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

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳Remaining


Dependency Pairs:

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


Rules:


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





The following dependency pair can be strictly oriented:

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


The following usable rule using the Ce-refinement can be oriented:

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


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

resulting in one new DP problem.
Used Argument Filtering System:
+'(x1, x2) -> +'(x1, x2)
+(x1, x2) -> +(x1, x2)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 3
             ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳Remaining


Dependency Pair:

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


Rules:


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





The following dependency pair can be strictly oriented:

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


The following usable rule using the Ce-refinement can be oriented:

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


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

resulting in one new DP problem.
Used Argument Filtering System:
+'(x1, x2) -> x1
+(x1, x2) -> +(x1, x2)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 3
             ↳AFS
             ...
               →DP Problem 4
                 ↳Dependency Graph
       →DP Problem 2
         ↳Remaining


Dependency Pair:


Rules:


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





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

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


Rules:


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




Termination of R could not be shown.
Duration:
0:00 minutes