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

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Negative Polynomial Order


Dependency Pairs:

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


Rules:


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


Strategy:

innermost




The following Dependency Pairs can be strictly oriented using the given order.

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


Moreover, the following usable rules (regarding the implicit AFS) are oriented.

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


Used ordering:
Polynomial Order with Interpretation:

POL( *'(x1, x2) ) = x2

POL( +(x1, x2) ) = x1 + x2 + 1

POL( +'(x1, x2) ) = x1 + x2

POL( *(x1, x2) ) = x2


This results in one new DP problem. 


   R
     ↳DPs
       →DP Problem 1
         ↳Neg POLO
           →DP Problem 2
             ↳Dependency Graph


Dependency Pairs:

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


Rules:


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


Strategy:

innermost




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


   R
     ↳DPs
       →DP Problem 1
         ↳Neg POLO
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 3
                 ↳Non Termination


Dependency Pairs:

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


Rules:


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


Strategy:

innermost




Found an infinite P-chain over R:
P =

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

R =

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

s = +'(+(x0', *(x'''', y'''')), *(x'''', z''''))
evaluates to t =+'(+(x0', *(x'''', y'''')), *(x'''', z''''))

Thus, s starts an infinite chain.

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