# Finding the complement using demorgans and involution (boolean alg)

## Homework Statement

Use only DeMorgan's relationships and Involution to find the complements of the following functions:
a.) f(A,B,C,D) = [A+(BCD)'][(AD)'+B(C'+A)]

## Homework Equations

Demorgans (x1 + x2 + ... + xn)' = x1'x2'...xn'

Involution (x')' = x

## The Attempt at a Solution

[[A+(BCD)'][(AD)'+B(C'+A)]]' to find the compliment, then using demorgans
[A+(BCD)']' + [(AD)'+B(C'+A)]'
[A'(BCD)] + (AD)[B(C'+A)]'
A'BCD + (AD)[B' + (C'+A)']
A'BCD + (AD)(B' + CA')

from here I don't know where to go, i would think the right side of the equation could turn to ADB' + ADCA' but i'm not sure, if it can ADCA' would just be 0 since AA' = 0. Don't know if I can do that though, just looking for some input and hopefully I didn't make a mistake towards the begining.

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## Answers and Replies

NascentOxygen
Staff Emeritus
Science Advisor
Hi buddyblakester, http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

## Homework Equations

Demorgans (x1 + x2 + ... + xn) = x1'x2'...xn'
That is not a correct expression for De Morgan's theorem.

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had it on my paper right but yea typed it in wrong, thanks

NascentOxygen
Staff Emeritus
Science Advisor
I hadn't noticed it was just a typo.

i would think the right side of the equation could turn to ADB' + ADCA' but i'm not sure, if it can ADCA' would just be 0 since AA' = 0.
Yes, that looks right.

You can check by constructing a Truth Table for the original expression and for your answer.

ok cool, seems like AA' = 0 and A + A' = 1 can really reduce some of these kinds of equations in my homework. thanks for the feedback