X1), X2) -> mark(leq(X1, X2))
leq(X1, mark(X2)) -> mark(leq(X1, X2))
leq(ok(X1), ok(X2)) -> ok(leq(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
diff(mark(X1), X2) -> mark(diff(X1, X2))
diff(X1, mark(X2)) -> mark(diff(X1, X2))
diff(ok(X1), ok(X2)) -> ok(diff(X1, X2))
proper(p(X)) -> p(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(leq(X1, X2)) -> leq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) -> diff(proper(X1), proper(X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

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


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

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

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
           →DP Problem 19
             ↳Polo
             ...
               →DP Problem 20
                 ↳Dependency Graph
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pair:


Rules:


active(p(0)) -> mark(0)
active(p(s(X))) -> mark(X)
active(leq(0, Y)) -> mark(true)
active(leq(s(X), 0)) -> mark(false)
active(leq(s(X), s(Y))) -> mark(leq(X, Y))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(diff(X, Y)) -> mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) -> p(active(X))
active(s(X)) -> s(active(X))
active(leq(X1, X2)) -> leq(active(X1), X2)
active(leq(X1, X2)) -> leq(X1, active(X2))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(diff(X1, X2)) -> diff(active(X1), X2)
active(diff(X1, X2)) -> diff(X1, active(X2))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
leq(mark(X1), X2) -> mark(leq(X1, X2))
leq(X1, mark(X2)) -> mark(leq(X1, X2))
leq(ok(X1), ok(X2)) -> ok(leq(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
diff(mark(X1), X2) -> mark(diff(X1, X2))
diff(X1, mark(X2)) -> mark(diff(X1, X2))
diff(ok(X1), ok(X2)) -> ok(diff(X1, X2))
proper(p(X)) -> p(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(leq(X1, X2)) -> leq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) -> diff(proper(X1), proper(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(diff(X1, X2)) -> ACTIVE(X2)
ACTIVE(diff(X1, X2)) -> ACTIVE(X1)
ACTIVE(if(X1, X2, X3)) -> ACTIVE(X1)
ACTIVE(leq(X1, X2)) -> ACTIVE(X2)
ACTIVE(leq(X1, X2)) -> ACTIVE(X1)
ACTIVE(s(X)) -> ACTIVE(X)
ACTIVE(p(X)) -> ACTIVE(X)


Rules:


active(p(0)) -> mark(0)
active(p(s(X))) -> mark(X)
active(leq(0, Y)) -> mark(true)
active(leq(s(X), 0)) -> mark(false)
active(leq(s(X), s(Y))) -> mark(leq(X, Y))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(diff(X, Y)) -> mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) -> p(active(X))
active(s(X)) -> s(active(X))
active(leq(X1, X2)) -> leq(active(X1), X2)
active(leq(X1, X2)) -> leq(X1, active(X2))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(diff(X1, X2)) -> diff(active(X1), X2)
active(diff(X1, X2)) -> diff(X1, active(X2))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
leq(mark(X1), X2) -> mark(leq(X1, X2))
leq(X1, mark(X2)) -> mark(leq(X1, X2))
leq(ok(X1), ok(X2)) -> ok(leq(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
diff(mark(X1), X2) -> mark(diff(X1, X2))
diff(X1, mark(X2)) -> mark(diff(X1, X2))
diff(ok(X1), ok(X2)) -> ok(diff(X1, X2))
proper(p(X)) -> p(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(leq(X1, X2)) -> leq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) -> diff(proper(X1), proper(X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pairs can be strictly oriented:

ACTIVE(leq(X1, X2)) -> ACTIVE(X2)
ACTIVE(leq(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(if(x1, x2, x3))=  x1  
  POL(leq(x1, x2))=  1 + x1 + x2  
  POL(s(x1))=  x1  
  POL(diff(x1, x2))=  x1 + x2  
  POL(p(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
           →DP Problem 21
             ↳Polynomial Ordering
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pairs:

ACTIVE(diff(X1, X2)) -> ACTIVE(X2)
ACTIVE(diff(X1, X2)) -> ACTIVE(X1)
ACTIVE(if(X1, X2, X3)) -> ACTIVE(X1)
ACTIVE(s(X)) -> ACTIVE(X)
ACTIVE(p(X)) -> ACTIVE(X)


Rules:


active(p(0)) -> mark(0)
active(p(s(X))) -> mark(X)
active(leq(0, Y)) -> mark(true)
active(leq(s(X), 0)) -> mark(false)
active(leq(s(X), s(Y))) -> mark(leq(X, Y))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(diff(X, Y)) -> mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) -> p(active(X))
active(s(X)) -> s(active(X))
active(leq(X1, X2)) -> leq(active(X1), X2)
active(leq(X1, X2)) -> leq(X1, active(X2))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(diff(X1, X2)) -> diff(active(X1), X2)
active(diff(X1, X2)) -> diff(X1, active(X2))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
leq(mark(X1), X2) -> mark(leq(X1, X2))
leq(X1, mark(X2)) -> mark(leq(X1, X2))
leq(ok(X1), ok(X2)) -> ok(leq(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
diff(mark(X1), X2) -> mark(diff(X1, X2))
diff(X1, mark(X2)) -> mark(diff(X1, X2))
diff(ok(X1), ok(X2)) -> ok(diff(X1, X2))
proper(p(X)) -> p(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(leq(X1, X2)) -> leq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) -> diff(proper(X1), proper(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(if(x1, x2, x3))=  x1  
  POL(s(x1))=  1 + x1  
  POL(diff(x1, x2))=  x1 + x2  
  POL(p(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
           →DP Problem 21
             ↳Polo
             ...
               →DP Problem 22
                 ↳Polynomial Ordering
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pairs:

ACTIVE(diff(X1, X2)) -> ACTIVE(X2)
ACTIVE(diff(X1, X2)) -> ACTIVE(X1)
ACTIVE(if(X1, X2, X3)) -> ACTIVE(X1)
ACTIVE(p(X)) -> ACTIVE(X)


Rules:


active(p(0)) -> mark(0)
active(p(s(X))) -> mark(X)
active(leq(0, Y)) -> mark(true)
active(leq(s(X), 0)) -> mark(false)
active(leq(s(X), s(Y))) -> mark(leq(X, Y))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(diff(X, Y)) -> mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) -> p(active(X))
active(s(X)) -> s(active(X))
active(leq(X1, X2)) -> leq(active(X1), X2)
active(leq(X1, X2)) -> leq(X1, active(X2))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(diff(X1, X2)) -> diff(active(X1), X2)
active(diff(X1, X2)) -> diff(X1, active(X2))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
leq(mark(X1), X2) -> mark(leq(X1, X2))
leq(X1, mark(X2)) -> mark(leq(X1, X2))
leq(ok(X1), ok(X2)) -> ok(leq(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
diff(mark(X1), X2) -> mark(diff(X1, X2))
diff(X1, mark(X2)) -> mark(diff(X1, X2))
diff(ok(X1), ok(X2)) -> ok(diff(X1, X2))
proper(p(X)) -> p(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(leq(X1, X2)) -> leq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) -> diff(proper(X1), proper(X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

ACTIVE(p(X)) -> ACTIVE(X)


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(ACTIVE(x1))=  x1  
  POL(if(x1, x2, x3))=  x1  
  POL(diff(x1, x2))=  x1 + x2  
  POL(p(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
           →DP Problem 21
             ↳Polo
             ...
               →DP Problem 23
                 ↳Polynomial Ordering
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pairs:

ACTIVE(diff(X1, X2)) -> ACTIVE(X2)
ACTIVE(diff(X1, X2)) -> ACTIVE(X1)
ACTIVE(if(X1, X2, X3)) -> ACTIVE(X1)


Rules:


active(p(0)) -> mark(0)
active(p(s(X))) -> mark(X)
active(leq(0, Y)) -> mark(true)
active(leq(s(X), 0)) -> mark(false)
active(leq(s(X), s(Y))) -> mark(leq(X, Y))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(diff(X, Y)) -> mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) -> p(active(X))
active(s(X)) -> s(active(X))
active(leq(X1, X2)) -> leq(active(X1), X2)
active(leq(X1, X2)) -> leq(X1, active(X2))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(diff(X1, X2)) -> diff(active(X1), X2)
active(diff(X1, X2)) -> diff(X1, active(X2))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
leq(mark(X1), X2) -> mark(leq(X1, X2))
leq(X1, mark(X2)) -> mark(leq(X1, X2))
leq(ok(X1), ok(X2)) -> ok(leq(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
diff(mark(X1), X2) -> mark(diff(X1, X2))
diff(X1, mark(X2)) -> mark(diff(X1, X2))
diff(ok(X1), ok(X2)) -> ok(diff(X1, X2))
proper(p(X)) -> p(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(leq(X1, X2)) -> leq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) -> diff(proper(X1), proper(X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pairs can be strictly oriented:

ACTIVE(diff(X1, X2)) -> ACTIVE(X2)
ACTIVE(diff(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(if(x1, x2, x3))=  x1  
  POL(diff(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 21
             ↳Polo
             ...
               →DP Problem 24
                 ↳Polynomial Ordering
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pair:

ACTIVE(if(X1, X2, X3)) -> ACTIVE(X1)


Rules:


active(p(0)) -> mark(0)
active(p(s(X))) -> mark(X)
active(leq(0, Y)) -> mark(true)
active(leq(s(X), 0)) -> mark(false)
active(leq(s(X), s(Y))) -> mark(leq(X, Y))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(diff(X, Y)) -> mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) -> p(active(X))
active(s(X)) -> s(active(X))
active(leq(X1, X2)) -> leq(active(X1), X2)
active(leq(X1, X2)) -> leq(X1, active(X2))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(diff(X1, X2)) -> diff(active(X1), X2)
active(diff(X1, X2)) -> diff(X1, active(X2))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
leq(mark(X1), X2) -> mark(leq(X1, X2))
leq(X1, mark(X2)) -> mark(leq(X1, X2))
leq(ok(X1), ok(X2)) -> ok(leq(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
diff(mark(X1), X2) -> mark(diff(X1, X2))
diff(X1, mark(X2)) -> mark(diff(X1, X2))
diff(ok(X1), ok(X2)) -> ok(diff(X1, X2))
proper(p(X)) -> p(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(leq(X1, X2)) -> leq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) -> diff(proper(X1), proper(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  

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 21
             ↳Polo
             ...
               →DP Problem 25
                 ↳Dependency Graph
       →DP Problem 7
         ↳Polo
       →DP Problem 8
         ↳Remaining


Dependency Pair:


Rules:


active(p(0)) -> mark(0)
active(p(s(X))) -> mark(X)
active(leq(0, Y)) -> mark(true)
active(leq(s(X), 0)) -> mark(false)
active(leq(s(X), s(Y))) -> mark(leq(X, Y))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(diff(X, Y)) -> mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) -> p(active(X))
active(s(X)) -> s(active(X))
active(leq(X1, X2)) -> leq(active(X1), X2)
active(leq(X1, X2)) -> leq(X1, active(X2))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(diff(X1, X2)) -> diff(active(X1), X2)
active(diff(X1, X2)) -> diff(X1, active(X2))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
leq(mark(X1), X2) -> mark(leq(X1, X2))
leq(X1, mark(X2)) -> mark(leq(X1, X2))
leq(ok(X1), ok(X2)) -> ok(leq(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
diff(mark(X1), X2) -> mark(diff(X1, X2))
diff(X1, mark(X2)) -> mark(diff(X1, X2))
diff(ok(X1), ok(X2)) -> ok(diff(X1, X2))
proper(p(X)) -> p(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(leq(X1, X2)) -> leq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) -> diff(proper(X1), proper(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(diff(X1, X2)) -> PROPER(X2)
PROPER(diff(X1, X2)) -> PROPER(X1)
PROPER(if(X1, X2, X3)) -> PROPER(X3)
PROPER(if(X1, X2, X3)) -> PROPER(X2)
PROPER(if(X1, X2, X3)) -> PROPER(X1)
PROPER(leq(X1, X2)) -> PROPER(X2)
PROPER(leq(X1, X2)) -> PROPER(X1)
PROPER(s(X)) -> PROPER(X)
PROPER(p(X)) -> PROPER(X)


Rules:


active(p(0)) -> mark(0)
active(p(s(X))) -> mark(X)
active(leq(0, Y)) -> mark(true)
active(leq(s(X), 0)) -> mark(false)
active(leq(s(X), s(Y))) -> mark(leq(X, Y))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(diff(X, Y)) -> mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) -> p(active(X))
active(s(X)) -> s(active(X))
active(leq(X1, X2)) -> leq(active(X1), X2)
active(leq(X1, X2)) -> leq(X1, active(X2))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(diff(X1, X2)) -> diff(active(X1), X2)
active(diff(X1, X2)) -> diff(X1, active(X2))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
leq(mark(X1), X2) -> mark(leq(X1, X2))
leq(X1, mark(X2)) -> mark(leq(X1, X2))
leq(ok(X1), ok(X2)) -> ok(leq(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
diff(mark(X1), X2) -> mark(diff(X1, X2))
diff(X1, mark(X2)) -> mark(diff(X1, X2))
diff(ok(X1), ok(X2)) -> ok(diff(X1, X2))
proper(p(X)) -> p(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(leq(X1, X2)) -> leq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) -> diff(proper(X1), proper(X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pairs can be strictly oriented:

PROPER(leq(X1, X2)) -> PROPER(X2)
PROPER(leq(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(if(x1, x2, x3))=  x1 + x2 + x3  
  POL(PROPER(x1))=  x1  
  POL(leq(x1, x2))=  1 + x1 + x2  
  POL(s(x1))=  x1  
  POL(diff(x1, x2))=  x1 + x2  
  POL(p(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Polo
           →DP Problem 26
             ↳Polynomial Ordering
       →DP Problem 8
         ↳Remaining


Dependency Pairs:

PROPER(diff(X1, X2)) -> PROPER(X2)
PROPER(diff(X1, X2)) -> PROPER(X1)
PROPER(if(X1, X2, X3)) -> PROPER(X3)
PROPER(if(X1, X2, X3)) -> PROPER(X2)
PROPER(if(X1, X2, X3)) -> PROPER(X1)
PROPER(s(X)) -> PROPER(X)
PROPER(p(X)) -> PROPER(X)


Rules:


active(p(0)) -> mark(0)
active(p(s(X))) -> mark(X)
active(leq(0, Y)) -> mark(true)
active(leq(s(X), 0)) -> mark(false)
active(leq(s(X), s(Y))) -> mark(leq(X, Y))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(diff(X, Y)) -> mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) -> p(active(X))
active(s(X)) -> s(active(X))
active(leq(X1, X2)) -> leq(active(X1), X2)
active(leq(X1, X2)) -> leq(X1, active(X2))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(diff(X1, X2)) -> diff(active(X1), X2)
active(diff(X1, X2)) -> diff(X1, active(X2))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
leq(mark(X1), X2) -> mark(leq(X1, X2))
leq(X1, mark(X2)) -> mark(leq(X1, X2))
leq(ok(X1), ok(X2)) -> ok(leq(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
diff(mark(X1), X2) -> mark(diff(X1, X2))
diff(X1, mark(X2)) -> mark(diff(X1, X2))
diff(ok(X1), ok(X2)) -> ok(diff(X1, X2))
proper(p(X)) -> p(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(leq(X1, X2)) -> leq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) -> diff(proper(X1), proper(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(if(x1, x2, x3))=  x1 + x2 + x3  
  POL(PROPER(x1))=  x1  
  POL(s(x1))=  1 + x1  
  POL(diff(x1, x2))=  x1 + x2  
  POL(p(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Polo
           →DP Problem 26
             ↳Polo
             ...
               →DP Problem 27
                 ↳Polynomial Ordering
       →DP Problem 8
         ↳Remaining


Dependency Pairs:

PROPER(diff(X1, X2)) -> PROPER(X2)
PROPER(diff(X1, X2)) -> PROPER(X1)
PROPER(if(X1, X2, X3)) -> PROPER(X3)
PROPER(if(X1, X2, X3)) -> PROPER(X2)
PROPER(if(X1, X2, X3)) -> PROPER(X1)
PROPER(p(X)) -> PROPER(X)


Rules:


active(p(0)) -> mark(0)
active(p(s(X))) -> mark(X)
active(leq(0, Y)) -> mark(true)
active(leq(s(X), 0)) -> mark(false)
active(leq(s(X), s(Y))) -> mark(leq(X, Y))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(diff(X, Y)) -> mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) -> p(active(X))
active(s(X)) -> s(active(X))
active(leq(X1, X2)) -> leq(active(X1), X2)
active(leq(X1, X2)) -> leq(X1, active(X2))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(diff(X1, X2)) -> diff(active(X1), X2)
active(diff(X1, X2)) -> diff(X1, active(X2))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
leq(mark(X1), X2) -> mark(leq(X1, X2))
leq(X1, mark(X2)) -> mark(leq(X1, X2))
leq(ok(X1), ok(X2)) -> ok(leq(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
diff(mark(X1), X2) -> mark(diff(X1, X2))
diff(X1, mark(X2)) -> mark(diff(X1, X2))
diff(ok(X1), ok(X2)) -> ok(diff(X1, X2))
proper(p(X)) -> p(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(leq(X1, X2)) -> leq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) -> diff(proper(X1), proper(X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

PROPER(p(X)) -> PROPER(X)


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(if(x1, x2, x3))=  x1 + x2 + x3  
  POL(PROPER(x1))=  x1  
  POL(diff(x1, x2))=  x1 + x2  
  POL(p(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Polo
       →DP Problem 6
         ↳Polo
       →DP Problem 7
         ↳Polo
           →DP Problem 26
             ↳Polo
             ...
               →DP Problem 28
                 ↳Polynomial Ordering
       →DP Problem 8
         ↳Remaining


Dependency Pairs:

PROPER(diff(X1, X2)) -> PROPER(X2)
PROPER(diff(X1, X2)) -> PROPER(X1)
PROPER(if(X1, X2, X3)) -> PROPER(X3)
PROPER(if(X1, X2, X3)) -> PROPER(X2)
PROPER(if(X1, X2, X3)) -> PROPER(X1)


Rules:


active(p(0)) -> mark(0)
active(p(s(X))) -> mark(X)
active(leq(0, Y)) -> mark(true)
active(leq(s(X), 0)) -> mark(false)
active(leq(s(X), s(Y))) -> mark(leq(X, Y))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(diff(X, Y)) -> mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) -> p(active(X))
active(s(X)) -> s(active(X))
active(leq(X1, X2)) -> leq(active(X1), X2)
active(leq(X1, X2)) -> leq(X1, active(X2))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(diff(X1, X2)) -> diff(active(X1), X2)
active(diff(X1, X2)) -> diff(X1, active(X2))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
leq(mark(X1), X2) -> mark(leq(X1, X2))
leq(X1, mark(X2)) -> mark(leq(X1, X2))
leq(ok(X1), ok(X2)) -> ok(leq(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
diff(mark(X1), X2) -> mark(diff(X1, X2))
diff(X1, mark(X2)) -> mark(diff(X1, X2))
diff(ok(X1), ok(X2)) -> ok(diff(X1, X2))
proper(p(X)) -> p(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(leq(X1, X2)) -> leq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) -> diff(proper(X1), proper(X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pairs can be strictly oriented:

PROPER(diff(X1, X2)) -> PROPER(X2)
PROPER(diff(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(if(x1, x2, x3))=  x1 + x2 + x3  
  POL(PROPER(x1))=  x1  
  POL(diff(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 26
             ↳Polo
             ...
               →DP Problem 29
                 ↳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)


Rules:


active(p(0)) -> mark(0)
active(p(s(X))) -> mark(X)
active(leq(0, Y)) -> mark(true)
active(leq(s(X), 0)) -> mark(false)
active(leq(s(X), s(Y))) -> mark(leq(X, Y))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(diff(X, Y)) -> mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) -> p(active(X))
active(s(X)) -> s(active(X))
active(leq(X1, X2)) -> leq(active(X1), X2)
active(leq(X1, X2)) -> leq(X1, active(X2))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(diff(X1, X2)) -> diff(active(X1), X2)
active(diff(X1, X2)) -> diff(X1, active(X2))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
leq(mark(X1), X2) -> mark(leq(X1, X2))
leq(X1, mark(X2)) -> mark(leq(X1, X2))
leq(ok(X1), ok(X2)) -> ok(leq(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
diff(mark(X1), X2) -> mark(diff(X1, X2))
diff(X1, mark(X2)) -> mark(diff(X1, X2))
diff(ok(X1), ok(X2)) -> ok(diff(X1, X2))
proper(p(X)) -> p(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(leq(X1, X2)) -> leq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) -> diff(proper(X1), proper(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(PROPER(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 26
             ↳Polo
             ...
               →DP Problem 30
                 ↳Dependency Graph
       →DP Problem 8
         ↳Remaining


Dependency Pair:


Rules:


active(p(0)) -> mark(0)
active(p(s(X))) -> mark(X)
active(leq(0, Y)) -> mark(true)
active(leq(s(X), 0)) -> mark(false)
active(leq(s(X), s(Y))) -> mark(leq(X, Y))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(diff(X, Y)) -> mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) -> p(active(X))
active(s(X)) -> s(active(X))
active(leq(X1, X2)) -> leq(active(X1), X2)
active(leq(X1, X2)) -> leq(X1, active(X2))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(diff(X1, X2)) -> diff(active(X1), X2)
active(diff(X1, X2)) -> diff(X1, active(X2))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
leq(mark(X1), X2) -> mark(leq(X1, X2))
leq(X1, mark(X2)) -> mark(leq(X1, X2))
leq(ok(X1), ok(X2)) -> ok(leq(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
diff(mark(X1), X2) -> mark(diff(X1, X2))
diff(X1, mark(X2)) -> mark(diff(X1, X2))
diff(ok(X1), ok(X2)) -> ok(diff(X1, X2))
proper(p(X)) -> p(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(leq(X1, X2)) -> leq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) -> diff(proper(X1), proper(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(p(0)) -> mark(0)
active(p(s(X))) -> mark(X)
active(leq(0, Y)) -> mark(true)
active(leq(s(X), 0)) -> mark(false)
active(leq(s(X), s(Y))) -> mark(leq(X, Y))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(diff(X, Y)) -> mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
active(p(X)) -> p(active(X))
active(s(X)) -> s(active(X))
active(leq(X1, X2)) -> leq(active(X1), X2)
active(leq(X1, X2)) -> leq(X1, active(X2))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(diff(X1, X2)) -> diff(active(X1), X2)
active(diff(X1, X2)) -> diff(X1, active(X2))
p(mark(X)) -> mark(p(X))
p(ok(X)) -> ok(p(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
leq(mark(X1), X2) -> mark(leq(X1, X2))
leq(X1, mark(X2)) -> mark(leq(X1, X2))
leq(ok(X1), ok(X2)) -> ok(leq(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
diff(mark(X1), X2) -> mark(diff(X1, X2))
diff(X1, mark(X2)) -> mark(diff(X1, X2))
diff(ok(X1), ok(X2)) -> ok(diff(X1, X2))
proper(p(X)) -> p(proper(X))
proper(0) -> ok(0)
proper(s(X)) -> s(proper(X))
proper(leq(X1, X2)) -> leq(proper(X1), proper(X2))
proper(true) -> ok(true)
proper(false) -> ok(false)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(diff(X1, X2)) -> diff(proper(X1), proper(X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))




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