Termination Proof using AProVE

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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

two new Dependency Pairs are created:

BIN(s(s(x'')), s(s(y''))) -> BIN(s(x''), s(y''))
BIN(s(s(s(x''''))), s(s(y''''))) -> BIN(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:

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

Rules:

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

Strategy:

innermost

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

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

three new Dependency Pairs are created:

BIN(s(s(s(x''''))), s(y'''')) -> BIN(s(s(x'''')), s(y''''))
BIN(s(s(s(x''''))), s(s(y''''))) -> BIN(s(s(x'''')), s(s(y'''')))
BIN(s(s(s(s(x'''''')))), s(s(y''''''))) -> BIN(s(s(s(x''''''))), s(s(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:

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

Rules:

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

Strategy:

innermost

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

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

four new Dependency Pairs are created:

BIN(s(s(s(x''''))), s(s(s(y'''')))) -> BIN(s(s(x'''')), s(s(y'''')))
BIN(s(s(s(s(x'''''')))), s(s(s(y'''''')))) -> BIN(s(s(s(x''''''))), s(s(y'''''')))
BIN(s(s(s(s(x'''''')))), s(s(y'''))) -> BIN(s(s(s(x''''''))), s(y'''))
BIN(s(s(s(s(s(x''''''''))))), s(s(s(y'''''''')))) -> BIN(s(s(s(s(x'''''''')))), s(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:

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

Rules:

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

Strategy:

innermost

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

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

seven new Dependency Pairs are created:

BIN(s(s(s(s(x'''''')))), s(s(s(y'''''')))) -> BIN(s(s(s(x''''''))), s(s(y'''''')))
BIN(s(s(s(s(x'''''')))), s(s(y''''''))) -> BIN(s(s(s(x''''''))), s(y''''''))
BIN(s(s(s(s(s(x''''''''))))), s(s(s(y'''''''')))) -> BIN(s(s(s(s(x'''''''')))), s(s(y'''''''')))
BIN(s(s(s(s(x'''''')))), s(s(s(s(y''''''))))) -> BIN(s(s(s(x''''''))), s(s(s(y''''''))))
BIN(s(s(s(s(s(x''''''''))))), s(s(s(s(y''''''''))))) -> BIN(s(s(s(s(x'''''''')))), s(s(s(y''''''''))))
BIN(s(s(s(s(s(x''''''''))))), s(s(s(y'''''')))) -> BIN(s(s(s(s(x'''''''')))), s(s(y'''''')))
BIN(s(s(s(s(s(s(x'''''''''')))))), s(s(s(s(y''''''''''))))) -> BIN(s(s(s(s(s(x''''''''''))))), s(s(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:

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

Rules:

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

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(s(x₁)) = 1 + x₁

POL(BIN(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 7

                 ↳Polynomial Ordering

Dependency Pairs:

Rules:

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

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(s(x₁)) = 1 + x₁

POL(BIN(x₁, x₂)) = 1 + x₁ + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 8

                 ↳Dependency Graph

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:04 minutes

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