Term Rewriting System R:
[x, y]
D(t) -> 1
D(constant) -> 0
D(+(x, y)) -> +(D(x), D(y))
D(*(x, y)) -> +(*(y, D(x)), *(x, D(y)))
D(-(x, y)) -> -(D(x), D(y))
D(minus(x)) -> minus(D(x))
D(div(x, y)) -> -(div(D(x), y), div(*(x, D(y)), pow(y, 2)))
D(ln(x)) -> div(D(x), x)
D(pow(x, y)) -> +(*(*(y, pow(x, -(y, 1))), D(x)), *(*(pow(x, y), ln(x)), D(y)))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

D'(+(x, y)) -> D'(x)
D'(+(x, y)) -> D'(y)
D'(*(x, y)) -> D'(x)
D'(*(x, y)) -> D'(y)
D'(-(x, y)) -> D'(x)
D'(-(x, y)) -> D'(y)
D'(minus(x)) -> D'(x)
D'(div(x, y)) -> D'(x)
D'(div(x, y)) -> D'(y)
D'(ln(x)) -> D'(x)
D'(pow(x, y)) -> D'(x)
D'(pow(x, y)) -> D'(y)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering


Dependency Pairs:

D'(pow(x, y)) -> D'(y)
D'(pow(x, y)) -> D'(x)
D'(ln(x)) -> D'(x)
D'(div(x, y)) -> D'(y)
D'(div(x, y)) -> D'(x)
D'(minus(x)) -> D'(x)
D'(-(x, y)) -> D'(y)
D'(-(x, y)) -> D'(x)
D'(*(x, y)) -> D'(y)
D'(*(x, y)) -> D'(x)
D'(+(x, y)) -> D'(y)
D'(+(x, y)) -> D'(x)


Rules:


D(t) -> 1
D(constant) -> 0
D(+(x, y)) -> +(D(x), D(y))
D(*(x, y)) -> +(*(y, D(x)), *(x, D(y)))
D(-(x, y)) -> -(D(x), D(y))
D(minus(x)) -> minus(D(x))
D(div(x, y)) -> -(div(D(x), y), div(*(x, D(y)), pow(y, 2)))
D(ln(x)) -> div(D(x), x)
D(pow(x, y)) -> +(*(*(y, pow(x, -(y, 1))), D(x)), *(*(pow(x, y), ln(x)), D(y)))





The following dependency pair can be strictly oriented:

D'(minus(x)) -> D'(x)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(D'(x1))=  x1  
  POL(pow(x1, x2))=  x1 + x2  
  POL(minus(x1))=  1 + x1  
  POL(*(x1, x2))=  x1 + x2  
  POL(-(x1, x2))=  x1 + x2  
  POL(div(x1, x2))=  x1 + x2  
  POL(+(x1, x2))=  x1 + x2  
  POL(ln(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳Polynomial Ordering


Dependency Pairs:

D'(pow(x, y)) -> D'(y)
D'(pow(x, y)) -> D'(x)
D'(ln(x)) -> D'(x)
D'(div(x, y)) -> D'(y)
D'(div(x, y)) -> D'(x)
D'(-(x, y)) -> D'(y)
D'(-(x, y)) -> D'(x)
D'(*(x, y)) -> D'(y)
D'(*(x, y)) -> D'(x)
D'(+(x, y)) -> D'(y)
D'(+(x, y)) -> D'(x)


Rules:


D(t) -> 1
D(constant) -> 0
D(+(x, y)) -> +(D(x), D(y))
D(*(x, y)) -> +(*(y, D(x)), *(x, D(y)))
D(-(x, y)) -> -(D(x), D(y))
D(minus(x)) -> minus(D(x))
D(div(x, y)) -> -(div(D(x), y), div(*(x, D(y)), pow(y, 2)))
D(ln(x)) -> div(D(x), x)
D(pow(x, y)) -> +(*(*(y, pow(x, -(y, 1))), D(x)), *(*(pow(x, y), ln(x)), D(y)))





The following dependency pairs can be strictly oriented:

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


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(D'(x1))=  x1  
  POL(pow(x1, x2))=  x1 + x2  
  POL(*(x1, x2))=  x1 + x2  
  POL(-(x1, x2))=  1 + x1 + x2  
  POL(div(x1, x2))=  x1 + x2  
  POL(+(x1, x2))=  x1 + x2  
  POL(ln(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳Polo
             ...
               →DP Problem 3
                 ↳Polynomial Ordering


Dependency Pairs:

D'(pow(x, y)) -> D'(y)
D'(pow(x, y)) -> D'(x)
D'(ln(x)) -> D'(x)
D'(div(x, y)) -> D'(y)
D'(div(x, y)) -> D'(x)
D'(*(x, y)) -> D'(y)
D'(*(x, y)) -> D'(x)
D'(+(x, y)) -> D'(y)
D'(+(x, y)) -> D'(x)


Rules:


D(t) -> 1
D(constant) -> 0
D(+(x, y)) -> +(D(x), D(y))
D(*(x, y)) -> +(*(y, D(x)), *(x, D(y)))
D(-(x, y)) -> -(D(x), D(y))
D(minus(x)) -> minus(D(x))
D(div(x, y)) -> -(div(D(x), y), div(*(x, D(y)), pow(y, 2)))
D(ln(x)) -> div(D(x), x)
D(pow(x, y)) -> +(*(*(y, pow(x, -(y, 1))), D(x)), *(*(pow(x, y), ln(x)), D(y)))





The following dependency pairs can be strictly oriented:

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


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(D'(x1))=  x1  
  POL(pow(x1, x2))=  x1 + x2  
  POL(*(x1, x2))=  1 + x1 + x2  
  POL(div(x1, x2))=  x1 + x2  
  POL(+(x1, x2))=  x1 + x2  
  POL(ln(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳Polo
             ...
               →DP Problem 4
                 ↳Polynomial Ordering


Dependency Pairs:

D'(pow(x, y)) -> D'(y)
D'(pow(x, y)) -> D'(x)
D'(ln(x)) -> D'(x)
D'(div(x, y)) -> D'(y)
D'(div(x, y)) -> D'(x)
D'(+(x, y)) -> D'(y)
D'(+(x, y)) -> D'(x)


Rules:


D(t) -> 1
D(constant) -> 0
D(+(x, y)) -> +(D(x), D(y))
D(*(x, y)) -> +(*(y, D(x)), *(x, D(y)))
D(-(x, y)) -> -(D(x), D(y))
D(minus(x)) -> minus(D(x))
D(div(x, y)) -> -(div(D(x), y), div(*(x, D(y)), pow(y, 2)))
D(ln(x)) -> div(D(x), x)
D(pow(x, y)) -> +(*(*(y, pow(x, -(y, 1))), D(x)), *(*(pow(x, y), ln(x)), D(y)))





The following dependency pairs can be strictly oriented:

D'(+(x, y)) -> D'(y)
D'(+(x, y)) -> D'(x)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(D'(x1))=  x1  
  POL(pow(x1, x2))=  x1 + x2  
  POL(div(x1, x2))=  x1 + x2  
  POL(+(x1, x2))=  1 + x1 + x2  
  POL(ln(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳Polo
             ...
               →DP Problem 5
                 ↳Polynomial Ordering


Dependency Pairs:

D'(pow(x, y)) -> D'(y)
D'(pow(x, y)) -> D'(x)
D'(ln(x)) -> D'(x)
D'(div(x, y)) -> D'(y)
D'(div(x, y)) -> D'(x)


Rules:


D(t) -> 1
D(constant) -> 0
D(+(x, y)) -> +(D(x), D(y))
D(*(x, y)) -> +(*(y, D(x)), *(x, D(y)))
D(-(x, y)) -> -(D(x), D(y))
D(minus(x)) -> minus(D(x))
D(div(x, y)) -> -(div(D(x), y), div(*(x, D(y)), pow(y, 2)))
D(ln(x)) -> div(D(x), x)
D(pow(x, y)) -> +(*(*(y, pow(x, -(y, 1))), D(x)), *(*(pow(x, y), ln(x)), D(y)))





The following dependency pairs can be strictly oriented:

D'(pow(x, y)) -> D'(y)
D'(pow(x, y)) -> D'(x)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(D'(x1))=  x1  
  POL(pow(x1, x2))=  1 + x1 + x2  
  POL(div(x1, x2))=  x1 + x2  
  POL(ln(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳Polo
             ...
               →DP Problem 6
                 ↳Polynomial Ordering


Dependency Pairs:

D'(ln(x)) -> D'(x)
D'(div(x, y)) -> D'(y)
D'(div(x, y)) -> D'(x)


Rules:


D(t) -> 1
D(constant) -> 0
D(+(x, y)) -> +(D(x), D(y))
D(*(x, y)) -> +(*(y, D(x)), *(x, D(y)))
D(-(x, y)) -> -(D(x), D(y))
D(minus(x)) -> minus(D(x))
D(div(x, y)) -> -(div(D(x), y), div(*(x, D(y)), pow(y, 2)))
D(ln(x)) -> div(D(x), x)
D(pow(x, y)) -> +(*(*(y, pow(x, -(y, 1))), D(x)), *(*(pow(x, y), ln(x)), D(y)))





The following dependency pair can be strictly oriented:

D'(ln(x)) -> D'(x)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(D'(x1))=  x1  
  POL(div(x1, x2))=  x1 + x2  
  POL(ln(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳Polo
             ...
               →DP Problem 7
                 ↳Polynomial Ordering


Dependency Pairs:

D'(div(x, y)) -> D'(y)
D'(div(x, y)) -> D'(x)


Rules:


D(t) -> 1
D(constant) -> 0
D(+(x, y)) -> +(D(x), D(y))
D(*(x, y)) -> +(*(y, D(x)), *(x, D(y)))
D(-(x, y)) -> -(D(x), D(y))
D(minus(x)) -> minus(D(x))
D(div(x, y)) -> -(div(D(x), y), div(*(x, D(y)), pow(y, 2)))
D(ln(x)) -> div(D(x), x)
D(pow(x, y)) -> +(*(*(y, pow(x, -(y, 1))), D(x)), *(*(pow(x, y), ln(x)), D(y)))





The following dependency pairs can be strictly oriented:

D'(div(x, y)) -> D'(y)
D'(div(x, y)) -> D'(x)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

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

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳Polo
             ...
               →DP Problem 8
                 ↳Dependency Graph


Dependency Pair:


Rules:


D(t) -> 1
D(constant) -> 0
D(+(x, y)) -> +(D(x), D(y))
D(*(x, y)) -> +(*(y, D(x)), *(x, D(y)))
D(-(x, y)) -> -(D(x), D(y))
D(minus(x)) -> minus(D(x))
D(div(x, y)) -> -(div(D(x), y), div(*(x, D(y)), pow(y, 2)))
D(ln(x)) -> div(D(x), x)
D(pow(x, y)) -> +(*(*(y, pow(x, -(y, 1))), D(x)), *(*(pow(x, y), ln(x)), D(y)))





Using the Dependency Graph resulted in no new DP problems.

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