OR-over-AND distributive law

Also called: OR over AND, OR distributes over AND, second distributive law, dual distributive law

The Boolean law A + BC = (A + B)(A + C). It has no match in ordinary arithmetic, but it always holds in Boolean algebra.

The second form of the distributive law distributes OR over AND:

In ordinary arithmetic this is false (2 + 3 × 4 = 14, but 5 × 6 = 30), so many students refuse to believe it. In Boolean algebra it is always true.

Why. Split on A:

If A = 1, the left side is 1, and both brackets on the right contain A, so the right side is 1 · 1 = 1.
If A = 0, the left side is , and the right side is . Same thing.

It is the dual of the familiar = .

When to use it:

Collapsing brackets. Read right to left, becomes in one step instead of multiplying out four terms.
Turning an SOP into a POS. = .
Proofs. It gives a neat proof that = .

The visual clue is two brackets that share a literal.

Worked examples

Example

Simplify .

Example

Rewrite as a product of two sums.

1.
Match with X = B, Y = , Z = C.
2.
Result: .
3.
Check A = 1, B = 0, C = 1: the original is 0 + 0 = 0, and (0 + 0)(0 + 1) = 0.

Rejecting the law because it fails in arithmetic. Boolean + is OR, not addition.
Writing = . Each bracket must contain A.
Missing the shared literal when collapsing. The two brackets must have one literal in common, with the same polarity.

Interactive questions with instant feedback and a worked solution for every wrong answer.

OR-over-AND distributive law is taught in Boolean Algebra.