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
Nar


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)


Additionally, the following rules can be oriented:

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))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(top(x1))=  0  
  POL(active(x1))=  x1  
  POL(m(x1))=  0  
  POL(proper(x1))=  1  
  POL(c)=  0  
  POL(b)=  0  
  POL(d)=  0  
  POL(mark(x1))=  0  
  POL(ok(x1))=  1 + x1  
  POL(f(x1, x2, x3))=  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
Nar


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)


Additionally, the following rules can be oriented:

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))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(top(x1))=  0  
  POL(active(x1))=  1 + x1  
  POL(m(x1))=  0  
  POL(proper(x1))=  0  
  POL(c)=  0  
  POL(b)=  0  
  POL(d)=  0  
  POL(mark(x1))=  1 + x1  
  POL(ok(x1))=  0  
  POL(f(x1, x2, x3))=  x3  
  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
Nar


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
Nar


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)


Additionally, the following rules can be oriented:

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))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(top(x1))=  0  
  POL(active(x1))=  x1  
  POL(m(x1))=  0  
  POL(proper(x1))=  x1  
  POL(ACTIVE(x1))=  x1  
  POL(c)=  0  
  POL(b)=  0  
  POL(d)=  0  
  POL(mark(x1))=  0  
  POL(ok(x1))=  0  
  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
Nar


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
Nar


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)


Additionally, the following rules can be oriented:

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))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(top(x1))=  0  
  POL(active(x1))=  x1  
  POL(m(x1))=  0  
  POL(proper(x1))=  x1  
  POL(c)=  0  
  POL(PROPER(x1))=  x1  
  POL(b)=  0  
  POL(d)=  0  
  POL(mark(x1))=  0  
  POL(ok(x1))=  0  
  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
Nar


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
Narrowing Transformation


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))





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

TOP(mark(x)) -> TOP(proper(x))
four new Dependency Pairs are created:

TOP(mark(b)) -> TOP(ok(b))
TOP(mark(c)) -> TOP(ok(c))
TOP(mark(d)) -> TOP(ok(d))
TOP(mark(f(x'', y', z'))) -> TOP(f(proper(x''), proper(y'), proper(z')))

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
Nar
           →DP Problem 9
Narrowing Transformation


Dependency Pairs:

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





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

TOP(ok(x)) -> TOP(active(x))
four new Dependency Pairs are created:

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

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
Nar
           →DP Problem 9
Nar
             ...
               →DP Problem 10
Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

TOP(ok(f(x'', y', z'))) -> TOP(f(x'', y', active(z')))
TOP(ok(f(b, c, x''))) -> TOP(mark(f(x'', x'', x'')))
TOP(mark(f(x'', y', z'))) -> TOP(f(proper(x''), proper(y'), proper(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))




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