Simplifying the boolean expression for leap year

Question

Suppose I have

p = year divisible by 4
q = year divisible by 100
r = year divisible by 400

and by constructing the truth table I get

 p | q | r | is_leap_year
---+---+---+--------------
 0 | * | * | 0
 1 | 0 | 0 | 1
 1 | 0 | 1 | 0
 1 | 1 | 0 | 0
 1 | 1 | 1 | 1

yield the following boolean expression

$$ \begin{align} & (\, p \land \neg q \land \neg r )\, \lor (\, p \land q \land r )\, \\ & \equiv p \land (\, (\, \neg q \land \neg r )\, \lor (\, q \land r )\, )\, & \text{(distributive law)} \\ & \equiv p \land (\, (\, (\, \neg q \land \neg r )\, \lor q )\, \land (\, (\, \neg q \land \neg r )\, \lor r )\, )\, & \text{(distributive law)} \\ & \equiv p \land (\, (\, q \lor (\, \neg q \land \neg r )\, )\, \land (\, r \lor (\, \neg q \land \neg r )\, )\, )\, & \text{(commutative law)} \\ & \equiv p \land (\, (\, (\, q \lor \neg q )\, \land (\, q \lor \neg r )\, )\, \land (\, (\, r \lor \neg q )\, \land (\, r \lor \neg r )\, )\, )\, & \text{(distributive law)} \\ & \equiv p \land (\, (\, \top \land (\, q \lor \neg r )\, )\, \land (\, (\, r \lor \neg q )\, \land \top )\, )\, & \text{(complemence law)} \\ & \equiv p \land (\, (\, q \lor \neg r )\, \land (\, r \lor \neg q )\, )\, & \text{(identity law)} \end{align} $$

it should be able to further simplified to simply

$$ r \lor \neg q \land p $$

however I couldn't proceed further from what I have now, am I missing something in between?

Answer 1

The truth table is:

 p | q | r | is_leap_year
---+---+---+--------------
 0 | 0 | 0 | 0
 1 | 0 | 0 | 1
 1 | 1 | 0 | 0
 1 | 1 | 1 | 1

which gives us the expression $$(p \land \lnot q \land \lnot r) \lor (p \land q \land r) $$

There is a much easier way to simplify it.

We can reduce $p \land \lnot q \land \lnot r$ to $p \land \lnot q $ as $100 \nmid a \implies 400 \nmid a $.

Also, we can reduce $p \land q \land r$ to just $r$ as $400 \mid a \implies 100 \mid a \implies 4 \mid a$.

Therefore, the expression simplifies to:

\begin{align} & (\, p \land \neg q \land \lnot r)\, \lor (\, p \land q \land r )\, \\ & \equiv (\, p \land \neg q )\, \lor (\, p \land q \land r )\, \\ & \equiv (\ p \land \lnot q )\ \lor (\ r )\, \\ & \equiv p \land \lnot q \lor r \end{align}