Truth tables construction: Worked examples — Hardware deep dive

Before anything, we only ever use two symbols for a signal:

The scenario matrix

Every truth-table problem you can be asked is one (or a blend) of these cells. Each worked example below is tagged with the cell(s) it covers.

#	Case class	What makes it tricky	Covered by
C1	2 inputs, single gate	the baseline, all-0s & all-1s corners	Ex 1
C2	NOT inside an OR/AND	evaluate the inside first (intermediate column)	Ex 2
C3	3 inputs, mixed AND+OR	precedence NOT→AND→OR, 8 rows	Ex 3
C4	Degenerate input (a variable that cancels out)	column that turns out irrelevant	Ex 4
C5	Constant / limiting output (always 0 or always 1)	tautology & contradiction	Ex 5
C6	Equivalence proof (two different-looking expressions, same table)	compare column-by-column	Ex 6
C7	Real-world word problem	translate English → Boolean → table	Ex 7
C8	Exam twist: precedence trap $A + B \cdot C$ vs $(A + B) \cdot C$	the classic wrong reading	Ex 8

Ex 1 — C1: baseline, 2 inputs, single OR gate

Steps:

Count inputs → rows. Two inputs, so $2^{2} = 4$ rows. Why this step? Fix the size before filling, so you can't skip or duplicate a row.
Lay the input columns by binary count $00, 01, 10, 11$ : right column $0, 1, 0, 1$ ; left column $0, 0, 1, 1$ . Why this step? Binary counting is a checksum that all combos appear exactly once.
Apply OR to each row. $F = 1$ unless both are $0$ . Why this step? OR only fails (gives 0) at the single all-0 corner.

$A$	$B$	$F = A + B$
0	0	0
0	1	1
1	0	1
1	1	1

Verify: Exactly one row is $0$ (the all-0s corner), three are $1$ . That matches "at least one 1". Row count $= 4 = 2^{2}$ . ✓

Ex 2 — C2: NOT living inside a bigger expression

Steps:

Rows: $2^{2} = 4$ . Binary columns as before.
Make an intermediate column $\overline{A}$ . Flip every $A$ value. Why this step? We solve the small inside piece first, so the outer AND becomes trivial ([divide-and-conquer](Boolean Algebra Laws)).
AND $\overline{A}$ with $B$ . Both must be $1$ . Why this step? AND lights up only where both intermediate values are $1$ .

$A$	$B$	$\overline{A}$	$F = \overline{A} \cdot B$
0	0	1	0
0	1	1	1
1	0	0	0
1	1	0	0

Verify: Exactly one row is $1$ — the row $A = 0, B = 1$ , matching "A false and B true". ✓

Ex 3 — C3: three inputs, mixed AND + OR (precedence)

Steps:

Rows: $2^{3} = 8$ . Right column $C$ toggles every row; $B$ every 2; $A$ every 4. Why this step? Guarantees all 8 combinations, no gaps.
Intermediate column $A \cdot B$ . $1$ only in the two rows where $A = B = 1$ . Why this step? AND binds tighter than OR, so it must be computed first — build it as its own column so you never mis-order.
OR that with $C$ . $F = 1$ if either $A \cdot B = 1$ or $C = 1$ .

$A$	$B$	$C$	$A \cdot B$	$F = A \cdot B + C$
0	0	0	0	0
0	0	1	0	1
0	1	0	0	0
0	1	1	0	1
1	0	0	0	0
1	0	1	0	1
1	1	0	1	1
1	1	1	1	1

Verify: Rows with $C = 1$ (four of them) are all $1$ . The extra row $A = B = 1, C = 0$ is also $1$ . Total = 5 ones. ✓ Read the $1$ -rows off directly and you have the SOP form.

Ex 4 — C4: a degenerate input (does the variable matter?)

Steps:

Rows: $2^{2} = 4$ .
Intermediates: $\overline{B}$ , then $A \cdot B$ , then $A \cdot \overline{B}$ . Why this step? Three small pieces beat one error-prone mental leap.
OR the two product terms.

$A$	$B$	$\overline{B}$	$A \cdot B$	$A \cdot \overline{B}$	$F$
0	0	1	0	0	0
0	1	0	0	0	0
1	0	1	0	1	1
1	1	0	1	0	1

Verify: The $F$ column equals the $A$ column exactly — $B$ never changes the answer. So $F = A$ . This is the law $A B + A \overline{B} = A (B + \overline{B}) = A \cdot 1 = A$ (see Boolean Algebra Laws). $B$ is a degenerate input here. ✓

Ex 5 — C5: limiting outputs — always 1, and always 0

Steps:

Rows: each has $1$ input → $2^{1} = 2$ rows. (The smallest non-trivial table.)
Intermediate $\overline{A}$ , then combine. Why this step? These two are the boundary cases every table lives between.

$A$	$\overline{A}$	$G = A + \overline{A}$	$H = A \cdot \overline{A}$
0	1	1	0
1	0	1	0

Verify:

$G$ is $1$ in every row → a tautology (always true). Output column has zero 0s.
$H$ is $0$ in every row → a contradiction (always false). Output column has zero 1s. These are the two limiting shapes of any output column. ✓

Ex 6 — C6: equivalence proof (De Morgan)

Steps:

One shared input block, $2^{2} = 4$ rows, so we compare fairly.
Build $F_{1}$ : intermediate $A \cdot B$ , then bar it.
Build $F_{2}$ : intermediates $\overline{A}, \overline{B}$ , then OR. Why this step? Two expressions are logically equal iff their output columns are identical row-by-row.

$A$	$B$	$A \cdot B$	$F_{1} = \overline{A \cdot B}$	$\overline{A}$	$\overline{B}$	$F_{2} = \overline{A} + \overline{B}$
0	0	0	1	1	1	1
0	1	0	1	1	0	1
1	0	0	1	0	1	1
1	1	1	0	0	0	0

Verify: $F_{1}$ and $F_{2}$ columns are identical ( $1, 1, 1, 0$ ). Therefore $\overline{A \cdot B} = \overline{A} + \overline{B}$ — this is De Morgan's Law, proved by table. ✓ (Boolean Algebra Laws)

Ex 7 — C7: real-world word problem

Steps:

Translate English → Boolean. "Engine on" $= E$ . "Unbuckled" $= \overline{B}$ . "And" $= \cdot$ . So $F = E \cdot \overline{B}$ . Why this step? Word problems are just Boolean expressions in disguise — nail the translation first.
Rows: two inputs $E, B$ → $2^{2} = 4$ .
Intermediate $\overline{B}$ , then AND with $E$ .

$E$	$B$	$\overline{B}$	$F = E \cdot \overline{B}$	Meaning
0	0	1	0	engine off → silent
0	1	0	0	engine off → silent
1	0	1	1	on & unbuckled → BUZZ
1	1	0	0	on & buckled → silent

Verify: The buzzer fires in exactly one row — engine on ( $E = 1$ ), unbuckled ( $B = 0$ ) — which is precisely the real safety rule. Units check: output is a bit (buzz/silent), inputs are bits. ✓

Ex 8 — C8: exam twist — the precedence trap

Steps:

Rows: three inputs → $2^{3} = 8$ .
Correct column $F = A + (B \cdot C)$ : intermediate $B \cdot C$ , then OR with $A$ .
Wrong column $G = (A + B) \cdot C$ : intermediate $A + B$ , then AND with $C$ . Why this step? Putting both side by side exposes exactly which rows the trap corrupts.

$A$	$B$	$C$	$B \cdot C$	$F = A + B \cdot C$	$A + B$	$G = (A + B) \cdot C$
0	0	0	0	0	0	0
0	0	1	0	0	0	0
0	1	0	0	0	1	0
0	1	1	1	1	1	1
1	0	0	0	1	1	0
1	0	1	0	1	1	1
1	1	0	0	1	1	0
1	1	1	1	1	1	1

Verify: Compare columns $F$ and $G$ . They differ in rows $A = 1, C = 0$ : rows $(1, 0, 0)$ and $(1, 1, 0)$ — 2 rows disagree. $F$ has 5 ones, $G$ has 3 ones. The correct answer is $F$ . ✓

Coverage check — did we hit every cell?

Every cell in the matrix has a fully worked, verified example. ✓

Active recall

Recall Test yourself

In Ex 4, why did $B$ vanish from the answer? ::: Because $B + \overline{B} = 1$ , so $A (B + \overline{B}) = A$ ; the variable cancels.
What is the output column of a tautology like $A + \overline{A}$ ? ::: All $1$ s (zero zeros).
How do you prove two expressions are equivalent? ::: Build both tables; they must match row-by-row.
Correct reading of $A + B \cdot C$ ? ::: $A + (B \cdot C)$ — AND binds tighter than OR.
In how many rows do $A + B \cdot C$ and $(A + B) \cdot C$ disagree? ::: 2 rows (both with $A = 1, C = 0$ ).

Connections

Logic Gates AND OR NOT — the atomic tables every example combines.
Boolean Algebra Laws — Ex 4 and Ex 6 are laws proved by table.
Sum of Products (SOP) — read the $1$ -rows of Ex 3 straight off.
Karnaugh Maps — the same 8 rows redrawn as a grid.
Binary Number System — the counting that orders every table.
Combinational Logic Circuits — the hardware Ex 7 describes.

Why can an input variable "vanish" from a truth table's output?

Because of

B + \overline{B} = 1

, so

A (B + \overline{B}) = A

; the variable's value never changes the output.

What output column marks a contradiction?

All zeros — the expression is never true (e.g.

A \cdot \overline{A}

How do you prove De Morgan's law

\overline{A B} = \overline{A} + \overline{B}

Build both truth tables; both give the column

1, 1, 1, 0

, so they are equivalent.