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Logic Gates and Boolean Algebra

Simplifying Boolean expressions

  • 3 things to know
  • AS Unit 1

Overview

Recall questions on why simplification is worth doing before a circuit is built, the absorption, complement and distributive laws applied to worked expressions, factorising, and what A ⊕ A, A ⊕ 0 and A ⊕ 1 reduce to.

What you need to know

Answer each one before you reveal it.

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Key term

Why is it worth simplifying a Boolean expression before building the circuit?

Key term

Simplify (A + B)·(A + ¬B).

Key term

What does A·¬A + B reduce to?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Why is it worth simplifying a Boolean expression before building the circuit?

    Model answer

    Fewer gates means a cheaper circuit that draws less power, takes up less space and settles faster because a signal passes through fewer stages.

  2. 2

    Simplify A·B + A.

    Model answer

    It simplifies to A, by the absorption law.

  3. 3

    Simplify A·(A + B).

    Model answer

    It simplifies to A, by the second absorption law.

  4. 4

    Simplify (A + B)·(A + ¬B).

    Model answer

    Using the distributive law it becomes A + B·¬B, and since B·¬B is 0 the result is A.

  5. 5

    What does A·¬A + B reduce to?

    Model answer

    A·¬A is 0 by the complement law, and 0 + B is B, so the expression is just B.

  6. 6

    Use the absorption law to simplify A + A·¬B.

    Model answer

    It simplifies to A, because absorption applies whatever the second factor of the AND term happens to be.

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