Term Rewriting System R:
[x, y]
-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
lt(x, 0) -> false
lt(0, s(y)) -> true
lt(s(x), s(y)) -> lt(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
div(x, 0) -> 0
div(0, y) -> 0
div(s(x), s(y)) -> if(lt(x, y), 0, s(div(-(x, y), s(y))))

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

-'(s(x), s(y)) -> -'(x, y)
LT(s(x), s(y)) -> LT(x, y)
DIV(s(x), s(y)) -> IF(lt(x, y), 0, s(div(-(x, y), s(y))))
DIV(s(x), s(y)) -> LT(x, y)
DIV(s(x), s(y)) -> DIV(-(x, y), s(y))
DIV(s(x), s(y)) -> -'(x, y)

Furthermore, R contains three SCCs.


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


Dependency Pair:

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


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
lt(x, 0) -> false
lt(0, s(y)) -> true
lt(s(x), s(y)) -> lt(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
div(x, 0) -> 0
div(0, y) -> 0
div(s(x), s(y)) -> if(lt(x, y), 0, s(div(-(x, y), 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), s(y)) -> -'(x, y)
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 4
Forward Instantiation Transformation
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar


Dependency Pair:

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


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
lt(x, 0) -> false
lt(0, s(y)) -> true
lt(s(x), s(y)) -> lt(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
div(x, 0) -> 0
div(0, y) -> 0
div(s(x), s(y)) -> if(lt(x, y), 0, s(div(-(x, y), 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'')), s(s(y''))) -> -'(s(x''), s(y''))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 4
FwdInst
             ...
               →DP Problem 5
Polynomial Ordering
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar


Dependency Pair:

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


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
lt(x, 0) -> false
lt(0, s(y)) -> true
lt(s(x), s(y)) -> lt(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
div(x, 0) -> 0
div(0, y) -> 0
div(s(x), s(y)) -> if(lt(x, y), 0, s(div(-(x, y), s(y))))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


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

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

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 4
FwdInst
             ...
               →DP Problem 6
Dependency Graph
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar


Dependency Pair:


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
lt(x, 0) -> false
lt(0, s(y)) -> true
lt(s(x), s(y)) -> lt(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
div(x, 0) -> 0
div(0, y) -> 0
div(s(x), s(y)) -> if(lt(x, y), 0, s(div(-(x, y), s(y))))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pair:

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


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
lt(x, 0) -> false
lt(0, s(y)) -> true
lt(s(x), s(y)) -> lt(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
div(x, 0) -> 0
div(0, y) -> 0
div(s(x), s(y)) -> if(lt(x, y), 0, s(div(-(x, y), s(y))))


Strategy:

innermost




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

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

LT(s(s(x'')), s(s(y''))) -> LT(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 7
Forward Instantiation Transformation
       →DP Problem 3
Nar


Dependency Pair:

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


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
lt(x, 0) -> false
lt(0, s(y)) -> true
lt(s(x), s(y)) -> lt(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
div(x, 0) -> 0
div(0, y) -> 0
div(s(x), s(y)) -> if(lt(x, y), 0, s(div(-(x, y), s(y))))


Strategy:

innermost




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

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

LT(s(s(s(x''''))), s(s(s(y'''')))) -> LT(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 7
FwdInst
             ...
               →DP Problem 8
Polynomial Ordering
       →DP Problem 3
Nar


Dependency Pair:

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


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
lt(x, 0) -> false
lt(0, s(y)) -> true
lt(s(x), s(y)) -> lt(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
div(x, 0) -> 0
div(0, y) -> 0
div(s(x), s(y)) -> if(lt(x, y), 0, s(div(-(x, y), s(y))))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


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

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

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 7
FwdInst
             ...
               →DP Problem 9
Dependency Graph
       →DP Problem 3
Nar


Dependency Pair:


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
lt(x, 0) -> false
lt(0, s(y)) -> true
lt(s(x), s(y)) -> lt(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
div(x, 0) -> 0
div(0, y) -> 0
div(s(x), s(y)) -> if(lt(x, y), 0, s(div(-(x, y), s(y))))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Narrowing Transformation


Dependency Pair:

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


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
lt(x, 0) -> false
lt(0, s(y)) -> true
lt(s(x), s(y)) -> lt(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
div(x, 0) -> 0
div(0, y) -> 0
div(s(x), s(y)) -> if(lt(x, y), 0, s(div(-(x, y), s(y))))


Strategy:

innermost




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

DIV(s(x), s(y)) -> DIV(-(x, y), s(y))
three new Dependency Pairs are created:

DIV(s(x''), s(0)) -> DIV(x'', s(0))
DIV(s(0), s(s(y''))) -> DIV(0, s(s(y'')))
DIV(s(s(x'')), s(s(y''))) -> DIV(-(x'', y''), s(s(y'')))

The transformation is resulting in two new DP problems:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 10
Forward Instantiation Transformation
           →DP Problem 11
Nar


Dependency Pair:

DIV(s(x''), s(0)) -> DIV(x'', s(0))


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
lt(x, 0) -> false
lt(0, s(y)) -> true
lt(s(x), s(y)) -> lt(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
div(x, 0) -> 0
div(0, y) -> 0
div(s(x), s(y)) -> if(lt(x, y), 0, s(div(-(x, y), s(y))))


Strategy:

innermost




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

DIV(s(x''), s(0)) -> DIV(x'', s(0))
one new Dependency Pair is created:

DIV(s(s(x'''')), s(0)) -> DIV(s(x''''), s(0))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 10
FwdInst
             ...
               →DP Problem 12
Polynomial Ordering
           →DP Problem 11
Nar


Dependency Pair:

DIV(s(s(x'''')), s(0)) -> DIV(s(x''''), s(0))


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
lt(x, 0) -> false
lt(0, s(y)) -> true
lt(s(x), s(y)) -> lt(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
div(x, 0) -> 0
div(0, y) -> 0
div(s(x), s(y)) -> if(lt(x, y), 0, s(div(-(x, y), s(y))))


Strategy:

innermost




The following dependency pair can be strictly oriented:

DIV(s(s(x'''')), s(0)) -> DIV(s(x''''), s(0))


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0)=  0  
  POL(DIV(x1, x2))=  1 + x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 10
FwdInst
             ...
               →DP Problem 15
Dependency Graph
           →DP Problem 11
Nar


Dependency Pair:


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
lt(x, 0) -> false
lt(0, s(y)) -> true
lt(s(x), s(y)) -> lt(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
div(x, 0) -> 0
div(0, y) -> 0
div(s(x), s(y)) -> if(lt(x, y), 0, s(div(-(x, y), s(y))))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 10
FwdInst
           →DP Problem 11
Narrowing Transformation


Dependency Pair:

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


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
lt(x, 0) -> false
lt(0, s(y)) -> true
lt(s(x), s(y)) -> lt(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
div(x, 0) -> 0
div(0, y) -> 0
div(s(x), s(y)) -> if(lt(x, y), 0, s(div(-(x, y), s(y))))


Strategy:

innermost




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

DIV(s(s(x'')), s(s(y''))) -> DIV(-(x'', y''), s(s(y'')))
three new Dependency Pairs are created:

DIV(s(s(x''')), s(s(0))) -> DIV(x''', s(s(0)))
DIV(s(s(0)), s(s(s(y')))) -> DIV(0, s(s(s(y'))))
DIV(s(s(s(x'))), s(s(s(y')))) -> DIV(-(x', y'), s(s(s(y'))))

The transformation is resulting in two new DP problems:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 10
FwdInst
           →DP Problem 11
Nar
             ...
               →DP Problem 13
Polynomial Ordering


Dependency Pair:

DIV(s(s(x''')), s(s(0))) -> DIV(x''', s(s(0)))


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
lt(x, 0) -> false
lt(0, s(y)) -> true
lt(s(x), s(y)) -> lt(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
div(x, 0) -> 0
div(0, y) -> 0
div(s(x), s(y)) -> if(lt(x, y), 0, s(div(-(x, y), s(y))))


Strategy:

innermost




The following dependency pair can be strictly oriented:

DIV(s(s(x''')), s(s(0))) -> DIV(x''', s(s(0)))


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0)=  0  
  POL(DIV(x1, x2))=  1 + x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 10
FwdInst
           →DP Problem 11
Nar
             ...
               →DP Problem 14
Polynomial Ordering


Dependency Pair:

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


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
lt(x, 0) -> false
lt(0, s(y)) -> true
lt(s(x), s(y)) -> lt(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
div(x, 0) -> 0
div(0, y) -> 0
div(s(x), s(y)) -> if(lt(x, y), 0, s(div(-(x, y), s(y))))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


Additionally, the following usable rules for innermost can be oriented:

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


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0)=  0  
  POL(DIV(x1, x2))=  1 + x1 + x2  
  POL(s(x1))=  1 + x1  
  POL(-(x1, x2))=  x1  

resulting in one new DP problem.


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