Boolean distributivity

Distributivity of OR

The distributive law of booleans ORing says whenever we have a boolean expression of the form A \cdot (B + C), we can distribute the AND over the OR to get A \cdot B + A \cdot C.

A \cdot (B + C) = A \cdot B + A \cdot C

This is because the AND operation requires both sides to be true for the result to be true, so for A \cdot (B + C) to be true, we need A to be true and either B or C to be true, which is the same as saying we need A and B to be true or A and C to be true, which is what A \cdot B + A \cdot C says.

Or, you can just think of it like how you ‘expand brackets’ in maths, it works basically the same way.

This also works the other way around, we can factor out the A from A \cdot B + A \cdot C to get A \cdot (B + C).

A \cdot B + A \cdot C = A \cdot (B + C)

The important thing is that we recognise the A \cdot is common in both of the terms which are ORed together, so that’s what goes on the outside of the brackets. This is the same as how we factorise in maths.

We put the + inside the brackets because, when we take away the two A \cdot from the two terms, we are left with B+C. That goes inside the brackets, again, just like in normal factorising.

This also works for more than 2 terms, for example:

A \cdot (B + C + D) = A \cdot B + A \cdot C + A \cdot D

Distributivity of AND

We can also distribute a slightly different way, more similar to how we would expand and factorise double brackets.

A + (B \cdot C) = (A + B) \cdot (A + C)

flashcards