div(0,

div(

quot(0, s(

quot(s(

quot(

R

↳Dependency Pair Analysis

DIV(x,y) -> QUOT(x,y,y)

QUOT(s(x), s(y),z) -> QUOT(x,y,z)

QUOT(x, 0, s(z)) -> DIV(x, s(z))

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

**QUOT( x, 0, s(z)) -> DIV(x, s(z))**

div(0,y) -> 0

div(x,y) -> quot(x,y,y)

quot(0, s(y),z) -> 0

quot(s(x), s(y),z) -> quot(x,y,z)

quot(x, 0, s(z)) -> s(div(x, s(z)))

innermost

The following dependency pair can be strictly oriented:

QUOT(s(x), s(y),z) -> QUOT(x,y,z)

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(QUOT(x)_{1}, x_{2}, x_{3})= x _{1}_{ }^{ }_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(DIV(x)_{1}, x_{2})= x _{1}_{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Dependency Graph

**QUOT( x, 0, s(z)) -> DIV(x, s(z))**

div(0,y) -> 0

div(x,y) -> quot(x,y,y)

quot(0, s(y),z) -> 0

quot(s(x), s(y),z) -> quot(x,y,z)

quot(x, 0, s(z)) -> s(div(x, s(z)))

innermost

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes