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