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
         ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS


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




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(*'(x1, x2))=  x1 + x2  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.
Used Argument Filtering System:
*'(x1, x2) -> *'(x1, x2)
s(x1) -> s(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 4
             ↳Dependency Graph
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS


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
         ↳AFS
       →DP Problem 2
         ↳Argument Filtering and Ordering
       →DP Problem 3
         ↳AFS


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




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(-'(x1, x2))=  x1 + x2  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.
Used Argument Filtering System:
-'(x1, x2) -> -'(x1, x2)
s(x1) -> s(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
           →DP Problem 5
             ↳Dependency Graph
       →DP Problem 3
         ↳AFS


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
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Argument Filtering and Ordering


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




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(EXP(x1, x2))=  x1 + x2  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.
Used Argument Filtering System:
EXP(x1, x2) -> EXP(x1, x2)
s(x1) -> s(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
           →DP Problem 6
             ↳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