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Home/ Questions/Q 231371
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Editorial Team
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Editorial Team
Asked: May 11, 20262026-05-11T19:55:38+00:00 2026-05-11T19:55:38+00:00

Let’s say you have a Boolean rule/expression like so (A OR B) AND (D

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Let’s say you have a Boolean rule/expression like so

(A OR B) AND (D OR E) AND F

You want to convert it into as many AND only expressions as possible, like so

A AND D AND F
A AND E AND F
B AND D AND F
B AND E AND F

You are just reducing the OR’s so it becomes

(A AND D AND F) OR (A AND E AND F) OR (...)

Is there a property in Boolean algebra that would do this?

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  1. Editorial Team
    Editorial Team
    2026-05-11T19:55:39+00:00Added an answer on May 11, 2026 at 7:55 pm

    Your example is exploiting the the distributivity of AND over OR, as shown here.

    All you need to do is apply that successively. For example, using x*(y+z) = (x*y)+(x*z) (where * denotes AND and + denotes OR):

    0. (A + B) * (D + E) * F
    1. Apply to the first 2 brackets results in ((A+B)*D)+((A+B)*E)
    2. Apply to content of each bracket results in (A*D+B*D)+(A*E+B*E)
    3. So now you have ((A*D+B*D)+(A*E+B*E))*F
    4. Applying the law again results in (A*D+B*D)*F+(A*E+B*E)*F
    5. Apply one more time results in A*D*F+B*D*F+A*E*F+B*E*F, QED
    
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