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March 3, 2015

March 3, 2015

Posted by **Anonymous** on Saturday, February 9, 2013 at 1:00pm.

(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 -
**Steve**, Saturday, February 9, 2013 at 2:23pmuse 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 -
**Anonymous**, Sunday, February 10, 2013 at 1:01amOOK, 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 -
**Steve**, Sunday, February 10, 2013 at 6:54pmno, the result is 0, since

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

each term is 0

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