Term Rewriting System R:
[x, y, z]
active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
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(b, c, x)) -> F(x, x, x)
ACTIVE(f(x, y, z)) -> F(x, y, active(z))
ACTIVE(f(x, y, z)) -> ACTIVE(z)
F(x, y, mark(z)) -> F(x, y, z)
F(ok(x), ok(y), ok(z)) -> F(x, y, z)
PROPER(f(x, y, z)) -> F(proper(x), proper(y), proper(z))
PROPER(f(x, y, z)) -> PROPER(x)
PROPER(f(x, y, z)) -> PROPER(y)
PROPER(f(x, y, z)) -> PROPER(z)
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 four SCCs. 


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


Dependency Pairs:

F(ok(x), ok(y), ok(z)) -> F(x, y, z)
F(x, y, mark(z)) -> F(x, y, z)


Rules:


active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
top(mark(x)) -> top(proper(x))
top(ok(x)) -> top(active(x))





The following dependency pair can be strictly oriented:

F(ok(x), ok(y), ok(z)) -> F(x, y, z)


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(F(x1, x2, x3))=  x1  

resulting in one new DP problem. 



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


Dependency Pair:

F(x, y, mark(z)) -> F(x, y, z)


Rules:


active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
top(mark(x)) -> top(proper(x))
top(ok(x)) -> top(active(x))





The following dependency pair can be strictly oriented:

F(x, y, mark(z)) -> F(x, y, z)


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, x2, x3))=  x3  

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
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
         ↳Remaining


Dependency Pair:

ACTIVE(f(x, y, z)) -> ACTIVE(z)


Rules:


active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
top(mark(x)) -> top(proper(x))
top(ok(x)) -> top(active(x))





The following dependency pair can be strictly oriented:

ACTIVE(f(x, y, z)) -> ACTIVE(z)


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, x2, x3))=  1 + x3  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 7
             ↳Dependency Graph
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Remaining


Dependency Pair:


Rules:


active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
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
         ↳Remaining


Dependency Pairs:

PROPER(f(x, y, z)) -> PROPER(z)
PROPER(f(x, y, z)) -> PROPER(y)
PROPER(f(x, y, z)) -> PROPER(x)


Rules:


active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
top(mark(x)) -> top(proper(x))
top(ok(x)) -> top(active(x))





The following dependency pairs can be strictly oriented:

PROPER(f(x, y, z)) -> PROPER(z)
PROPER(f(x, y, z)) -> PROPER(y)
PROPER(f(x, y, z)) -> 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, x2, x3))=  1 + x1 + x2 + x3  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
           →DP Problem 8
             ↳Dependency Graph
       →DP Problem 4
         ↳Remaining


Dependency Pair:


Rules:


active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
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
         ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

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


Rules:


active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
top(mark(x)) -> top(proper(x))
top(ok(x)) -> top(active(x))




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