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

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

EXP(x, s(y)) -> *'(x, exp(x, y))
EXP(x, s(y)) -> EXP(x, y)
*'(s(x), y) -> *'(x, y)
-'(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
FwdInst


Dependency Pair:

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


Rules:


exp(x, 0) -> s(0)
exp(x, s(y)) -> *(x, exp(x, y))
*(0, y) -> 0
*(s(x), y) -> +(y, *(x, y))
-(0, y) -> 0
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)


Strategy:

innermost




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

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

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


Dependency Pair:

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


Rules:


exp(x, 0) -> s(0)
exp(x, s(y)) -> *(x, exp(x, y))
*(0, y) -> 0
*(s(x), y) -> +(y, *(x, y))
-(0, y) -> 0
-(x, 0) -> x
-(s(x), s(y)) -> -(x, 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'')), y'') -> *'(s(x''), y'')
one new Dependency Pair is created:

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


Dependency Pair:

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


Rules:


exp(x, 0) -> s(0)
exp(x, s(y)) -> *(x, exp(x, y))
*(0, y) -> 0
*(s(x), y) -> +(y, *(x, y))
-(0, y) -> 0
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


Dependency Pair:


Rules:


exp(x, 0) -> s(0)
exp(x, s(y)) -> *(x, exp(x, y))
*(0, y) -> 0
*(s(x), y) -> +(y, *(x, y))
-(0, y) -> 0
-(x, 0) -> x
-(s(x), s(y)) -> -(x, 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
FwdInst


Dependency Pair:

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


Rules:


exp(x, 0) -> s(0)
exp(x, s(y)) -> *(x, exp(x, y))
*(0, y) -> 0
*(s(x), y) -> +(y, *(x, y))
-(0, y) -> 0
-(x, 0) -> x
-(s(x), s(y)) -> -(x, 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 2
FwdInst
           →DP Problem 7
Forward Instantiation Transformation
       →DP Problem 3
FwdInst


Dependency Pair:

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


Rules:


exp(x, 0) -> s(0)
exp(x, s(y)) -> *(x, exp(x, y))
*(0, y) -> 0
*(s(x), y) -> +(y, *(x, y))
-(0, y) -> 0
-(x, 0) -> x
-(s(x), s(y)) -> -(x, 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 2
FwdInst
           →DP Problem 7
FwdInst
             ...
               →DP Problem 8
Polynomial Ordering
       →DP Problem 3
FwdInst


Dependency Pair:

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


Rules:


exp(x, 0) -> s(0)
exp(x, s(y)) -> *(x, exp(x, y))
*(0, y) -> 0
*(s(x), y) -> +(y, *(x, y))
-(0, y) -> 0
-(x, 0) -> x
-(s(x), s(y)) -> -(x, 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 w.r.t. to the implicit AFS 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 2
FwdInst
           →DP Problem 7
FwdInst
             ...
               →DP Problem 9
Dependency Graph
       →DP Problem 3
FwdInst


Dependency Pair:


Rules:


exp(x, 0) -> s(0)
exp(x, s(y)) -> *(x, exp(x, y))
*(0, y) -> 0
*(s(x), y) -> +(y, *(x, y))
-(0, y) -> 0
-(x, 0) -> x
-(s(x), s(y)) -> -(x, 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
Forward Instantiation Transformation


Dependency Pair:

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


Rules:


exp(x, 0) -> s(0)
exp(x, s(y)) -> *(x, exp(x, y))
*(0, y) -> 0
*(s(x), y) -> +(y, *(x, y))
-(0, y) -> 0
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)


Strategy:

innermost




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

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

EXP(x'', s(s(y''))) -> EXP(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
FwdInst
           →DP Problem 10
Forward Instantiation Transformation


Dependency Pair:

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


Rules:


exp(x, 0) -> s(0)
exp(x, s(y)) -> *(x, exp(x, y))
*(0, y) -> 0
*(s(x), y) -> +(y, *(x, y))
-(0, y) -> 0
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)


Strategy:

innermost




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

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

EXP(x'''', s(s(s(y'''')))) -> EXP(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 3
FwdInst
           →DP Problem 10
FwdInst
             ...
               →DP Problem 11
Polynomial Ordering


Dependency Pair:

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


Rules:


exp(x, 0) -> s(0)
exp(x, s(y)) -> *(x, exp(x, y))
*(0, y) -> 0
*(s(x), y) -> +(y, *(x, y))
-(0, y) -> 0
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)


Strategy:

innermost




The following dependency pair can be strictly oriented:

EXP(x'''', s(s(s(y'''')))) -> EXP(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(EXP(x1, x2))=  1 + x2  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
           →DP Problem 10
FwdInst
             ...
               →DP Problem 12
Dependency Graph


Dependency Pair:


Rules:


exp(x, 0) -> s(0)
exp(x, s(y)) -> *(x, exp(x, y))
*(0, y) -> 0
*(s(x), y) -> +(y, *(x, y))
-(0, y) -> 0
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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