Termination Proof using AProVE

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.

     ↳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.

     ↳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 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(minus(x₁)) = 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.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳Polynomial Ordering

Dependency Pairs:

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'(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(minus(x₁)) = 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)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳Polo

...

               →DP Problem 3

                 ↳Polynomial Ordering

Dependency Pairs:

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 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(minus(x₁)) = x₁

POL(*(x₁, x₂)) = x₁ + x₂

POL(-(x₁, x₂)) = x₁ + x₂

POL(div(x₁, x₂)) = 1 + x₁ + x₂

POL(+(x₁, x₂)) = x₁ + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳Polo

...

               →DP Problem 4

                 ↳Polynomial Ordering

Dependency Pairs:

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(minus(x₁)) = 1 + x₁

POL(*(x₁, x₂)) = x₁ + x₂

POL(-(x₁, x₂)) = x₁ + x₂

POL(+(x₁, x₂)) = x₁ + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳Polo

...

               →DP Problem 5

                 ↳Polynomial Ordering

Dependency Pairs:

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(*(x₁, x₂)) = x₁ + x₂

POL(-(x₁, x₂)) = 1 + x₁ + x₂

POL(+(x₁, x₂)) = x₁ + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳Polo

...

               →DP Problem 6

                 ↳Polynomial Ordering

Dependency Pairs:

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(*(x₁, x₂)) = 1 + x₁ + x₂

POL(+(x₁, x₂)) = x₁ + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳Polo

...

               →DP Problem 7

                 ↳Polynomial Ordering

Dependency Pairs:

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(+(x₁, x₂)) = 1 + x₁ + x₂

resulting in one new DP problem.

     ↳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

POL(D'(x₁))	= x₁
POL(pow(x₁, x₂))	= 1 + x₁ + x₂
POL(minus(x₁))	= 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