Termination Proof using AProVE

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.

     ↳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.

     ↳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:

     ↳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:

     ↳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:

     ↳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:

     ↳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:

     ↳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.

     ↳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