I have simplified a Boolean expression to $$(\lnot a \land \lnot b \land \lnot c) \lor (a \land (b \lor c)).$$ Is there any way to simplify this further, e.g. using De Morgan's or anything?
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$\begingroup$Your expression can no further be simplified, using only $\land,\lor, \;\text{and/or}\;\lnot$. So how you choose to present the expression depends on context and/or your preference. If you want to minimize the repetition of variables, your posted expression does so.
However, we can put the expression into one of a number of standard forms, disjunctive normal form, by merely expanding the second term, using the distributivity of conjunction over disjunction, to get:
$$(\lnot a \land \lnot b \land \lnot c) \lor (a \land (b\lor c))\equiv (\lnot a \land \lnot b \land \lnot c)\lor (a\land b) \lor (a \land c)$$
We could also reconfigurate your original post to present it in conjunctive normal form: $$(\lnot a \land \lnot b \land \lnot c) \lor (a \land (b\lor c))\equiv (\lnot a \lor b \lor c) \land (a \lor \lnot b) \land (a \lor \lnot c)$$
$\endgroup$ 0 $\begingroup$Since at the time I'm posting this, there may be a typo in the question, I'm taking the expression to be simplified to be $ (\neg a\land\neg b\land\neg c)\lor\big(a\land(b\lor c)\big) $.
Doing the prior simplification:$(\neg a\land\neg b\land\neg c)\lor\big(a\land(b\lor c)\big)=(\neg a\land\neg b\land\neg c)\lor\big((a\land b)\lor (a \land c)\big)$ and then using a karnaugh map:
$$ \begin{array}{c|c|c|c|c} a, bc & 00 & 01 & 11 & 10 \\ \hline 0 & 1 & & & \\ \hline 1 & & 1 & 1 & 1\\ \hline \end{array}=a \oplus (\neg b \land \neg c) $$
$\endgroup$ 2 $\begingroup$The simplest way to rewrite this, if you're allowed to use something other than $\;\land,\lor,\lnot\;$, is the following: \begin{align} & (\lnot a \land \lnot b \land \lnot c) \;\lor\; \big(a \land(b \lor c)\big) \\ = & \;\;\;\;\;\text{"DeMorgan -- to make the left and right parts similar"} \\ & \big(\lnot a\land \lnot (b \lor c)\big) \;\lor\; \big(a \land(b \lor c)\big) \\ = & \;\;\;\;\;\text{"$\;(p \land q) \lor (\lnot p \land \lnot q)\;$ is one of the ways to write $\;p \equiv q\;$"} \\ & a \;\equiv\; b \lor c \\ \end{align}
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