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SECTION 03 / 6

Boolean Algebra — The Rules

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Start here In school maths, you learned rules like a×1=aa \times 1 = a and a+0=aa + 0 = a. They let you simplify long expressions. Boolean algebra has its own rules for 0s and 1s. A simpler expression means fewer gates, so a cheaper, faster, cooler chip.

Where the rules come from

  • Axioms: basic facts we accept as true at the start (like “0 ≠ 1”).
  • Laws and theorems: rules we prove from the axioms, and then use to simplify.

The laws (with their “duals”)

Every law comes in a pair. In the second one (the dual), AND ↔︎ OR and 0 ↔︎ 1 are swapped.

Each law and its dual: swap AND with OR, and 0 with 1.
Each law and its dual: swap AND with OR, and 0 with 1.
Open image ↗
Laws in pairs
Law OR version AND version (dual) In words
Identity X+0=XX + 0 = X X⋅1=XX \cdot 1 = X 0 doesn’t change an OR; 1 doesn’t change an AND
Null (domination) X+1=1X + 1 = 1 X⋅0=0X \cdot 0 = 0 OR with 1 is always 1; AND with 0 is always 0
Idempotent X+X=XX + X = X X⋅X=XX \cdot X = X Repeating doesn’t change anything
Involution (X‾)‾=X\overline{\left(\overline{X}\right)} = X Flip twice = original
Complement X+X‾=1X + \overline{X} = 1 X⋅X‾=0X \cdot \overline{X} = 0 Something OR its opposite = 1; AND⁡=0\operatorname{AND} = 0
Commutative X+Y=Y+XX + Y = Y + X X⋅Y=Y⋅XX \cdot Y = Y \cdot X Order doesn’t matter
Associative (X+Y)+Z=X+(Y+Z)\left(X + Y\right) + Z = X + \left(Y + Z\right) (X⋅Y)⋅Z=X⋅(Y⋅Z)\left(X \cdot Y\right) \cdot Z = X \cdot \left(Y \cdot Z\right) Brackets don’t matter
Distributive X⋅(Y+Z)=XY+XZX \cdot \left(Y + Z\right) = X Y + X Z X+Y⋅Z=(X+Y)(X+Z)X + Y \cdot Z = \left(X + Y\right) \left(X + Z\right) The second one is new! It doesn’t work in normal maths

Duality rule: if something is true, its dual is also true. To find the dual, swap AND ↔︎ OR and 0 ↔︎ 1. Don’t change the variables or their bars.

Three simplification shortcuts (use these a lot)

Theorem Rule Why
Combining X⋅Y+X⋅Y‾=XX \cdot Y + X \cdot \overline{Y} = X X⋅(Y+Y‾)=X⋅1=XX \cdot \left(Y + \overline{Y}\right) = X \cdot 1 = X
Absorption X+X⋅Y=XX + X \cdot Y = X X⋅(1+Y)=X⋅1=XX \cdot \left(1 + Y\right) = X \cdot 1 = X
Elimination (X+Y‾)⋅Y=X⋅Y\left(X + \overline{Y}\right) \cdot Y = X \cdot Y XY+Y‾Y=XY+0=XYX Y + \overline{Y} Y = X Y + 0 = X Y

How to prove a theorem

Example 1: prove X⋅Y+X⋅Y‾=XX \cdot Y + X \cdot \overline{Y} = X

Step Expression Law used
1 X⋅(Y+Y‾)X \cdot \left(Y + \overline{Y}\right) Distributive
2 X⋅1X \cdot 1 Complement
3 X Identity

Example 2: prove X+X⋅Y=XX + X \cdot Y = X

Step Expression Law used
1 X⋅1+X⋅YX \cdot 1 + X \cdot Y Identity (X=X⋅1)\left(X = X \cdot 1\right)
2 X⋅(1+Y)X \cdot \left(1 + Y\right) Distributive
3 X⋅1X \cdot 1 Null (1+Y=1)\left(1 + Y = 1\right)
4 X Identity

Another way to check: fill in a truth table for both sides. If every row matches, they’re equal.

In one line: Boolean laws (identity, null, idempotent, complement, distributive and more) come in dual pairs, and shortcuts like XY+XY‾=XX Y + X \overline{Y} = X and X+XY=XX + X Y = X make circuits smaller. Exam words: axiom, theorem, duality, identity, null, idempotent, involution, complement, commutative, associative, distributive, combining, absorption, simplification.

Quick check

Write the dual of X+0=XX + 0 = X. X⋅1=XX \cdot 1 = X.
Simplify A⋅B+A⋅B‾A \cdot B + A \cdot \overline{B}. A (combining theorem).
Simplify A+A⋅B⋅CA + A \cdot B \cdot C. A (absorption: A⋅(1+BC)=AA \cdot \left(1 + B C\right) = A).
Is X+Y⋅Z=(X+Y)(X+Z)X + Y \cdot Z = \left(X + Y\right) \left(X + Z\right) true in Boolean algebra? Yes, it’s the second distributive law (it doesn’t hold in normal arithmetic).
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