Post a New Question

Boolean Algebra

posted by .

I need help starting this, I don't want a plain answer just a what to do.

(bc'+a'd)(ab'+cd')-Simplify

My best guess is to use the distributive
property.

(a+b)(a+c)= a(b+c). But now that there is a 4th variable what do I do?

  • Boolean Algebra -

    use it twice

    (bc'+a'd)(ab'+cd')
    = bc'(ab'+cd') + a'd(ab'+cd')
    = abb'c' + bcc'd' + aa'b'd + a'cdd'

    Now, since xx' = 0, the result is 0

  • Boolean Algebra -

    OOK, thank you, I see how you did that step. that makes perfect sense.

    so simplified you get

    ac'+bd'+b'd+a'c

  • Boolean Algebra -

    no, the result is 0, since

    abb'c = a(bb')c = a0c = 0

    each term is 0

Answer This Question

First Name
School Subject
Your Answer

Related Questions

More Related Questions

Post a New Question