# Simplifying $(A + B)' \cdot (C + D + F)' + (A + B)$

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!

• What do you mean by "simplify"? How do you tell a "simple" formula from a non-"simple" one? To me it looks reasonably simple. Commented Nov 12, 2015 at 11:38
• 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. Commented Nov 12, 2015 at 11:42
• is $\oplus$ exlusive disjunction? Commented Nov 12, 2015 at 11:43
• It is an or (we normally use '+" in our class, is ⊕ an incorrect representation?) Commented Nov 12, 2015 at 11:46
• @qwerty $\oplus$ is usually reserved for XOR as opposed to OR. So, for example, $1 \oplus 1 = 0$. Commented Nov 12, 2015 at 11:46

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}