Term Rewriting System R:
[x]
active(f(x)) -> mark(f(f(x)))
chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
chk(no(c)) -> active(c)
mat(f(x), f(y)) -> f(mat(x, y))
mat(f(x), c) -> no(c)
f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))
tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

ACTIVE(f(x)) -> F(f(x))
CHK(no(f(x))) -> F(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
CHK(no(f(x))) -> CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))
CHK(no(f(x))) -> MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)
CHK(no(f(x))) -> F(f(f(f(f(f(f(f(f(f(X))))))))))
CHK(no(f(x))) -> F(f(f(f(f(f(f(f(f(X)))))))))
CHK(no(f(x))) -> F(f(f(f(f(f(f(f(X))))))))
CHK(no(f(x))) -> F(f(f(f(f(f(f(X)))))))
CHK(no(f(x))) -> F(f(f(f(f(f(X))))))
CHK(no(f(x))) -> F(f(f(f(f(X)))))
CHK(no(f(x))) -> F(f(f(f(X))))
CHK(no(f(x))) -> F(f(f(X)))
CHK(no(f(x))) -> F(f(X))
CHK(no(f(x))) -> F(X)
CHK(no(c)) -> ACTIVE(c)
MAT(f(x), f(y)) -> F(mat(x, y))
MAT(f(x), f(y)) -> MAT(x, y)
F(active(x)) -> ACTIVE(f(x))
F(active(x)) -> F(x)
F(no(x)) -> F(x)
F(mark(x)) -> F(x)
TP(mark(x)) -> TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
TP(mark(x)) -> CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))
TP(mark(x)) -> MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)
TP(mark(x)) -> F(f(f(f(f(f(f(f(f(f(X))))))))))
TP(mark(x)) -> F(f(f(f(f(f(f(f(f(X)))))))))
TP(mark(x)) -> F(f(f(f(f(f(f(f(X))))))))
TP(mark(x)) -> F(f(f(f(f(f(f(X)))))))
TP(mark(x)) -> F(f(f(f(f(f(X))))))
TP(mark(x)) -> F(f(f(f(f(X)))))
TP(mark(x)) -> F(f(f(f(X))))
TP(mark(x)) -> F(f(f(X)))
TP(mark(x)) -> F(f(X))
TP(mark(x)) -> F(X)

Furthermore, R contains three SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Modular Removal of Rules
       →DP Problem 2
         ↳Nar
       →DP Problem 3
         ↳Nar


Dependency Pairs:

F(mark(x)) -> F(x)
F(no(x)) -> F(x)
F(active(x)) -> F(x)
F(active(x)) -> ACTIVE(f(x))
ACTIVE(f(x)) -> F(f(x))


Rules:


active(f(x)) -> mark(f(f(x)))
chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
chk(no(c)) -> active(c)
mat(f(x), f(y)) -> f(mat(x, y))
mat(f(x), c) -> no(c)
f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))
tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))





We have the following set of usable rules:

f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))
active(f(x)) -> mark(f(f(x)))
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(active(x1))=  x1  
  POL(ACTIVE(x1))=  x1  
  POL(no(x1))=  x1  
  POL(mark(x1))=  x1  
  POL(F(x1))=  x1  
  POL(f(x1))=  x1  

We have the following set D of usable symbols: {ACTIVE, active, no, mark, F, f}
No Dependency Pairs can be deleted.
5 non usable rules have been deleted.

The result of this processor delivers one new DP problem.



   R
     ↳DPs
       →DP Problem 1
         ↳MRR
           →DP Problem 4
             ↳Modular Removal of Rules
       →DP Problem 2
         ↳Nar
       →DP Problem 3
         ↳Nar


Dependency Pairs:

F(mark(x)) -> F(x)
F(no(x)) -> F(x)
F(active(x)) -> F(x)
F(active(x)) -> ACTIVE(f(x))
ACTIVE(f(x)) -> F(f(x))


Rules:


f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))
active(f(x)) -> mark(f(f(x)))





We have the following set of usable rules:

f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))
active(f(x)) -> mark(f(f(x)))
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(active(x1))=  1 + x1  
  POL(ACTIVE(x1))=  x1  
  POL(no(x1))=  x1  
  POL(mark(x1))=  x1  
  POL(F(x1))=  x1  
  POL(f(x1))=  x1  

We have the following set D of usable symbols: {ACTIVE, active, no, mark, F, f}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

F(active(x)) -> F(x)
F(active(x)) -> ACTIVE(f(x))

The following rules can be deleted as the lhs is strictly greater than the corresponding rhs:

active(f(x)) -> mark(f(f(x)))


The result of this processor delivers one new DP problem.



   R
     ↳DPs
       →DP Problem 1
         ↳MRR
           →DP Problem 4
             ↳MRR
             ...
               →DP Problem 5
                 ↳Modular Removal of Rules
       →DP Problem 2
         ↳Nar
       →DP Problem 3
         ↳Nar


Dependency Pairs:

F(mark(x)) -> F(x)
F(no(x)) -> F(x)
ACTIVE(f(x)) -> F(f(x))


Rules:


f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))





We have the following set of usable rules:

f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(ACTIVE(x1))=  x1  
  POL(active(x1))=  x1  
  POL(no(x1))=  x1  
  POL(mark(x1))=  x1  
  POL(F(x1))=  x1  
  POL(f(x1))=  x1  

We have the following set D of usable symbols: {active, no, mark, F, f}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

ACTIVE(f(x)) -> F(f(x))

No Rules can be deleted.

The result of this processor delivers one new DP problem.



   R
     ↳DPs
       →DP Problem 1
         ↳MRR
           →DP Problem 4
             ↳MRR
             ...
               →DP Problem 6
                 ↳Modular Removal of Rules
       →DP Problem 2
         ↳Nar
       →DP Problem 3
         ↳Nar


Dependency Pairs:

F(mark(x)) -> F(x)
F(no(x)) -> F(x)


Rules:


f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))





We have the following set of usable rules: none
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(no(x1))=  x1  
  POL(mark(x1))=  x1  
  POL(F(x1))=  x1  

We have the following set D of usable symbols: {F}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

F(mark(x)) -> F(x)
F(no(x)) -> F(x)

No Rules can be deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved. 



   R
     ↳DPs
       →DP Problem 1
         ↳MRR
       →DP Problem 2
         ↳Narrowing Transformation
       →DP Problem 3
         ↳Nar


Dependency Pair:

CHK(no(f(x))) -> CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))


Rules:


active(f(x)) -> mark(f(f(x)))
chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
chk(no(c)) -> active(c)
mat(f(x), f(y)) -> f(mat(x, y))
mat(f(x), c) -> no(c)
f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))
tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))





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

CHK(no(f(x))) -> CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))
two new Dependency Pairs are created:

CHK(no(f(f(y)))) -> CHK(f(mat(f(f(f(f(f(f(f(f(f(X))))))))), y)))
CHK(no(f(c))) -> CHK(no(c))

The transformation is resulting in no new DP problems. 



   R
     ↳DPs
       →DP Problem 1
         ↳MRR
       →DP Problem 2
         ↳Nar
       →DP Problem 3
         ↳Narrowing Transformation


Dependency Pair:

TP(mark(x)) -> TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))


Rules:


active(f(x)) -> mark(f(f(x)))
chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
chk(no(c)) -> active(c)
mat(f(x), f(y)) -> f(mat(x, y))
mat(f(x), c) -> no(c)
f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))
tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))





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

TP(mark(x)) -> TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
two new Dependency Pairs are created:

TP(mark(f(y))) -> TP(chk(f(mat(f(f(f(f(f(f(f(f(f(X))))))))), y))))
TP(mark(c)) -> TP(chk(no(c)))

The transformation is resulting in no new DP problems.


Termination of R successfully shown.
Duration:
0:02 minutes