Boolean algebra laws (commutative, associative, distributive): Visual walkthrough

Step 0 — The only two symbols we are allowed

Before any law, agree on what the marks mean. A variable like $A$ is a switch that is either off (we write $0$ ) or on (we write $1$ ). There is nothing in between — no $2$ , no half-on.

$A \cdot B$ (read " $A$ AND $B$ ") turns on only when both switches are on. Think: two switches wired in series on one wire — current passes only if both are closed.
$A + B$ (read " $A$ OR $B$ ") turns on when at least one switch is on. Think: two switches wired in parallel — either one alone lets current through.

Figure — Boolean algebra laws (commutative, associative, distributive)

Look at the picture: series = AND (both must close), parallel = OR (either closes). Hold these two shapes in your head — the whole proof is about rearranging wires between series and parallel.

Step 1 — Draw the left-hand side as a circuit

WHAT. We build a lamp circuit for $A + B \cdot C$ .

WHY. A formula is abstract; a circuit is something you can trace with your finger. If the lamp lights, the expression is $1$ .

PICTURE. Reading the formula outside-in: it is an OR ( $+$ ) between two things — the single switch $A$ , and the block $B \cdot C$ . The $B \cdot C$ block is an AND, so $B$ and $C$ sit in series. Then that whole series block sits in parallel with $A$ .

$one lone switch A + B, C in series B \cdot C$

There are exactly two ways for current to reach the lamp: the top wire through $A$ alone, or the bottom wire through $B$ -then- $C$ . That is the entire meaning of the left side.

Step 2 — Draw the right-hand side as a circuit

WHAT. We build a lamp circuit for $(A + B) \cdot (A + C)$ .

WHY. We want to compare shapes, so we draw the target the same way.

PICTURE. Outside-in this is an AND ( $\cdot$ ) between two brackets — so the two brackets sit in series. Each bracket is an OR, so inside each one the two switches sit in parallel.

$A, B in parallel (A + B) \cdot A, C in parallel (A + C)$

Notice something odd already: switch $A$ appears twice — once in the left gate, once in the right gate. That duplication is exactly what the law is claiming is necessary, and Step 5 will show why.

Step 3 — Expand algebraically, term by term

WHAT. Multiply the two brackets out, using the ordinary distributive law $X (Y + Z) = X Y + X Z$ (the "AND-over-OR" form, which behaves just like school algebra).

WHY. We can't yet see the two circuits are equal. Algebra lets us grind one side until it turns into the other. We start from the right side because it is the bushier one.

$(A + B) (A + C) = term 1 A \cdot A + term 2 A \cdot C + term 3 B \cdot A + term 4 B \cdot C$

Each term is one way to pick one switch from the left bracket and one from the right bracket:

term 1 $A \cdot A$ — took $A$ from both,
term 2 $A \cdot C$ — took $A$ from left, $C$ from right,
term 3 $B \cdot A$ — took $B$ from left, $A$ from right,
term 4 $B \cdot C$ — took $B$ from left, $C$ from right.

The grid picture shows all four pairings as cells. Keep every cell for now — the next step deletes and merges them.

Step 4 — Collapse using two Boolean facts

WHAT. Simplify the four terms.

WHY. In ordinary algebra $A \cdot A = A^{2}$ and you'd be stuck. In Boolean algebra a switch ANDed with itself is just itself, and OR-ing $1$ with anything gives $1$ . These two facts (from Boolean Algebra Identity & Null Laws) are what make Boolean algebra shrink instead of grow.

Fact 1 — Idempotence: $A \cdot A = A$ . A switch in series with a copy of itself is still just " $A$ is on". So term 1 becomes plain $A$ : $A + A \cdot C + A \cdot B + B \cdot C$ (We also used commutativity, $B \cdot A = A \cdot B$ , to line up the $A$ 's.)

Factor $A$ out of the first three terms: $= A \cdot (1 + C + B) + B \cdot C$ Here $A$ is pulled outside; the leftover inside is $1 + C + B$ , where the $1$ is really $A \cdot A / A$ ... no — the $1$ appears because $A = A \cdot 1$ , so factoring $A$ from the bare $A$ leaves a $1$ .

Fact 2 — Null / domination: $1 + X = 1$ . An OR with a permanently-on switch is always on. So the bracket collapses: $1 + C + B = 1$

Therefore: $A \cdot (1) + B \cdot C = A + B \cdot C ■$

The right side turned into the left side. The law is proven — but why did $A$ needing to appear twice matter? That is the picture we still owe you.

Step 5 — WHY the $A$ must be copied into both brackets

WHAT. We test the wrong version $A + B C = ? (A + B) \cdot C$ — the version where a student "distributes once" like in normal algebra and forgets the second $A$ .

WHY. Seeing a law fail is what fixes it in memory. We hunt for a single row of inputs where the two sides disagree.

PICTURE. Try $A = 1, B = 0, C = 0$ :

Correct left side: $A + B \cdot C = 1 + 0 \cdot 0 = 1 + 0 = 1$ . Lamp ON (the top wire through $A$ alone carries current).
Wrong form: $(A + B) \cdot C = (1 + 0) \cdot 0 = 1 \cdot 0 = 0$ . Lamp OFF — because $C$ is in series and $C = 0$ blocks everything, even though $A$ is on.

In the wrong circuit, $C = 0$ acts like a broken wire on the only path. The correct $(A + B) (A + C)$ form survives because when $C = 0$ , the second gate becomes $(A + 0) = A = 1$ , and the first gate $(A + B) = (1 + 0) = 1$ , so both series gates pass and the lamp lights. The second copy of $A$ is the escape hatch that keeps $A$ 's power alive when $B$ and $C$ both fail. Drop it, and $A$ loses its voice.

Step 6 — All the degenerate cases at once

WHAT. Check the "boring" corner inputs so no scenario surprises the reader.

WHY. Contract rule: cover every case. A law you only tested on one row is a guess.

$A$	$B$	$C$	Left $A + B C$	Right $(A + B) (A + C)$	Agree?
0	0	0	$0$	$0 \cdot 0 = 0$	✓
0	0	1	$0$	$0 \cdot 1 = 0$	✓
0	1	0	$0$	$1 \cdot 0 = 0$	✓
0	1	1	$1$	$1 \cdot 1 = 1$	✓
1	0	0	$1$	$1 \cdot 1 = 1$	✓
1	0	1	$1$	$1 \cdot 1 = 1$	✓
1	1	0	$1$	$1 \cdot 1 = 1$	✓
1	1	1	$1$	$1 \cdot 1 = 1$	✓

Two degenerate readings worth naming:

When $A = 1$ : left $= 1 + B C = 1$ (the $1$ dominates), right $= (1) (1) = 1$ . The whole thing is stuck on — $A$ overrides $B, C$ entirely.
When $A = 0$ : left $= 0 + B C = B C$ , right $= (B) (C) = B C$ . With $A$ gone, both sides quietly reduce to the same plain AND of $B$ and $C$ .

Every row agrees, every corner explained. Nothing can ambush the reader now.

The one-picture summary

One figure, the whole journey: the parallel-of-(series) circuit on the left morphs into the series-of-(parallels) circuit on the right, with $A$ shown splitting into two copies — the duplication the law demands. Same lamp, same truth table, different wiring. This trade is the seed of Combinational Circuit Simplification and gets automated by Karnaugh Maps; the "flip AND/OR" flavour continues in De Morgan's Theorems.

Recall Feynman retelling in plain words

I want a lamp that lights if switch A is on, OR if both B and C are on. The easy build: run A on one wire, and B-then-C (in a row) on another wire, both feeding the lamp. Now a friend claims the same lamp can be built a totally different way: two little gates in a row, and in each gate you place A next to one of the others — first gate is "A or B", second is "A or C". At first that looks wasteful — why put A twice? Because if I ever forget the second A, then whenever the second gate's partner (C) is off, the whole row goes dead and my lamp won't light even when A is on — which is wrong. The second copy of A is a spare doorway that keeps A's on-signal flowing no matter what B and C do. Put A in both gates, and the two wildly different-looking circuits light on exactly the same inputs. That freedom — same behaviour, cheaper wiring — is why chip designers love this rule.

Active Recall

Circuit shape of

A + B C

A

in parallel with the series pair

B

–

C

; the parenthesised right side is two OR-parallels wired in series.

Why

A \cdot A = A

(idempotence)

A switch in series with a copy of itself still just means "

A

is on"; there is no

A^{2}

because values are only

0

1

and

0 \cdot 0 = 0, 1 \cdot 1 = 1

Row that breaks

(A + B) \cdot C

A = 1, B = 0, C = 0

gives correct

= 1

but wrong form

= 0

; the lamp should be ON but the missing second

A

lets

C = 0

kill the only path.

A + B C

and

(A + B) (A + C)

when

A = 0

Both collapse to

B \cdot C

The two collapsing facts used in the proof

Idempotence

A \cdot A = A

and domination

1 + X = 1

#flashcards/hardware

Connections

Parent topic — Boolean algebra laws
Boolean Algebra Identity & Null Laws (supplies $A \cdot A = A$ and $1 + X = 1$ used in Step 4)
Logic Gates AND OR NOT (the physical series/parallel meaning of Step 0)
De Morgan's Theorems (next AND/OR flipping tool)
Karnaugh Maps (automates the factoring seen here)
Combinational Circuit Simplification (where the cheaper wiring pays off)

Visual walkthrough — Boolean algebra laws (commutative, associative, distributive)