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

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC.


   R
DPs
       →DP Problem 1
Forward Instantiation Transformation


Dependency Pairs:

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


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(s(x), y) -> +(x, s(y))


Strategy:

innermost




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

+'(s(x), y) -> +'(x, y)
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 2
Forward Instantiation Transformation


Dependency Pairs:

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


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(s(x), y) -> +(x, s(y))


Strategy:

innermost




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

+'(s(x), y) -> +'(x, s(y))
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 2
FwdInst
             ...
               →DP Problem 3
Forward Instantiation Transformation


Dependency Pairs:

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


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(s(x), y) -> +(x, s(y))


Strategy:

innermost




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

+'(s(s(x'')), y'') -> +'(s(x''), y'')
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 2
FwdInst
             ...
               →DP Problem 4
Forward Instantiation Transformation


Dependency Pairs:

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


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(s(x), y) -> +(x, s(y))


Strategy:

innermost




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

+'(s(s(x'')), y'') -> +'(s(x''), s(y''))
four new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 2
FwdInst
             ...
               →DP Problem 5
Forward Instantiation Transformation


Dependency Pairs:

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


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(s(x), y) -> +(x, s(y))


Strategy:

innermost




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

+'(s(s(s(x''''))), y') -> +'(s(s(x'''')), s(y'))
four new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 2
FwdInst
             ...
               →DP Problem 6
Polynomial Ordering


Dependency Pairs:

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


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(s(x), y) -> +(x, s(y))


Strategy:

innermost




The following dependency pairs can be strictly oriented:

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


There are no usable rules for innermost that need to be oriented.

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

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 2
FwdInst
             ...
               →DP Problem 7
Dependency Graph


Dependency Pair:


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(s(x), y) -> +(x, s(y))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:01 minutes