How would I go from DNF to a simplified formula with less symbols? Here's a DNF:
$$(\neg A_1 \land \neg A_2 \land \neg A_3 ) \lor (A_1 \land \neg A_2 \land \neg A_3 ) \lor (\neg A_1 \land \neg A_2 \land A_3 ) \lor (\neg A_1 \land  A_2 \land \neg A_3 )$$
And the problem states "Find a wff for this DNF in which there are at most 5 connective symbols." 
I've spent the past hour distributing & using De Morgan's laws and I'm not really getting anywhere. Is there some obvious process that I am missing here? 
 A: *

*The formula is true when at most one of $A_1, A_2$ and $A_3$ is true (and only then).

*Hence it is the negation of 'At least two of $A_1, A_2$ and $A_3$ are true'.

*Write 'at least two of $A_1, A_2$ and $A_3$ are true' with five connectives. There is a very natural way of doing this.

*Now rewrite what you got above with four connectives.

*Negate what you got on 4.

A: For reasons of space we shall represent the three entities $A_1,A_2,A_3$ by $a, b, c$, and their negations using the bar notation, $\bar a, \bar b, \bar c$.
$\small{\begin{align}
& 
(\bar a\wedge\bar b\wedge \bar c)\vee(a\wedge\bar b\wedge\bar c)\vee(\bar a\wedge \bar b\wedge c)\vee(\bar a\wedge b\wedge \bar c) 
& \text{as given}
\\ \iff & 
\bigl((\bar a\wedge\bar b\wedge \bar c)\vee(\bar a\wedge\bar b\wedge \bar c)\bigr)\vee(a\wedge\bar b\wedge\bar c)\vee(\bar a\wedge \bar b\wedge c)\vee(\bar a\wedge b\wedge \bar c) 
& \text{idempotent}
\\ \iff & 
\bigl((\bar a\wedge\bar b\wedge \bar c)\vee(a\wedge\bar b\wedge\bar c)\bigr)\vee\bigl((\bar a\wedge\bar b\wedge \bar c)\vee(\bar a\wedge \bar b\wedge c)\vee(\bar a\wedge b\wedge \bar c)\bigr) 
& \text{commute and associate}
\\ \iff & 
(\bar b\wedge \bar c)\vee\bigl((\bar a\wedge\bar b\wedge \bar c)\vee(\bar a\wedge \bar b\wedge c)\vee(\bar a\wedge b\wedge \bar c)\bigr) 
& \text{resolve}
\\ \iff & 
(\bar b\wedge\bar c)\vee\Bigl(\bar a\wedge \bigl((\bar b\wedge \bar c)\vee(\bar b\wedge c)\vee(b\wedge \bar c)\bigr)\Bigr)
& \text{distribute}
\\ \iff & 
(\bar b\wedge\bar c)\vee\Bigl(\bar a\wedge \bigl(\bar b\vee(b\wedge \bar c)\bigr)\Bigr)
& \text{resolve}
\\ \iff & 
(\bar b\wedge\bar c)\vee\bigl(\bar a\wedge (\bar b\vee \bar c)\bigr)
& \text{distribute and absorb}
\\ \iff & 
(\bar b\wedge\bar c)\vee(\bar a\wedge \bar b)\vee(\bar a\wedge \bar c)
& \text{distribute}
\end{align}}$
