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!
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 $$ \begin{align} P'Q+P &= P'Q + P(Q+Q') \\&= P'Q + PQ + PQ' \\&= P'Q + PQ + PQ + PQ' \\&= (P'+P)Q + P(Q+Q') \\&= P + Q \end{align} $$