sum(0) -> 0

sum(s(

sum1(0) -> 0

sum1(s(

R

↳Dependency Pair Analysis

SUM(s(x)) -> SUM(x)

SUM1(s(x)) -> SUM1(x)

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Polo

**SUM(s( x)) -> SUM(x)**

sum(0) -> 0

sum(s(x)) -> +(sum(x), s(x))

sum1(0) -> 0

sum1(s(x)) -> s(+(sum1(x), +(x,x)))

innermost

The following dependency pair can be strictly oriented:

SUM(s(x)) -> SUM(x)

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

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(SUM(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 3

↳Dependency Graph

→DP Problem 2

↳Polo

sum(0) -> 0

sum(s(x)) -> +(sum(x), s(x))

sum1(0) -> 0

sum1(s(x)) -> s(+(sum1(x), +(x,x)))

innermost

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polynomial Ordering

**SUM1(s( x)) -> SUM1(x)**

sum(0) -> 0

sum(s(x)) -> +(sum(x), s(x))

sum1(0) -> 0

sum1(s(x)) -> s(+(sum1(x), +(x,x)))

innermost

The following dependency pair can be strictly oriented:

SUM1(s(x)) -> SUM1(x)

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

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(SUM1(x)_{1})= x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 4

↳Dependency Graph

sum(0) -> 0

sum(s(x)) -> +(sum(x), s(x))

sum1(0) -> 0

sum1(s(x)) -> s(+(sum1(x), +(x,x)))

innermost

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes