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Boolean Algebra — worked example

Digital Logic 101 · Boolean Algebra · Example

Simplify F = A·B + A·B′ + A′·B

Three product terms, two variables. We’ll factor, apply complement and identity, then absorption. Each step cites exactly one rule.

  1. Factor A out of the first two terms. A⋅B+A⋅B′=A⋅(B+B′)A \cdot B + A \cdot B' = A \cdot (B + B') by the distributive law.
    F = A·(B + B′) + A′·B
  2. Apply complement. B+B′=1B + B' = 1 by the complement identity.
    F = A·1 + A′·B
  3. Apply identity. A⋅1=AA \cdot 1 = A.
    F = A + A′·B
  4. Apply extended absorption. A+A′⋅B=A+BA + A' \cdot B = A + B. (Proof: factor as A⋅(1+B)+A′⋅BA \cdot (1 + B) + A' \cdot B, expand, then collapseB⋅(A+A′)=BB \cdot (A + A') = B.)
    F = A + B
Final answer: F=A+BF = A + B. Same result as a 2-variable Karnaugh map (Topic 5) or Quine–McCluskey (Topic 6) on the minterm set {1, 2, 3}.

Want to try this hands-on? Open the Simulate stage and type the same expression — each identity card lights up exactly when its rule applies.