Term Rewriting System R:
[X, Y]
minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

MINUS(s(X), s(Y)) -> P(minus(X, Y))
MINUS(s(X), s(Y)) -> MINUS(X, Y)
DIV(s(X), s(Y)) -> DIV(minus(X, Y), s(Y))
DIV(s(X), s(Y)) -> MINUS(X, Y)

Furthermore, R contains two SCCs.


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


Dependency Pair:

MINUS(s(X), s(Y)) -> MINUS(X, Y)


Rules:


minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> s(div(minus(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

MINUS(s(X), s(Y)) -> MINUS(X, Y)
one new Dependency Pair is created:

MINUS(s(s(X'')), s(s(Y''))) -> MINUS(s(X''), s(Y''))

The transformation is resulting in one new DP problem:



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


Dependency Pair:

MINUS(s(s(X'')), s(s(Y''))) -> MINUS(s(X''), s(Y''))


Rules:


minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> s(div(minus(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

MINUS(s(s(X'')), s(s(Y''))) -> MINUS(s(X''), s(Y''))
one new Dependency Pair is created:

MINUS(s(s(s(X''''))), s(s(s(Y'''')))) -> MINUS(s(s(X'''')), s(s(Y'''')))

The transformation is resulting in one new DP problem:



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


Dependency Pair:

MINUS(s(s(s(X''''))), s(s(s(Y'''')))) -> MINUS(s(s(X'''')), s(s(Y'''')))


Rules:


minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

MINUS(s(s(s(X''''))), s(s(s(Y'''')))) -> MINUS(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(MINUS(x1, x2))=  1 + x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.



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


Dependency Pair:


Rules:


minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> s(div(minus(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
Narrowing Transformation


Dependency Pair:

DIV(s(X), s(Y)) -> DIV(minus(X, Y), s(Y))


Rules:


minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> s(div(minus(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(minus(X, Y), s(Y))
two new Dependency Pairs are created:

DIV(s(X''), s(0)) -> DIV(X'', s(0))
DIV(s(s(X'')), s(s(Y''))) -> DIV(p(minus(X'', Y'')), s(s(Y'')))

The transformation is resulting in two new DP problems:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
           →DP Problem 6
Forward Instantiation Transformation
           →DP Problem 7
Nar


Dependency Pair:

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


Rules:


minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> s(div(minus(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
Nar
           →DP Problem 6
FwdInst
             ...
               →DP Problem 8
Polynomial Ordering
           →DP Problem 7
Nar


Dependency Pair:

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


Rules:


minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> s(div(minus(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
Nar
           →DP Problem 6
FwdInst
             ...
               →DP Problem 11
Dependency Graph
           →DP Problem 7
Nar


Dependency Pair:


Rules:


minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> s(div(minus(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
Nar
           →DP Problem 6
FwdInst
           →DP Problem 7
Narrowing Transformation


Dependency Pair:

DIV(s(s(X'')), s(s(Y''))) -> DIV(p(minus(X'', Y'')), s(s(Y'')))


Rules:


minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> s(div(minus(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(p(minus(X'', Y'')), s(s(Y'')))
two new Dependency Pairs are created:

DIV(s(s(X''')), s(s(0))) -> DIV(p(X'''), s(s(0)))
DIV(s(s(s(X'))), s(s(s(Y')))) -> DIV(p(p(minus(X', Y'))), s(s(s(Y'))))

The transformation is resulting in two new DP problems:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
           →DP Problem 6
FwdInst
           →DP Problem 7
Nar
             ...
               →DP Problem 9
Polynomial Ordering


Dependency Pair:

DIV(s(s(X''')), s(s(0))) -> DIV(p(X'''), s(s(0)))


Rules:


minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

DIV(s(s(X''')), s(s(0))) -> DIV(p(X'''), s(s(0)))


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

p(s(X)) -> X


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

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
           →DP Problem 6
FwdInst
           →DP Problem 7
Nar
             ...
               →DP Problem 10
Polynomial Ordering


Dependency Pair:

DIV(s(s(s(X'))), s(s(s(Y')))) -> DIV(p(p(minus(X', Y'))), s(s(s(Y'))))


Rules:


minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

DIV(s(s(s(X'))), s(s(s(Y')))) -> DIV(p(p(minus(X', Y'))), s(s(s(Y'))))


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

p(s(X)) -> X
minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))


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

resulting in one new DP problem.


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