Who can enter?

The problem looks for the most efficient ruling to satisfy all the given conditions allowing the entry of the people to the said establishment. Note that we have four binary conditions (Having firearms, Being 18 and above, Wearing sleeveless tops, Wearing slippers). If we treat these as boolean variables, we can construct a very efficient truth function for that consisting of the least number of disjunct minterms (or products of absolute conditions). We will construct a $K$ -map to do that.

We first label these conditions $A$ , $B$ , $C$ , $D$ .

$A$ = Having an age of $18$ and above

$B$ = Possession of firearms

$C$ = Wearing a sleeveless top

$D$ = Wearing slippers

Now we have the combinations of $AB$ as the columns, and those of $CD$ as the rows.

$\begin{array}{|c|c|c|c|c|} \hline \ \ CD|AB \ \ &\ \ 00\ \ &\ \ 01\ \ &\ \ 11\ \ &\ \ 10\ \ \\ \hline \hline \ \ 00\ \ &\ \ \ \ &\ \ \ \ &\ \ \ \ &\ \ \ \ \\ \hline \ \ 01\ \ &\ \ \ \ &\ \ \ \ &\ \ \ \ &\ \ \ \ \\ \hline \ \ 11\ \ &\ \ \ \ &\ \ \ \ &\ \ \ \ &\ \ \ \ \\ \hline \ \ 10\ \ &\ \ \ \ &\ \ \ \ &\ \ \ \ &\ \ \ \ \\ \hline \end{array}$

Now, we place those $7$ conditions here, and see whether it simplifies to a more condensed ruling.

Condition 1: Those who are 18 and above ( $A$ ) that have firearms ( $B$ ), but are wearing sleeveless tops ( $C$ ) and slippers ( $D$ ).

So we put a $1$ on where $ABCD$ is.

$\begin{array}{|c|c|c|c|c|} \hline \ \ CD|AB \ \ &\ \ 00\ \ &\ \ 01\ \ &\ \ 11\ \ &\ \ 10\ \ \\ \hline \hline \ \ 00\ \ &\ \ \ \ &\ \ \ \ &\ \ \ \ &\ \ \ \ \\ \hline \ \ 01\ \ &\ \ \ \ &\ \ \ \ &\ \ \ \ &\ \ \ \ \\ \hline \ \ 11\ \ &\ \ \ \ &\ \ \ \ &\ \ 1\ \ &\ \ \ \ \\ \hline \hline \ \ 10\ \ &\ \ \ \ &\ \ \ \ &\ \ \ \ &\ \ \ \ \\ \hline \end{array}$

Well, fill this in for the rest, and we will get

$\begin{array}{|c|c|c|c|c|} \hline \ \ CD|AB \ \ &\ \ 00\ \ &\ \ 01\ \ &\ \ 11\ \ &\ \ 10\ \ \\ \hline \hline \ \ 00\ \ &\ \ 1\ \ &\ \ 1\ \ &\ \ \ \ &\ \ \ \ \\ \hline \ \ 01\ \ &\ \ 1\ \ &\ \ 1\ \ &\ \ 1\ \ &\ \ \ \ \\ \hline \ \ 11\ \ &\ \ \ \ &\ \ 1\ \ &\ \ 1\ \ &\ \ \ \ \\ \hline \hline \ \ 10\ \ &\ \ \ \ &\ \ \ \ &\ \ \ \ &\ \ \ \ \\ \hline \end{array}$

Note that there are two squares of 2 units per side. This means that we can condense the seven rules into two disjunct rules.

For the first square closest to the edge, note the that for all four instances, $\{ A'B'C'D', A'BC'D', A'B'C'D, A'BC'D \}$ , it is always true that $A=0$ and $C=0$ , or that $A'C'$ .

For the other square, a similar reasoning would lead us to $BD$ .

So, the condensed truth function for the ruling of entry to the establishment is

$f( A,B,C,D) = A'C' + BD$

meaning that there are only two disjunct rules. These are the ones who are allowed to enter.

1. Those who are not $18$ and above and don't wear sleeveless tops.

2. Those who have firearms and wear slippers.