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Firstly, please forgive me for my lack of experience in boolean algebra - I have not touched it in years. Also, as this is a coursework assignment I am only hoping for a little nudge in a right direction :)

I am required to simplify the following boolean expression for logical circuit:

$E= (A + B)’ \bullet (C + D + F)’ + (A + B)$

I applied De Morgan for ($A + B$)' and for ($C + D + F$)', however I am left with the following, which 'feels' rather long:

$E = A’\bullet B’ \bullet C’ \bullet D’ \bullet F’ + (A + B)$

Is there anything else I can possibly do with what I am left with?

Thank you!

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  • $\begingroup$ What do you mean by "simplify"? How do you tell a "simple" formula from a non-"simple" one? To me it looks reasonably simple. $\endgroup$
    – Alex M.
    Commented Nov 12, 2015 at 11:38
  • $\begingroup$ Thank you for your reply! I am basically wondering whether there is any other law that I cannot see that I could still apply for the existing formula, as I am not too certain myself how to tell whether a formula is in its "simplest" form. $\endgroup$
    – qwerty
    Commented Nov 12, 2015 at 11:42
  • $\begingroup$ is $\oplus$ exlusive disjunction? $\endgroup$
    – Alex M.
    Commented Nov 12, 2015 at 11:43
  • $\begingroup$ It is an or (we normally use '+" in our class, is ⊕ an incorrect representation?) $\endgroup$
    – qwerty
    Commented Nov 12, 2015 at 11:46
  • $\begingroup$ @qwerty $\oplus$ is usually reserved for XOR as opposed to OR. So, for example, $1 \oplus 1 = 0$. $\endgroup$
    – Ben Grossmann
    Commented Nov 12, 2015 at 11:46

1 Answer 1

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Note that $P'\cdot Q + P = Q+P$; this rule applies to your initial formula.

This is easily seen from Venn diagrams or truth tables; algebraically one can do $$ \begin{align} P'Q+P &= P'Q + P(Q+Q') \\&= P'Q + PQ + PQ' \\&= P'Q + PQ + PQ + PQ' \\&= (P'+P)Q + P(Q+Q') \\&= P + Q \end{align} $$

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  • $\begingroup$ Ah, I see! So therefore, my result should be C'⋅D'⋅F+ (A+B), right? $\endgroup$
    – qwerty
    Commented Nov 12, 2015 at 12:14
  • $\begingroup$ Sorry, I'll stop myself there - I was not meant to ask for an answer :-) Thank you very much for such a great explanation - there is a lot of work ahead of me! :) $\endgroup$
    – qwerty
    Commented Nov 12, 2015 at 12:15
  • $\begingroup$ @qwerty you're allowed to ask questions, you know. You're right; that's exactly the answer you should get. $\endgroup$
    – Ben Grossmann
    Commented Nov 12, 2015 at 12:37

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