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

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Forward Instantiation Transformation


Dependency Pair:

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


Rule:


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


Strategy:

innermost




On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

.'(.(x, y), z) -> .'(y, z)
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem: 



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


Dependency Pair:

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


Rule:


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


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


Additionally, the following usable rule for innermost 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  

resulting in one new DP problem. 



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


Dependency Pair:


Rule:


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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