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.



   R
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.


   R
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:



   R
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:



   R
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:



   R
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:



   R
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:



   R
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'''')))
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  

resulting in one new DP problem.



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