Termination Proof using AProVE

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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

F(s(x), y) -> F(f(x, y), y)
F(s(x), y) -> F(x, y)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

Dependency Pairs:

F(s(x), y) -> F(x, y)
F(s(x), y) -> F(f(x, y), y)

Rules:

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

Strategy:

innermost

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

F(s(x), y) -> F(f(x, y), y)

two new Dependency Pairs are created:

F(s(0), y'') -> F(0, y'')
F(s(s(x'')), y'') -> F(f(f(x'', y''), y''), y'')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Narrowing Transformation

Dependency Pairs:

F(s(s(x'')), y'') -> F(f(f(x'', y''), y''), y'')
F(s(x), y) -> F(x, y)

Rules:

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

Strategy:

innermost

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

F(s(s(x'')), y'') -> F(f(f(x'', y''), y''), y'')

two new Dependency Pairs are created:

F(s(s(0)), y''') -> F(f(0, y'''), y''')
F(s(s(s(x'))), y''') -> F(f(f(f(x', y'''), y'''), y'''), y''')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 3

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(s(x'))), y''') -> F(f(f(f(x', y'''), y'''), y'''), y''')
F(s(s(0)), y''') -> F(f(0, y'''), y''')
F(s(x), y) -> F(x, y)

Rules:

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

Strategy:

innermost

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

F(s(s(0)), y''') -> F(f(0, y'''), y''')

one new Dependency Pair is created:

F(s(s(0)), y''') -> F(0, y''')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 4

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(s(x), y) -> F(x, y)
F(s(s(s(x'))), y''') -> F(f(f(f(x', y'''), y'''), y'''), y''')

Rules:

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

Strategy:

innermost

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

F(s(x), y) -> F(x, y)

two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 5

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(s(s(s(s(x''')))), y') -> F(s(s(s(x'''))), y')
F(s(s(x'')), y'') -> F(s(x''), y'')
F(s(s(s(x'))), y''') -> F(f(f(f(x', y'''), y'''), y'''), y''')

Rules:

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

Strategy:

innermost

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

F(s(s(s(x'))), y''') -> F(f(f(f(x', y'''), y'''), y'''), y''')

no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 6

                 ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 7

                 ↳Argument Filtering and Ordering

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

F(x₁, x₂) -> F(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 8

                 ↳Dependency Graph

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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