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

↳Remaining Obligation(s)

The following remains to be proven:

**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)))

Duration:

0:00 minutes