Term Rewriting System R:
[X, X1, X2, X3]
active(f(a, b, X)) -> mark(f(X, X, X))
active(c) -> mark(a)
active(c) -> mark(b)
active(f(X1, X2, X3)) -> f(active(X1), X2, X3)
active(f(X1, X2, X3)) -> f(X1, X2, active(X3))
f(mark(X1), X2, X3) -> mark(f(X1, X2, X3))
f(X1, X2, mark(X3)) -> mark(f(X1, X2, X3))
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(a) -> ok(a)
proper(b) -> ok(b)
proper(c) -> ok(c)
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(a, b, X)) -> F(X, X, X)
ACTIVE(f(X1, X2, X3)) -> F(active(X1), X2, X3)
ACTIVE(f(X1, X2, X3)) -> ACTIVE(X1)
ACTIVE(f(X1, X2, X3)) -> F(X1, X2, active(X3))
ACTIVE(f(X1, X2, X3)) -> ACTIVE(X3)
F(mark(X1), X2, X3) -> F(X1, X2, X3)
F(X1, X2, mark(X3)) -> F(X1, X2, X3)
F(ok(X1), ok(X2), ok(X3)) -> F(X1, X2, X3)
PROPER(f(X1, X2, X3)) -> F(proper(X1), proper(X2), proper(X3))
PROPER(f(X1, X2, X3)) -> PROPER(X1)
PROPER(f(X1, X2, X3)) -> PROPER(X2)
PROPER(f(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 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(X1), ok(X2), ok(X3)) -> F(X1, X2, X3)
F(X1, X2, mark(X3)) -> F(X1, X2, X3)
F(mark(X1), X2, X3) -> F(X1, X2, X3)


Rules:


active(f(a, b, X)) -> mark(f(X, X, X))
active(c) -> mark(a)
active(c) -> mark(b)
active(f(X1, X2, X3)) -> f(active(X1), X2, X3)
active(f(X1, X2, X3)) -> f(X1, X2, active(X3))
f(mark(X1), X2, X3) -> mark(f(X1, X2, X3))
f(X1, X2, mark(X3)) -> mark(f(X1, X2, X3))
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(a) -> ok(a)
proper(b) -> ok(b)
proper(c) -> ok(c)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

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


There are no usable rules using the Ce-refinement 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))=  x2  

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 Pairs:

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


Rules:


active(f(a, b, X)) -> mark(f(X, X, X))
active(c) -> mark(a)
active(c) -> mark(b)
active(f(X1, X2, X3)) -> f(active(X1), X2, X3)
active(f(X1, X2, X3)) -> f(X1, X2, active(X3))
f(mark(X1), X2, X3) -> mark(f(X1, X2, X3))
f(X1, X2, mark(X3)) -> mark(f(X1, X2, X3))
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(a) -> ok(a)
proper(b) -> ok(b)
proper(c) -> ok(c)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

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


There are no usable rules using the Ce-refinement 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
Polynomial Ordering
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Remaining


Dependency Pair:

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


Rules:


active(f(a, b, X)) -> mark(f(X, X, X))
active(c) -> mark(a)
active(c) -> mark(b)
active(f(X1, X2, X3)) -> f(active(X1), X2, X3)
active(f(X1, X2, X3)) -> f(X1, X2, active(X3))
f(mark(X1), X2, X3) -> mark(f(X1, X2, X3))
f(X1, X2, mark(X3)) -> mark(f(X1, X2, X3))
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(a) -> ok(a)
proper(b) -> ok(b)
proper(c) -> ok(c)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pair can be strictly oriented:

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


There are no usable rules using the Ce-refinement that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(mark(x1))=  1 + x1  
  POL(F(x1, x2, x3))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
           →DP Problem 5
Polo
             ...
               →DP Problem 7
Dependency Graph
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Remaining


Dependency Pair:


Rules:


active(f(a, b, X)) -> mark(f(X, X, X))
active(c) -> mark(a)
active(c) -> mark(b)
active(f(X1, X2, X3)) -> f(active(X1), X2, X3)
active(f(X1, X2, X3)) -> f(X1, X2, active(X3))
f(mark(X1), X2, X3) -> mark(f(X1, X2, X3))
f(X1, X2, mark(X3)) -> mark(f(X1, X2, X3))
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(a) -> ok(a)
proper(b) -> ok(b)
proper(c) -> ok(c)
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 Pairs:

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


Rules:


active(f(a, b, X)) -> mark(f(X, X, X))
active(c) -> mark(a)
active(c) -> mark(b)
active(f(X1, X2, X3)) -> f(active(X1), X2, X3)
active(f(X1, X2, X3)) -> f(X1, X2, active(X3))
f(mark(X1), X2, X3) -> mark(f(X1, X2, X3))
f(X1, X2, mark(X3)) -> mark(f(X1, X2, X3))
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(a) -> ok(a)
proper(b) -> ok(b)
proper(c) -> ok(c)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pairs can be strictly oriented:

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


There are no usable rules using the Ce-refinement that need to be oriented.

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

resulting in one new DP problem.



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


Dependency Pair:


Rules:


active(f(a, b, X)) -> mark(f(X, X, X))
active(c) -> mark(a)
active(c) -> mark(b)
active(f(X1, X2, X3)) -> f(active(X1), X2, X3)
active(f(X1, X2, X3)) -> f(X1, X2, active(X3))
f(mark(X1), X2, X3) -> mark(f(X1, X2, X3))
f(X1, X2, mark(X3)) -> mark(f(X1, X2, X3))
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(a) -> ok(a)
proper(b) -> ok(b)
proper(c) -> ok(c)
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(X1, X2, X3)) -> PROPER(X3)
PROPER(f(X1, X2, X3)) -> PROPER(X2)
PROPER(f(X1, X2, X3)) -> PROPER(X1)


Rules:


active(f(a, b, X)) -> mark(f(X, X, X))
active(c) -> mark(a)
active(c) -> mark(b)
active(f(X1, X2, X3)) -> f(active(X1), X2, X3)
active(f(X1, X2, X3)) -> f(X1, X2, active(X3))
f(mark(X1), X2, X3) -> mark(f(X1, X2, X3))
f(X1, X2, mark(X3)) -> mark(f(X1, X2, X3))
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(a) -> ok(a)
proper(b) -> ok(b)
proper(c) -> ok(c)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))





The following dependency pairs can be strictly oriented:

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


There are no usable rules using the Ce-refinement 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 9
Dependency Graph
       →DP Problem 4
Remaining


Dependency Pair:


Rules:


active(f(a, b, X)) -> mark(f(X, X, X))
active(c) -> mark(a)
active(c) -> mark(b)
active(f(X1, X2, X3)) -> f(active(X1), X2, X3)
active(f(X1, X2, X3)) -> f(X1, X2, active(X3))
f(mark(X1), X2, X3) -> mark(f(X1, X2, X3))
f(X1, X2, mark(X3)) -> mark(f(X1, X2, X3))
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(a) -> ok(a)
proper(b) -> ok(b)
proper(c) -> ok(c)
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(a, b, X)) -> mark(f(X, X, X))
active(c) -> mark(a)
active(c) -> mark(b)
active(f(X1, X2, X3)) -> f(active(X1), X2, X3)
active(f(X1, X2, X3)) -> f(X1, X2, active(X3))
f(mark(X1), X2, X3) -> mark(f(X1, X2, X3))
f(X1, X2, mark(X3)) -> mark(f(X1, X2, X3))
f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
proper(a) -> ok(a)
proper(b) -> ok(b)
proper(c) -> ok(c)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))




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