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

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

GE(s(x), s(y)) -> GE(x, y)
MINUS(s(x), s(y)) -> MINUS(x, y)
DIV(x, y) -> IFY(ge(y, s(0)), x, y)
DIV(x, y) -> GE(y, s(0))
IFY(true, x, y) -> IF(ge(x, y), x, y)
IFY(true, x, y) -> GE(x, y)
IF(true, x, y) -> DIV(minus(x, y), y)
IF(true, x, y) -> MINUS(x, y)

Furthermore, R contains three SCCs.


   R
DPs
       →DP Problem 1
Polynomial Ordering
       →DP Problem 2
Polo
       →DP Problem 3
Nar


Dependency Pair:

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


Rules:


ge(x, 0) -> true
ge(0, s(x)) -> false
ge(s(x), s(y)) -> ge(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
div(x, y) -> ify(ge(y, s(0)), x, y)
ify(false, x, y) -> divByZeroError
ify(true, x, y) -> if(ge(x, y), x, y)
if(false, x, y) -> 0
if(true, x, y) -> s(div(minus(x, y), y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


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

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

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
           →DP Problem 4
Dependency Graph
       →DP Problem 2
Polo
       →DP Problem 3
Nar


Dependency Pair:


Rules:


ge(x, 0) -> true
ge(0, s(x)) -> false
ge(s(x), s(y)) -> ge(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
div(x, y) -> ify(ge(y, s(0)), x, y)
ify(false, x, y) -> divByZeroError
ify(true, x, y) -> if(ge(x, y), x, y)
if(false, x, y) -> 0
if(true, x, y) -> s(div(minus(x, y), y))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polynomial Ordering
       →DP Problem 3
Nar


Dependency Pair:

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


Rules:


ge(x, 0) -> true
ge(0, s(x)) -> false
ge(s(x), s(y)) -> ge(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
div(x, y) -> ify(ge(y, s(0)), x, y)
ify(false, x, y) -> divByZeroError
ify(true, x, y) -> if(ge(x, y), x, y)
if(false, x, y) -> 0
if(true, x, y) -> s(div(minus(x, y), y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


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

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

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
           →DP Problem 5
Dependency Graph
       →DP Problem 3
Nar


Dependency Pair:


Rules:


ge(x, 0) -> true
ge(0, s(x)) -> false
ge(s(x), s(y)) -> ge(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
div(x, y) -> ify(ge(y, s(0)), x, y)
ify(false, x, y) -> divByZeroError
ify(true, x, y) -> if(ge(x, y), x, y)
if(false, x, y) -> 0
if(true, x, y) -> s(div(minus(x, y), y))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pairs:

IF(true, x, y) -> DIV(minus(x, y), y)
IFY(true, x, y) -> IF(ge(x, y), x, y)
DIV(x, y) -> IFY(ge(y, s(0)), x, y)


Rules:


ge(x, 0) -> true
ge(0, s(x)) -> false
ge(s(x), s(y)) -> ge(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
div(x, y) -> ify(ge(y, s(0)), x, y)
ify(false, x, y) -> divByZeroError
ify(true, x, y) -> if(ge(x, y), x, y)
if(false, x, y) -> 0
if(true, x, y) -> s(div(minus(x, y), y))


Strategy:

innermost




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

DIV(x, y) -> IFY(ge(y, s(0)), x, y)
two new Dependency Pairs are created:

DIV(x, 0) -> IFY(false, x, 0)
DIV(x, s(x'')) -> IFY(ge(x'', 0), x, s(x''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Nar
           →DP Problem 6
Rewriting Transformation


Dependency Pairs:

IFY(true, x, y) -> IF(ge(x, y), x, y)
DIV(x, s(x'')) -> IFY(ge(x'', 0), x, s(x''))
IF(true, x, y) -> DIV(minus(x, y), y)


Rules:


ge(x, 0) -> true
ge(0, s(x)) -> false
ge(s(x), s(y)) -> ge(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
div(x, y) -> ify(ge(y, s(0)), x, y)
ify(false, x, y) -> divByZeroError
ify(true, x, y) -> if(ge(x, y), x, y)
if(false, x, y) -> 0
if(true, x, y) -> s(div(minus(x, y), y))


Strategy:

innermost




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

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

DIV(x, s(x'')) -> IFY(true, x, s(x''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Nar
           →DP Problem 6
Rw
             ...
               →DP Problem 7
Narrowing Transformation


Dependency Pairs:

DIV(x, s(x'')) -> IFY(true, x, s(x''))
IF(true, x, y) -> DIV(minus(x, y), y)
IFY(true, x, y) -> IF(ge(x, y), x, y)


Rules:


ge(x, 0) -> true
ge(0, s(x)) -> false
ge(s(x), s(y)) -> ge(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
div(x, y) -> ify(ge(y, s(0)), x, y)
ify(false, x, y) -> divByZeroError
ify(true, x, y) -> if(ge(x, y), x, y)
if(false, x, y) -> 0
if(true, x, y) -> s(div(minus(x, y), y))


Strategy:

innermost




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

IFY(true, x, y) -> IF(ge(x, y), x, y)
three new Dependency Pairs are created:

IFY(true, x'', 0) -> IF(true, x'', 0)
IFY(true, 0, s(x'')) -> IF(false, 0, s(x''))
IFY(true, s(x''), s(y'')) -> IF(ge(x'', y''), s(x''), s(y''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Nar
           →DP Problem 6
Rw
             ...
               →DP Problem 8
Instantiation Transformation


Dependency Pairs:

IF(true, x, y) -> DIV(minus(x, y), y)
IFY(true, s(x''), s(y'')) -> IF(ge(x'', y''), s(x''), s(y''))
DIV(x, s(x'')) -> IFY(true, x, s(x''))


Rules:


ge(x, 0) -> true
ge(0, s(x)) -> false
ge(s(x), s(y)) -> ge(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
div(x, y) -> ify(ge(y, s(0)), x, y)
ify(false, x, y) -> divByZeroError
ify(true, x, y) -> if(ge(x, y), x, y)
if(false, x, y) -> 0
if(true, x, y) -> s(div(minus(x, y), y))


Strategy:

innermost




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

IF(true, x, y) -> DIV(minus(x, y), y)
one new Dependency Pair is created:

IF(true, s(x'''''), s(y'''')) -> DIV(minus(s(x'''''), s(y'''')), s(y''''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Nar
           →DP Problem 6
Rw
             ...
               →DP Problem 9
Rewriting Transformation


Dependency Pairs:

DIV(x, s(x'')) -> IFY(true, x, s(x''))
IF(true, s(x'''''), s(y'''')) -> DIV(minus(s(x'''''), s(y'''')), s(y''''))
IFY(true, s(x''), s(y'')) -> IF(ge(x'', y''), s(x''), s(y''))


Rules:


ge(x, 0) -> true
ge(0, s(x)) -> false
ge(s(x), s(y)) -> ge(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
div(x, y) -> ify(ge(y, s(0)), x, y)
ify(false, x, y) -> divByZeroError
ify(true, x, y) -> if(ge(x, y), x, y)
if(false, x, y) -> 0
if(true, x, y) -> s(div(minus(x, y), y))


Strategy:

innermost




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

IF(true, s(x'''''), s(y'''')) -> DIV(minus(s(x'''''), s(y'''')), s(y''''))
one new Dependency Pair is created:

IF(true, s(x'''''), s(y'''')) -> DIV(minus(x''''', y''''), s(y''''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Nar
           →DP Problem 6
Rw
             ...
               →DP Problem 10
Forward Instantiation Transformation


Dependency Pairs:

IF(true, s(x'''''), s(y'''')) -> DIV(minus(x''''', y''''), s(y''''))
IFY(true, s(x''), s(y'')) -> IF(ge(x'', y''), s(x''), s(y''))
DIV(x, s(x'')) -> IFY(true, x, s(x''))


Rules:


ge(x, 0) -> true
ge(0, s(x)) -> false
ge(s(x), s(y)) -> ge(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
div(x, y) -> ify(ge(y, s(0)), x, y)
ify(false, x, y) -> divByZeroError
ify(true, x, y) -> if(ge(x, y), x, y)
if(false, x, y) -> 0
if(true, x, y) -> s(div(minus(x, y), y))


Strategy:

innermost




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

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

DIV(s(x'''''), s(x'''')) -> IFY(true, s(x'''''), s(x''''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Nar
           →DP Problem 6
Rw
             ...
               →DP Problem 11
Narrowing Transformation


Dependency Pairs:

IFY(true, s(x''), s(y'')) -> IF(ge(x'', y''), s(x''), s(y''))
DIV(s(x'''''), s(x'''')) -> IFY(true, s(x'''''), s(x''''))
IF(true, s(x'''''), s(y'''')) -> DIV(minus(x''''', y''''), s(y''''))


Rules:


ge(x, 0) -> true
ge(0, s(x)) -> false
ge(s(x), s(y)) -> ge(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
div(x, y) -> ify(ge(y, s(0)), x, y)
ify(false, x, y) -> divByZeroError
ify(true, x, y) -> if(ge(x, y), x, y)
if(false, x, y) -> 0
if(true, x, y) -> s(div(minus(x, y), y))


Strategy:

innermost




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

IF(true, s(x'''''), s(y'''')) -> DIV(minus(x''''', y''''), s(y''''))
two new Dependency Pairs are created:

IF(true, s(x''''''), s(0)) -> DIV(x'''''', s(0))
IF(true, s(s(x')), s(s(y'))) -> DIV(minus(x', y'), s(s(y')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Nar
           →DP Problem 6
Rw
             ...
               →DP Problem 12
Instantiation Transformation


Dependency Pairs:

IF(true, s(s(x')), s(s(y'))) -> DIV(minus(x', y'), s(s(y')))
DIV(s(x'''''), s(x'''')) -> IFY(true, s(x'''''), s(x''''))
IF(true, s(x''''''), s(0)) -> DIV(x'''''', s(0))
IFY(true, s(x''), s(y'')) -> IF(ge(x'', y''), s(x''), s(y''))


Rules:


ge(x, 0) -> true
ge(0, s(x)) -> false
ge(s(x), s(y)) -> ge(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
div(x, y) -> ify(ge(y, s(0)), x, y)
ify(false, x, y) -> divByZeroError
ify(true, x, y) -> if(ge(x, y), x, y)
if(false, x, y) -> 0
if(true, x, y) -> s(div(minus(x, y), y))


Strategy:

innermost




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

DIV(s(x'''''), s(x'''')) -> IFY(true, s(x'''''), s(x''''))
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Nar
           →DP Problem 6
Rw
             ...
               →DP Problem 13
Instantiation Transformation


Dependency Pairs:

DIV(s(x''''''), s(0)) -> IFY(true, s(x''''''), s(0))
IF(true, s(x''''''), s(0)) -> DIV(x'''''', s(0))
IFY(true, s(x''), s(y'')) -> IF(ge(x'', y''), s(x''), s(y''))
DIV(s(x''''''), s(s(y'''))) -> IFY(true, s(x''''''), s(s(y''')))
IF(true, s(s(x')), s(s(y'))) -> DIV(minus(x', y'), s(s(y')))


Rules:


ge(x, 0) -> true
ge(0, s(x)) -> false
ge(s(x), s(y)) -> ge(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
div(x, y) -> ify(ge(y, s(0)), x, y)
ify(false, x, y) -> divByZeroError
ify(true, x, y) -> if(ge(x, y), x, y)
if(false, x, y) -> 0
if(true, x, y) -> s(div(minus(x, y), y))


Strategy:

innermost




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

IFY(true, s(x''), s(y'')) -> IF(ge(x'', y''), s(x''), s(y''))
two new Dependency Pairs are created:

IFY(true, s(x'''), s(0)) -> IF(ge(x''', 0), s(x'''), s(0))
IFY(true, s(x'''), s(s(y'''''))) -> IF(ge(x''', s(y''''')), s(x'''), s(s(y''''')))

The transformation is resulting in two new DP problems:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Nar
           →DP Problem 6
Rw
             ...
               →DP Problem 14
Rewriting Transformation


Dependency Pairs:

IF(true, s(x''''''), s(0)) -> DIV(x'''''', s(0))
IFY(true, s(x'''), s(0)) -> IF(ge(x''', 0), s(x'''), s(0))
DIV(s(x''''''), s(0)) -> IFY(true, s(x''''''), s(0))


Rules:


ge(x, 0) -> true
ge(0, s(x)) -> false
ge(s(x), s(y)) -> ge(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
div(x, y) -> ify(ge(y, s(0)), x, y)
ify(false, x, y) -> divByZeroError
ify(true, x, y) -> if(ge(x, y), x, y)
if(false, x, y) -> 0
if(true, x, y) -> s(div(minus(x, y), y))


Strategy:

innermost




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

IFY(true, s(x'''), s(0)) -> IF(ge(x''', 0), s(x'''), s(0))
one new Dependency Pair is created:

IFY(true, s(x'''), s(0)) -> IF(true, s(x'''), s(0))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Nar
           →DP Problem 6
Rw
             ...
               →DP Problem 16
Forward Instantiation Transformation


Dependency Pairs:

IFY(true, s(x'''), s(0)) -> IF(true, s(x'''), s(0))
DIV(s(x''''''), s(0)) -> IFY(true, s(x''''''), s(0))
IF(true, s(x''''''), s(0)) -> DIV(x'''''', s(0))


Rules:


ge(x, 0) -> true
ge(0, s(x)) -> false
ge(s(x), s(y)) -> ge(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
div(x, y) -> ify(ge(y, s(0)), x, y)
ify(false, x, y) -> divByZeroError
ify(true, x, y) -> if(ge(x, y), x, y)
if(false, x, y) -> 0
if(true, x, y) -> s(div(minus(x, y), y))


Strategy:

innermost




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

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

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Nar
           →DP Problem 6
Rw
             ...
               →DP Problem 18
Forward Instantiation Transformation


Dependency Pairs:

DIV(s(x''''''), s(0)) -> IFY(true, s(x''''''), s(0))
IF(true, s(s(x'''''''')), s(0)) -> DIV(s(x''''''''), s(0))
IFY(true, s(x'''), s(0)) -> IF(true, s(x'''), s(0))


Rules:


ge(x, 0) -> true
ge(0, s(x)) -> false
ge(s(x), s(y)) -> ge(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
div(x, y) -> ify(ge(y, s(0)), x, y)
ify(false, x, y) -> divByZeroError
ify(true, x, y) -> if(ge(x, y), x, y)
if(false, x, y) -> 0
if(true, x, y) -> s(div(minus(x, y), y))


Strategy:

innermost




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

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

IFY(true, s(s(x'''''''''')), s(0)) -> IF(true, s(s(x'''''''''')), s(0))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Nar
           →DP Problem 6
Rw
             ...
               →DP Problem 20
Polynomial Ordering


Dependency Pairs:

IF(true, s(s(x'''''''')), s(0)) -> DIV(s(x''''''''), s(0))
IFY(true, s(s(x'''''''''')), s(0)) -> IF(true, s(s(x'''''''''')), s(0))
DIV(s(x''''''), s(0)) -> IFY(true, s(x''''''), s(0))


Rules:


ge(x, 0) -> true
ge(0, s(x)) -> false
ge(s(x), s(y)) -> ge(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
div(x, y) -> ify(ge(y, s(0)), x, y)
ify(false, x, y) -> divByZeroError
ify(true, x, y) -> if(ge(x, y), x, y)
if(false, x, y) -> 0
if(true, x, y) -> s(div(minus(x, y), y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

IF(true, 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(IFY(x1, x2, x3))=  x2  
  POL(0)=  0  
  POL(DIV(x1, x2))=  x1  
  POL(true)=  0  
  POL(s(x1))=  1 + x1  
  POL(IF(x1, x2, x3))=  x2  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Nar
           →DP Problem 6
Rw
             ...
               →DP Problem 22
Dependency Graph


Dependency Pairs:

IFY(true, s(s(x'''''''''')), s(0)) -> IF(true, s(s(x'''''''''')), s(0))
DIV(s(x''''''), s(0)) -> IFY(true, s(x''''''), s(0))


Rules:


ge(x, 0) -> true
ge(0, s(x)) -> false
ge(s(x), s(y)) -> ge(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
div(x, y) -> ify(ge(y, s(0)), x, y)
ify(false, x, y) -> divByZeroError
ify(true, x, y) -> if(ge(x, y), x, y)
if(false, x, y) -> 0
if(true, x, y) -> s(div(minus(x, y), y))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Nar
           →DP Problem 6
Rw
             ...
               →DP Problem 15
Forward Instantiation Transformation


Dependency Pairs:

IF(true, s(s(x')), s(s(y'))) -> DIV(minus(x', y'), s(s(y')))
IFY(true, s(x'''), s(s(y'''''))) -> IF(ge(x''', s(y''''')), s(x'''), s(s(y''''')))
DIV(s(x''''''), s(s(y'''))) -> IFY(true, s(x''''''), s(s(y''')))


Rules:


ge(x, 0) -> true
ge(0, s(x)) -> false
ge(s(x), s(y)) -> ge(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
div(x, y) -> ify(ge(y, s(0)), x, y)
ify(false, x, y) -> divByZeroError
ify(true, x, y) -> if(ge(x, y), x, y)
if(false, x, y) -> 0
if(true, x, y) -> s(div(minus(x, y), y))


Strategy:

innermost




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

IFY(true, s(x'''), s(s(y'''''))) -> IF(ge(x''', s(y''''')), s(x'''), s(s(y''''')))
one new Dependency Pair is created:

IFY(true, s(s(x'0')), s(s(y''''''))) -> IF(ge(s(x'0'), s(y'''''')), s(s(x'0')), s(s(y'''''')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Nar
           →DP Problem 6
Rw
             ...
               →DP Problem 17
Rewriting Transformation


Dependency Pairs:

IFY(true, s(s(x'0')), s(s(y''''''))) -> IF(ge(s(x'0'), s(y'''''')), s(s(x'0')), s(s(y'''''')))
DIV(s(x''''''), s(s(y'''))) -> IFY(true, s(x''''''), s(s(y''')))
IF(true, s(s(x')), s(s(y'))) -> DIV(minus(x', y'), s(s(y')))


Rules:


ge(x, 0) -> true
ge(0, s(x)) -> false
ge(s(x), s(y)) -> ge(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
div(x, y) -> ify(ge(y, s(0)), x, y)
ify(false, x, y) -> divByZeroError
ify(true, x, y) -> if(ge(x, y), x, y)
if(false, x, y) -> 0
if(true, x, y) -> s(div(minus(x, y), y))


Strategy:

innermost




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

IFY(true, s(s(x'0')), s(s(y''''''))) -> IF(ge(s(x'0'), s(y'''''')), s(s(x'0')), s(s(y'''''')))
one new Dependency Pair is created:

IFY(true, s(s(x'0')), s(s(y''''''))) -> IF(ge(x'0', y''''''), s(s(x'0')), s(s(y'''''')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Nar
           →DP Problem 6
Rw
             ...
               →DP Problem 19
Polynomial Ordering


Dependency Pairs:

IF(true, s(s(x')), s(s(y'))) -> DIV(minus(x', y'), s(s(y')))
IFY(true, s(s(x'0')), s(s(y''''''))) -> IF(ge(x'0', y''''''), s(s(x'0')), s(s(y'''''')))
DIV(s(x''''''), s(s(y'''))) -> IFY(true, s(x''''''), s(s(y''')))


Rules:


ge(x, 0) -> true
ge(0, s(x)) -> false
ge(s(x), s(y)) -> ge(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
div(x, y) -> ify(ge(y, s(0)), x, y)
ify(false, x, y) -> divByZeroError
ify(true, x, y) -> if(ge(x, y), x, y)
if(false, x, y) -> 0
if(true, x, y) -> s(div(minus(x, y), y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

IF(true, s(s(x')), s(s(y'))) -> DIV(minus(x', y'), s(s(y')))


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

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
ge(x, 0) -> true
ge(0, s(x)) -> false
ge(s(x), s(y)) -> ge(x, y)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(IFY(x1, x2, x3))=  x2  
  POL(0)=  0  
  POL(false)=  0  
  POL(DIV(x1, x2))=  x1  
  POL(minus(x1, x2))=  x1  
  POL(true)=  0  
  POL(s(x1))=  1 + x1  
  POL(ge(x1, x2))=  0  
  POL(IF(x1, x2, x3))=  x2  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Nar
           →DP Problem 6
Rw
             ...
               →DP Problem 21
Dependency Graph


Dependency Pairs:

IFY(true, s(s(x'0')), s(s(y''''''))) -> IF(ge(x'0', y''''''), s(s(x'0')), s(s(y'''''')))
DIV(s(x''''''), s(s(y'''))) -> IFY(true, s(x''''''), s(s(y''')))


Rules:


ge(x, 0) -> true
ge(0, s(x)) -> false
ge(s(x), s(y)) -> ge(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
div(x, y) -> ify(ge(y, s(0)), x, y)
ify(false, x, y) -> divByZeroError
ify(true, x, y) -> if(ge(x, y), x, y)
if(false, x, y) -> 0
if(true, x, y) -> s(div(minus(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