Term Rewriting System R:
[x, y, z]
+(x, 0) -> x
+(minus(x), x) -> 0
minus(0) -> 0
minus(minus(x)) -> x
minus(+(x, y)) -> +(minus(y), minus(x))
*(x, 1) -> x
*(x, 0) -> 0
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(x, minus(y)) -> minus(*(x, y))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering
       →DP Problem 2
         ↳Polo


Dependency Pairs:

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


Rules:


+(x, 0) -> x
+(minus(x), x) -> 0
minus(0) -> 0
minus(minus(x)) -> x
minus(+(x, y)) -> +(minus(y), minus(x))
*(x, 1) -> x
*(x, 0) -> 0
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(x, minus(y)) -> minus(*(x, y))





The following dependency pairs can be strictly oriented:

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


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(MINUS(x1))=  x1  
  POL(+(x1, x2))=  1 + x1 + x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 3
             ↳Dependency Graph
       →DP Problem 2
         ↳Polo


Dependency Pair:


Rules:


+(x, 0) -> x
+(minus(x), x) -> 0
minus(0) -> 0
minus(minus(x)) -> x
minus(+(x, y)) -> +(minus(y), minus(x))
*(x, 1) -> x
*(x, 0) -> 0
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(x, minus(y)) -> minus(*(x, y))





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polynomial Ordering


Dependency Pairs:

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


Rules:


+(x, 0) -> x
+(minus(x), x) -> 0
minus(0) -> 0
minus(minus(x)) -> x
minus(+(x, y)) -> +(minus(y), minus(x))
*(x, 1) -> x
*(x, 0) -> 0
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(x, minus(y)) -> minus(*(x, y))





The following dependency pair can be strictly oriented:

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


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(*'(x1, x2))=  x2  
  POL(minus(x1))=  1 + x1  
  POL(+(x1, x2))=  x1 + x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 4
             ↳Polynomial Ordering


Dependency Pairs:

*'(x, +(y, z)) -> *'(x, z)
*'(x, +(y, z)) -> *'(x, y)


Rules:


+(x, 0) -> x
+(minus(x), x) -> 0
minus(0) -> 0
minus(minus(x)) -> x
minus(+(x, y)) -> +(minus(y), minus(x))
*(x, 1) -> x
*(x, 0) -> 0
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(x, minus(y)) -> minus(*(x, y))





The following dependency pairs can be strictly oriented:

*'(x, +(y, z)) -> *'(x, z)
*'(x, +(y, z)) -> *'(x, y)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(*'(x1, x2))=  x2  
  POL(+(x1, x2))=  1 + x1 + x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 4
             ↳Polo
             ...
               →DP Problem 5
                 ↳Dependency Graph


Dependency Pair:


Rules:


+(x, 0) -> x
+(minus(x), x) -> 0
minus(0) -> 0
minus(minus(x)) -> x
minus(+(x, y)) -> +(minus(y), minus(x))
*(x, 1) -> x
*(x, 0) -> 0
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(x, minus(y)) -> minus(*(x, y))





Using the Dependency Graph resulted in no new DP problems.

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