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))))

Innermost 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 one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Usable Rules (Innermost)


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))))


Strategy:

innermost




As we are in the innermost case, we can delete all 1 non-usable-rules.


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


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


Strategy:

innermost




We have the following set of usable rules:

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


Innermost Termination of R successfully shown.
Duration:
0:04 minutes