circuitboolean-algebrakarnaugh-map

Simplifying a Karnaugh Map


In this image it says the Output should be A' + B, but to me, it seems like it should be A' + AB. If I'm wrong, could someone explain to me how A' + B is the correct answer? The reason I ask this is because in the second group of ones, you get ABC + ABC'. Since A and B areboth common in the last two groups, shouldn't they both be included? Thanks in advance.

Karnaugh Map Simplification Example

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Solution

  • The key bit you are missing is that A' + B is equal to A' + AB

    Because of the Distributive Property of boolean algebra we can rewrite

    A' + AB
    

    as the product

    (A' + A)(A' + B)
    

    and since the first term is the same as just saying 1, we are left with only our second term.