D(t) -> 1

D(constant) -> 0

D(+(

D(*(

D(-(

R

↳Dependency Pair Analysis

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)

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

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

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'(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(*(x)_{1}, x_{2})= x _{1}+ x_{2}_{ }^{ }_{ }^{ }POL(-(x)_{1}, x_{2})= 1 + x _{1}+ x_{2}_{ }^{ }_{ }^{ }POL(+(x)_{1}, x_{2})= x _{1}+ x_{2}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polynomial Ordering

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

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'(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(*(x)_{1}, x_{2})= 1 + x _{1}+ x_{2}_{ }^{ }_{ }^{ }POL(+(x)_{1}, x_{2})= x _{1}+ x_{2}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

...

→DP Problem 3

↳Polynomial Ordering

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

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'(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(+(x)_{1}, x_{2})= 1 + x _{1}+ x_{2}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

...

→DP Problem 4

↳Dependency Graph

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))

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes