Boolean variables and operations (AND, OR, NOT): Worked examples

This page is the drill ground for Boolean variables and operations. The parent note taught you the three operations. Here we make sure you have seen every kind of situation they can appear in, so nothing on an exam can surprise you. Work each example's Forecast in your head before reading on — that guessing is where the learning happens.

Before we start, one reminder of the alphabet we are allowed to use, so no symbol sneaks in undefined:

The scenario matrix

Boolean algebra has no "quadrants" or "signs" like trigonometry does — its entire universe is ${0, 1}$ . So "every scenario" means every shape of problem the topic can throw at you. Here is the full list, and which worked example covers each cell:

Cell	Case class	What is tricky about it	Covered by
C1	Single-operator, both extreme inputs	The pure definition, no precedence	Ex 1
C2	Precedence collision (NOT + AND + OR together)	Doing them in the wrong order	Ex 2
C3	Degenerate / redundant variable	A variable that does nothing to the answer	Ex 3
C4	All-zero and all-one inputs (limiting cases)	Confirming the "floor" and "ceiling" behaviour	Ex 4
C5	Full truth table, 2 inputs ( $2^{2}$ rows)	Enumerating every combination without gaps	Ex 5
C6	Full truth table, 3 inputs ( $2^{3}$ rows)	Same, but $8$ rows — easy to miss one	Ex 6
C7	The De Morgan trap (NOT over OR/AND)	The bar does not distribute naively	Ex 7
C8	Real-world word problem	Turning English into Boolean	Ex 8
C9	Exam-style twist: expression that is always the same	Recognising a constant / tautology	Ex 9

We will visit all nine cells. Two of them get a picture, because seeing the "flip / gate / saturate" as geometry locks it in.

Example 1 — Cell C1 · pure single operators

The next picture shows why these three feel so different — each operator is a little machine with its own personality.

Figure — Boolean variables and operations (AND, OR, NOT)

Example 2 — Cell C2 · precedence collision

Example 3 — Cell C3 · a redundant (degenerate) variable

Example 4 — Cell C4 · all-zero and all-one limiting cases

Worked example For

$G = A \cdot B + \overline{C}$ , evaluate the two extreme inputs: $(A, B, C) = (0, 0, 0)$ and $(1, 1, 1)$ . Forecast: One extreme gives $0$ , the other gives $1$ — but which is which? The $\overline{C}$ term is the twist.

All-zero, $(0, 0, 0)$ : AND part $0 \cdot 0 = 0$ ; NOT part $\overline{0} = 1$ ; OR them: $0 + 1 = 1$ . Why? Even with everything "off", the $\overline{C}$ flips $C = 0$ into a $1$ , and OR passes it through.
All-one, $(1, 1, 1)$ : AND part $1 \cdot 1 = 1$ ; NOT part $\overline{1} = 0$ ; OR them: $1 + 0 = 1$ . Why? Now the AND term supplies the $1$ instead.
Notice both extremes gave $1$ but for different reasons. Why check both? Extremes are where sign/flip errors hide; confirming the "floor" $(0, 0, 0)$ and "ceiling" $(1, 1, 1)$ catches most bugs.

Verify: Is $G$ ever $0$ ? It needs $A \cdot B = 0$ and $\overline{C} = 0$ (i.e. $C = 1$ ). Try $(0, 1, 1)$ : $0 \cdot 1 + \overline{1} = 0 + 0 = 0$ ✓ — so $G$ is not constant; the extremes just happened to both be $1$ . Answers: both extremes $= 1$ .

Example 5 — Cell C5 · full 2-input truth table

Worked example Build the complete truth table of

$H = \overline{A} + B$ . Forecast: With $2$ inputs there are $2^{2} = 4$ rows. Which single row do you think gives $0$ ?

List all $4$ input combinations. Why? A truth table must be exhaustive — miss a row and you have not defined the function.
Compute $\overline{A}$ (flip column $A$ ), then OR with $B$ . Why in that order? NOT before OR (precedence).

$A$	$B$	$\overline{A}$	$H = \overline{A} + B$
0	0	1	1
0	1	1	1
1	0	0	0
1	1	0	1

Only row $(A = 1, B = 0)$ gives $0$ . Why? That is the one row where $\overline{A} = 0$ and $B = 0$ , so OR has nothing to grab.

Verify: $H = \overline{A} + B$ is exactly the logical statement " $A$ implies $B$ ". It should be true (=1) everywhere except when the premise $A = 1$ holds but the conclusion $B = 0$ fails — which is precisely our single $0$ row. Meaning matches table. Answer: table above; sole $0$ at $(1, 0)$ .

Example 6 — Cell C6 · full 3-input truth table (8 rows)

Worked example Build the truth table of

$M = A \cdot B + \overline{C}$ (the function from Ex 4), all $2^{3} = 8$ rows. Forecast: How many of the $8$ rows are $0$ ? Guess before counting.

Write $C$ changing fastest, then $B$ , then $A$ — the standard binary counting order $000, 001, \dots, 111$ . Why? Counting in binary guarantees you list each combination once, no gaps.
For each row: compute $A \cdot B$ , compute $\overline{C}$ , then OR.

$A$	$B$	$C$	$A \cdot B$	$\overline{C}$	$M$
0	0	0	0	1	1
0	0	1	0	0	0
0	1	0	0	1	1
0	1	1	0	0	0
1	0	0	0	1	1
1	0	1	0	0	0
1	1	0	1	1	1
1	1	1	1	0	1

Count the $0$ rows: three of them, all with $C = 1$ and $A \cdot B = 0$ . Why? $M = 0$ needs both the AND part off and $\overline{C}$ off ( $C = 1$ ) — matching Ex 4's algebra.

Verify: The rows $(0, 0, 0)$ and $(1, 1, 1)$ both read $1$ — consistent with Example 4. And $(0, 1, 1)$ reads $0$ — consistent with the " $G$ is not constant" check there. Answer: $3$ rows are $0$ .

Example 7 — Cell C7 · the De Morgan trap

Worked example Evaluate

$\overline{A + B}$ for $A = 1, B = 0$ , then compare to the tempting wrong answer $\overline{A} + \overline{B}$ . Forecast: Will "bar over the whole thing" equal "bar over each piece"? Guess before computing.

Correct: do the inside first. $A + B = 1 + 0 = 1$ , then $\overline{1} = 0$ . Why? The bar covers the whole expression, so the OR is inside the NOT and must finish first.
Tempting wrong route: $\overline{A} + \overline{B} = \overline{1} + \overline{0} = 0 + 1 = 1$ . Why show it? To expose the trap — this gives $1$ , but the correct answer was $0$ . They disagree.
The right transformation flips the operator too: $\overline{A + B} = \overline{A} \cdot \overline{B}$ . Check: $\overline{1} \cdot \overline{0} = 0 \cdot 1 = 0$ ✓. Why? This is De Morgan's law — NOT does not distribute over OR unless you also swap OR to AND.

Verify: $\overline{A + B} = 0$ (correct) vs $\overline{A} + \overline{B} = 1$ (wrong) vs $\overline{A} \cdot \overline{B} = 0$ (correct De Morgan). The picture below shows why the bar "crossing" an OR turns it into an AND.

Answer: $\overline{A + B} = 0$ ; the naive $\overline{A} + \overline{B}$ is wrong.

Example 8 — Cell C8 · real-world word problem

Worked example A door unlocks when:

the keycard is valid AND (it is working hours OR the manager override is on). Let $K, H, V$ be $1$ when true. Should the door open for $K = 1, H = 0, V = 1$ ? Forecast: The employee has a valid card, it is after hours, but the manager flipped the override. Open or stay locked?

Translate English to Boolean: $Open = K \cdot (H + V)$ . Why the brackets? The word "AND" joins $K$ to the whole "(H OR V)" phrase; without brackets, precedence would wrongly split it.
Inside first: $H + V = 0 + 1 = 1$ . Why? OR needs just one $1$ — the override supplies it.
Then AND: $K \cdot 1 = 1 \cdot 1 = 1$ . Why? Both operands are $1$ , so the strict AND is satisfied.

Verify: Sanity by story: a valid card plus an active override should open the door even after hours — and it does ( $1$ ). Counter-check $V = 0$ (no override, after hours): $K \cdot (0 + 0) = 1 \cdot 0 = 0$ , correctly locked. Answer: the door opens ( $1$ ).

Example 9 — Cell C9 · exam twist: an expression that is always the same

Active Recall

Recall Which cell was hardest? Test the traps.

In $\overline{A} + B \cdot C$ , what runs first? ::: NOT, then AND, then OR
$\overline{A + B}$ equals which of $\overline{A} + \overline{B}$ or $\overline{A} \cdot \overline{B}$ ? ::: == $\overline{A} \cdot \overline{B}$ (De Morgan)==
How many rows for $3$ inputs? ::: == $2^{3} = 8$ ==
What does $A + \overline{A}$ always equal? ::: == $1$ (tautology)==
In $A + A \cdot B$ , is $B$ needed? ::: ==No — absorption gives $A$ ; $B$ is redundant==
Door logic $K \cdot (H + V)$ with $K = 1, H = 0, V = 1$ ? ::: == $1$ , door opens==

Connections

Parent: Boolean variables and operations — the definitions drilled here.
Truth Tables — the $2^{n}$ -row enumeration used in Ex 5 and Ex 6.
Logic Gates — the circuit machines pictured in figure s01.
De Morgan's Laws — the trap defused in Ex 7.
Boolean Algebra Laws — absorption (Ex 3), complement/tautology (Ex 9).
NAND and NOR gates — combinations built from these primitives.
Binary Number System — the counting order that keeps truth tables gap-free.

Worked examples — Boolean variables and operations (AND, OR, NOT)