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

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains two SCCs. 


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


Dependency Pair:

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


Rules:


+(x, 0) -> x
+(x, i(x)) -> 0
+(+(x, y), z) -> +(x, +(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(+(x, y), z) -> +(*(x, z), *(y, z))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


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

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

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
+(x, i(x)) -> 0
+(+(x, y), z) -> +(x, +(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(+(x, y), z) -> +(*(x, z), *(y, z))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pairs:

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


Rules:


+(x, 0) -> x
+(x, i(x)) -> 0
+(+(x, y), z) -> +(x, +(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(+(x, y), z) -> +(*(x, z), *(y, z))


Strategy:

innermost




The following dependency pairs can be strictly oriented:

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


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(*'(x1, x2))=  x1  
  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
             ↳Dependency Graph


Dependency Pairs:

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


Rules:


+(x, 0) -> x
+(x, i(x)) -> 0
+(+(x, y), z) -> +(x, +(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(+(x, y), z) -> +(*(x, z), *(y, z))


Strategy:

innermost




Using the Dependency Graph the DP problem was split into 1 DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 4
             ↳DGraph
             ...
               →DP Problem 5
                 ↳Polynomial Ordering


Dependency Pair:

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


Rules:


+(x, 0) -> x
+(x, i(x)) -> 0
+(+(x, y), z) -> +(x, +(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(+(x, y), z) -> +(*(x, z), *(y, z))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost 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 + x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 4
             ↳DGraph
             ...
               →DP Problem 6
                 ↳Dependency Graph


Dependency Pair:


Rules:


+(x, 0) -> x
+(x, i(x)) -> 0
+(+(x, y), z) -> +(x, +(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(+(x, y), z) -> +(*(x, z), *(y, z))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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