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

↳Instantiation Transformation

**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)))

On this DP problem, an Instantiation SCC transformation can be performed.

As a result of transforming the rule

one new Dependency Pair is created:

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Inst

→DP Problem 2

↳Argument Filtering and Ordering

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

The following dependency pair can be strictly oriented:

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

There are no usable rules using the Ce-refinement that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

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

resulting in one new DP problem.

Used Argument Filtering System:

QUOT(x,_{1}x,_{2}x) ->_{3}x_{1}

s(x) -> s(_{1}x)_{1}

DIV(x,_{1}x) ->_{2}x_{1}

R

↳DPs

→DP Problem 1

↳Inst

→DP Problem 2

↳AFS

...

→DP Problem 3

↳Dependency Graph

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

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes