Term Rewriting System R:
[X, Y, X1, X2, X3]
active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

ACTIVE(f(X)) -> IF(X, c, f(true))
ACTIVE(f(X)) -> F(true)
ACTIVE(f(X)) -> F(active(X))
ACTIVE(f(X)) -> ACTIVE(X)
ACTIVE(if(X1, X2, X3)) -> IF(active(X1), X2, X3)
ACTIVE(if(X1, X2, X3)) -> ACTIVE(X1)
ACTIVE(if(X1, X2, X3)) -> IF(X1, active(X2), X3)
ACTIVE(if(X1, X2, X3)) -> ACTIVE(X2)
F(mark(X)) -> F(X)
F(ok(X)) -> F(X)
IF(mark(X1), X2, X3) -> IF(X1, X2, X3)
IF(X1, mark(X2), X3) -> IF(X1, X2, X3)
IF(ok(X1), ok(X2), ok(X3)) -> IF(X1, X2, X3)
PROPER(f(X)) -> F(proper(X))
PROPER(f(X)) -> PROPER(X)
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)
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 five SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Nar


Dependency Pairs:

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


Rules:


active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
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))=  x3  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 6
             ↳Polynomial Ordering
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Nar


Dependency Pairs:

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


Rules:


active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
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 6
             ↳Polo
             ...
               →DP Problem 7
                 ↳Polynomial Ordering
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Nar


Dependency Pair:

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


Rules:


active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

IF(X1, mark(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))=  x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 6
             ↳Polo
             ...
               →DP Problem 8
                 ↳Dependency Graph
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Nar


Dependency Pair:


Rules:


active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
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
         ↳Nar


Dependency Pairs:

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


Rules:


active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

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


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

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

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 9
             ↳Polynomial Ordering
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Nar


Dependency Pair:

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


Rules:


active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

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


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

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

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 9
             ↳Polo
             ...
               →DP Problem 10
                 ↳Dependency Graph
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Nar


Dependency Pair:


Rules:


active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
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
         ↳Nar


Dependency Pairs:

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


Rules:


active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pairs can be strictly oriented:

ACTIVE(if(X1, X2, X3)) -> ACTIVE(X2)
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 + x2  
  POL(f(x1))=  x1  

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
         ↳Nar


Dependency Pair:

ACTIVE(f(X)) -> ACTIVE(X)


Rules:


active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

ACTIVE(f(X)) -> ACTIVE(X)


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(ACTIVE(x1))=  x1  
  POL(f(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 11
             ↳Polo
             ...
               →DP Problem 12
                 ↳Dependency Graph
       →DP Problem 4
         ↳Polo
       →DP Problem 5
         ↳Nar


Dependency Pair:


Rules:


active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
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
         ↳Nar


Dependency Pairs:

PROPER(if(X1, X2, X3)) -> PROPER(X3)
PROPER(if(X1, X2, X3)) -> PROPER(X2)
PROPER(if(X1, X2, X3)) -> PROPER(X1)
PROPER(f(X)) -> PROPER(X)


Rules:


active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
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  
  POL(f(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 13
             ↳Polynomial Ordering
       →DP Problem 5
         ↳Nar


Dependency Pair:

PROPER(f(X)) -> PROPER(X)


Rules:


active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

PROPER(f(X)) -> PROPER(X)


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(PROPER(x1))=  x1  
  POL(f(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
         ↳Nar


Dependency Pair:


Rules:


active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
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
         ↳Narrowing Transformation


Dependency Pairs:

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


Rules:


active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TOP(mark(X)) -> TOP(proper(X))
five new Dependency Pairs are created:

TOP(mark(f(X''))) -> TOP(f(proper(X'')))
TOP(mark(if(X1', X2', X3'))) -> TOP(if(proper(X1'), proper(X2'), proper(X3')))
TOP(mark(c)) -> TOP(ok(c))
TOP(mark(true)) -> TOP(ok(true))
TOP(mark(false)) -> TOP(ok(false))

The transformation is 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
         ↳Nar
           →DP Problem 15
             ↳Narrowing Transformation


Dependency Pairs:

TOP(mark(false)) -> TOP(ok(false))
TOP(mark(true)) -> TOP(ok(true))
TOP(mark(c)) -> TOP(ok(c))
TOP(mark(if(X1', X2', X3'))) -> TOP(if(proper(X1'), proper(X2'), proper(X3')))
TOP(mark(f(X''))) -> TOP(f(proper(X'')))
TOP(ok(X)) -> TOP(active(X))


Rules:


active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TOP(ok(X)) -> TOP(active(X))
six new Dependency Pairs are created:

TOP(ok(f(X''))) -> TOP(mark(if(X'', c, f(true))))
TOP(ok(if(true, X'', Y'))) -> TOP(mark(X''))
TOP(ok(if(false, X'', Y'))) -> TOP(mark(Y'))
TOP(ok(f(X''))) -> TOP(f(active(X'')))
TOP(ok(if(X1', X2', X3'))) -> TOP(if(active(X1'), X2', X3'))
TOP(ok(if(X1', X2', X3'))) -> TOP(if(X1', active(X2'), X3'))

The transformation is 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
         ↳Nar
           →DP Problem 15
             ↳Nar
             ...
               →DP Problem 16
                 ↳Polynomial Ordering


Dependency Pairs:

TOP(ok(if(X1', X2', X3'))) -> TOP(if(X1', active(X2'), X3'))
TOP(ok(if(X1', X2', X3'))) -> TOP(if(active(X1'), X2', X3'))
TOP(ok(f(X''))) -> TOP(f(active(X'')))
TOP(ok(if(false, X'', Y'))) -> TOP(mark(Y'))
TOP(ok(if(true, X'', Y'))) -> TOP(mark(X''))
TOP(ok(f(X''))) -> TOP(mark(if(X'', c, f(true))))
TOP(mark(f(X''))) -> TOP(f(proper(X'')))
TOP(mark(if(X1', X2', X3'))) -> TOP(if(proper(X1'), proper(X2'), proper(X3')))


Rules:


active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

TOP(ok(if(false, X'', Y'))) -> TOP(mark(Y'))


Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(active(x1))=  x1  
  POL(proper(x1))=  x1  
  POL(if(x1, x2, x3))=  x1 + x2 + x3  
  POL(c)=  0  
  POL(false)=  1  
  POL(true)=  0  
  POL(mark(x1))=  x1  
  POL(ok(x1))=  x1  
  POL(TOP(x1))=  1 + x1  
  POL(f(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
         ↳Nar
           →DP Problem 15
             ↳Nar
             ...
               →DP Problem 17
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

TOP(ok(if(X1', X2', X3'))) -> TOP(if(X1', active(X2'), X3'))
TOP(ok(if(X1', X2', X3'))) -> TOP(if(active(X1'), X2', X3'))
TOP(ok(f(X''))) -> TOP(f(active(X'')))
TOP(ok(if(true, X'', Y'))) -> TOP(mark(X''))
TOP(ok(f(X''))) -> TOP(mark(if(X'', c, f(true))))
TOP(mark(f(X''))) -> TOP(f(proper(X'')))
TOP(mark(if(X1', X2', X3'))) -> TOP(if(proper(X1'), proper(X2'), proper(X3')))


Rules:


active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))




Termination of R could not be shown.
Duration:
0:04 minutes