De-Morgan's theorem for 3 variables?

Solution 1:

DeMorgan's Theorem applied to $(A + B + C)'$ is as follows:

$$(A + B + C)' = A'B'C'{}{}{}{}$$

We have $\;\;$NOT(A or B or C) $\;\equiv\;$ Not(A) and Not(B) and Not(C),

which in boolean-algebra equates to $A'B'C'$

Both these extensions from DeMorgan's defined for two variables can be justified precisely because we can apply DeMorgan's in a nested manner, and in so doing, reapply, etc, in the end, it is equivalent to an immediate extension of it's application to three variables (or more) variables, provided they are connected by the same connective, $\land, \lor$.

For example, we can write $(A+B+C)' \equiv \big(A + (B+C)\big)' \equiv \big(A' \cdot (B+C)'\big) \equiv A'\cdot (B'C') \equiv A'B'C'$.

Indeed, provided we have a negated series of multiple variables all connected by the SAME connective (all and'ed or all or'ed), we can generalize DeMorgan's to even more than three variables, again, due to the associativity of AND and OR connectives. For any arbitrary finite number of connected variables:

So, $$(ABCDEFGHIJ)' = A' + B' + C' + \cdots + H' + I' + J'$$

And $$(A + B + C + \cdots + H + I + J)' = A'B'C'D'E'F'G'H'I'J'$$

Solution 2:

This is one instance where introducing another variable provides some insight. Let $D = B\lor C$.

Then, we have: $$\begin{align} \neg(A\lor B\lor C) &= \neg(A\lor D)\\ &=\neg A \land \neg D \\ &=\neg A \land \neg(B\lor C) \\ &=\neg A \land \neg B \land \neg C \end{align}$$

Thus: $$\neg(A\lor B \lor C) = \neg A \land \neg B \land \neg C$$

The idea is effectively the same for even more terms. Thus we can have: $$\neg(P_1 \lor P_2 \lor \cdots \lor P_n) = \neg P_1 \land \neg P_2 \land \cdots \land \neg P_n$$ ...and... $$\neg(P_1 \land P_2 \land \cdots \land P_n) = \neg P_1 \lor \neg P_2 \lor \cdots \lor \neg P_n$$ (Note: I'm more familiar with this notation for logic, so I'm using it. $\lor$ is or, $\land$ is and, and $\neg$ is not.)