Term Rewriting System R:
[x, y]
minus(minus(x)) -> x
minus(+(x, y)) -> *(minus(minus(minus(x))), minus(minus(minus(y))))
minus(*(x, y)) -> +(minus(minus(minus(x))), minus(minus(minus(y))))
f(minus(x)) -> minus(minus(minus(f(x))))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

MINUS(+(x, y)) -> MINUS(minus(minus(x)))
MINUS(+(x, y)) -> MINUS(minus(x))
MINUS(+(x, y)) -> MINUS(x)
MINUS(+(x, y)) -> MINUS(minus(minus(y)))
MINUS(+(x, y)) -> MINUS(minus(y))
MINUS(+(x, y)) -> MINUS(y)
MINUS(*(x, y)) -> MINUS(minus(minus(x)))
MINUS(*(x, y)) -> MINUS(minus(x))
MINUS(*(x, y)) -> MINUS(x)
MINUS(*(x, y)) -> MINUS(minus(minus(y)))
MINUS(*(x, y)) -> MINUS(minus(y))
MINUS(*(x, y)) -> MINUS(y)
F(minus(x)) -> MINUS(minus(minus(f(x))))
F(minus(x)) -> MINUS(minus(f(x)))
F(minus(x)) -> MINUS(f(x))
F(minus(x)) -> F(x)

Furthermore, R contains two SCCs. 


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


Dependency Pairs:

MINUS(*(x, y)) -> MINUS(y)
MINUS(*(x, y)) -> MINUS(minus(y))
MINUS(*(x, y)) -> MINUS(minus(minus(y)))
MINUS(*(x, y)) -> MINUS(x)
MINUS(*(x, y)) -> MINUS(minus(x))
MINUS(*(x, y)) -> MINUS(minus(minus(x)))
MINUS(+(x, y)) -> MINUS(y)
MINUS(+(x, y)) -> MINUS(minus(y))
MINUS(+(x, y)) -> MINUS(minus(minus(y)))
MINUS(+(x, y)) -> MINUS(x)
MINUS(+(x, y)) -> MINUS(minus(x))
MINUS(+(x, y)) -> MINUS(minus(minus(x)))


Rules:


minus(minus(x)) -> x
minus(+(x, y)) -> *(minus(minus(minus(x))), minus(minus(minus(y))))
minus(*(x, y)) -> +(minus(minus(minus(x))), minus(minus(minus(y))))
f(minus(x)) -> minus(minus(minus(f(x))))





We have the following set of usable rules:

minus(minus(x)) -> x
minus(+(x, y)) -> *(minus(minus(minus(x))), minus(minus(minus(y))))
minus(*(x, y)) -> +(minus(minus(minus(x))), minus(minus(minus(y))))
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(MINUS(x1))=  1 + x1  
  POL(*(x1, x2))=  x1 + x2  
  POL(minus(x1))=  x1  
  POL(+(x1, x2))=  x1 + x2  

We have the following set D of usable symbols: {MINUS, minus, *, +}
No Dependency Pairs can be deleted.
1 non usable rules have been deleted.

The result of this processor delivers one new DP problem.



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


Dependency Pairs:

MINUS(*(x, y)) -> MINUS(y)
MINUS(*(x, y)) -> MINUS(minus(y))
MINUS(*(x, y)) -> MINUS(minus(minus(y)))
MINUS(*(x, y)) -> MINUS(x)
MINUS(*(x, y)) -> MINUS(minus(x))
MINUS(*(x, y)) -> MINUS(minus(minus(x)))
MINUS(+(x, y)) -> MINUS(y)
MINUS(+(x, y)) -> MINUS(minus(y))
MINUS(+(x, y)) -> MINUS(minus(minus(y)))
MINUS(+(x, y)) -> MINUS(x)
MINUS(+(x, y)) -> MINUS(minus(x))
MINUS(+(x, y)) -> MINUS(minus(minus(x)))


Rules:


minus(minus(x)) -> x
minus(+(x, y)) -> *(minus(minus(minus(x))), minus(minus(minus(y))))
minus(*(x, y)) -> +(minus(minus(minus(x))), minus(minus(minus(y))))





We have the following set of usable rules:

minus(minus(x)) -> x
minus(+(x, y)) -> *(minus(minus(minus(x))), minus(minus(minus(y))))
minus(*(x, y)) -> +(minus(minus(minus(x))), minus(minus(minus(y))))
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(MINUS(x1))=  x1  
  POL(*(x1, x2))=  1 + x1 + x2  
  POL(minus(x1))=  x1  
  POL(+(x1, x2))=  1 + x1 + x2  

We have the following set D of usable symbols: {MINUS, minus, *, +}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

MINUS(*(x, y)) -> MINUS(y)
MINUS(*(x, y)) -> MINUS(minus(y))
MINUS(*(x, y)) -> MINUS(minus(minus(y)))
MINUS(*(x, y)) -> MINUS(x)
MINUS(*(x, y)) -> MINUS(minus(x))
MINUS(*(x, y)) -> MINUS(minus(minus(x)))
MINUS(+(x, y)) -> MINUS(y)
MINUS(+(x, y)) -> MINUS(minus(y))
MINUS(+(x, y)) -> MINUS(minus(minus(y)))
MINUS(+(x, y)) -> MINUS(x)
MINUS(+(x, y)) -> MINUS(minus(x))
MINUS(+(x, y)) -> MINUS(minus(minus(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
         ↳Size-Change Principle


Dependency Pair:

F(minus(x)) -> F(x)


Rules:


minus(minus(x)) -> x
minus(+(x, y)) -> *(minus(minus(minus(x))), minus(minus(minus(y))))
minus(*(x, y)) -> +(minus(minus(minus(x))), minus(minus(minus(y))))
f(minus(x)) -> minus(minus(minus(f(x))))





We number the DPs as follows:
  1. F(minus(x)) -> F(x)
and get the following Size-Change Graph(s):
{1} , {1}
1>1

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1>1

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
minus(x1) -> minus(x1)

We obtain no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes