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Boolean Algebra Axioms

Identity, complement, commutative, associative laws.

Darshan N
Updated: 7 April 2026
6 min read

Every digital circuit obeys a handful of rules that never break. These rules, called Boolean algebra axioms, are the bedrock of every ALU, decoder, and multiplexer built on silicon.

Boolean Algebra Axioms — Reference TableAxiomAND formOR formIdentityA · 1 = AA + 0 = ANull (Annihilator)A · 0 = 0A + 1 = 1IdempotentA · A = AA + A = AComplementA · A' = 0A + A' = 1Involution(A')' = A(A')' = AClosureA,B ∈ {0,1} → A·B ∈ {0,1}sameApplies to all Boolean systems — 74LS/74HC families included
Figure 1: The six foundational axioms of Boolean algebra with dual AND/OR forms

Core Concept

Boolean algebra works only with two values: 0 (false) and 1 (true). This property is called closure. Every gate output stays within that set, no matter the inputs.

The identity axioms say that ANDing any variable with 1 returns it unchanged, and ORing with 0 leaves it unchanged. The null axioms are the opposite extreme: AND with 0 always gives 0, OR with 1 always gives 1. These four rules explain why unused gate inputs tied to VCC or GND behave predictably.

The complement axiom underlies every inverter. A 74HC04 hex inverter running at 5 V has a propagation delay of about 7 ns and a fan-out of 10 in the 74HC family. Its entire job is producing A' so that A · A' = 0 and A + A' = 1 hold in hardware.

Boolean Expression

The six axioms in compact form: A·1=A, A+0=A, A·0=0, A+1=1, A·A=A, A+A=A, A·A'=0, A+A'=1, (A')'=A. Each axiom has a dual: swap AND with OR and swap 0 with 1 to get the partner axiom. This duality runs through all of Boolean algebra.

Example
Given:
  A = 1, B = 0

Formula / Rule:
  Evaluate F = (A · 1) + (B · A') using axioms

Step by step:
  Step 1: A · 1 = A = 1          (Identity axiom)
  Step 2: A' = 1' = 0           (Complement + Involution)
  Step 3: B · A' = 0 · 0 = 0    (Null axiom: anything AND 0 = 0)
  Step 4: F = 1 + 0 = 1           (Identity axiom: A + 0 = A)

Final Answer:
  F = 1
Exam Tip: GATE often asks which axiom justifies a simplification step. Remember that A+1=1 (not A) — the null axiom for OR always returns 1, not the variable. Also, the dual of every axiom is equally valid; swapping AND/OR and 0/1 always produces another correct axiom. Involution is tested as a trap: applying complement twice returns the original variable.

Key Properties

  • Closure: output is always 0 or 1; no intermediate voltage states in ideal logic
  • Identity elements: 1 for AND, 0 for OR — distinct values, commonly swapped in exams
  • Null elements: 0 for AND, 1 for OR — one input forces the output regardless of others
  • Idempotent law holds only in Boolean algebra, not ordinary algebra (2+2 ≠ 2)
  • 74HC04 inverter: Vcc = 2–6 V, t_pd ≈ 7 ns at 5 V, fan-out = 10, I_cc ≈ 0.08 mA static
  • Involution (double complement) is used in logic minimization to restore a cancelled inversion
  • Duality principle: every valid axiom or theorem has a valid dual — swap AND↔OR, 0↔1

Quick Revision

  • Six axioms: closure, identity, null, idempotent, complement, involution
  • Identity for AND is 1; identity for OR is 0
  • Null for AND is 0; null for OR is 1
  • A·A'=0 and A+A'=1 always, regardless of what A is
  • (A')'=A — double inversion cancels out
  • Duality: swap AND↔OR and 0↔1 to get partner axiom
  • Exam trap: students write A+1=A instead of A+1=1; the null axiom overrides the variable

Boolean Algebra Axioms

Verify your command of the fundamental axioms that define Boolean algebra structure.

Question 1 of 3

Q1.Which of the following correctly states the complement axiom in Boolean algebra?