QED
Boolean Algebra · step 2 of 13

The duality principle

Every Boolean identity has a dual, obtained by swapping + with · and 0 with 1 while leaving the variables and complements alone. The duality principle says the dual of a theorem is also a theorem — because the axioms come in dual pairs. So proving x + xy = x automatically gives x(x + y) = x for free.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

  1. Swap the operationsEvery + becomes · and every · becomes +. Take care with implicit multiplication: xy means x·y and becomes x + y.
  2. Swap the constants0 ↔ 1 everywhere they appear.
  3. Leave variables and complements alonex stays x and x′ stays x′. Complementing the variables gives a different transformation.
  4. Re-bracket carefullyPrecedence changes when operations swap, so brackets that were unnecessary may become essential.

Worked example

State the dual of x + x′y = x + y and verify it is also an identity.

  1. Swap + and ·: x·(x′ + y) = x·y.
  2. Verify: x(x′ + y) = xx′ + xy by distributivity.
  3. = 0 + xy by the complement law.
  4. = xy by identity.

Answer. The dual x(x′ + y) = xy holds, as duality guarantees.

Where marks get dropped

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The duality principle — frequently asked questions

Why does duality work?

Because the axiom list is self-dual: dualising every axiom produces the axiom list again. So dualising a proof step by step yields a valid proof of the dual statement.

Is duality the same as complementation?

No. Complementing an expression gives De Morgan’s transformation, which also complements the variables. Duality leaves the variables untouched.

How is duality useful in practice?

It halves the work: prove one of each dual pair. It also converts a sum-of-products minimisation technique into a product-of-sums one automatically.

The rest of Boolean Algebra

Axioms, laws, simplification and Boolean functions. Each subtopic below has its own method, worked example and mark-losing traps.

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