He is calculating the cumulative distribution first. That is, he is finding, for instance, the probability that the minimum die roll is greater than or equal to 2. This is an easier quantity to calculate: the minimum die roll is greater than or equal to 2 if and only if all three die rolls were greater than or equal to 2. So, the probability that the minimum die roll is greater than or equal to 2 is $p(Y\ge 2)=({5\over6})^3$ (we know each die roll was 2 or greater. The rolls are presumed independent and the probability that a particular die is roll is equal to 2 or greater is $5/6$).
Then the probability that the minimum die roll is equal to $i$ is calculated using the cumulative distribution. So, for instance $p(Y=1)=p(Y\ge 1)-p(Y\ge 2)$.
The last formula can be thought of as follows: take the set of outcomes $A=[Y\ge 1]$. Then this set of outcomes differs from the set of outcomes $B=[Y=1]$ by exactly the set of outcomes $C=P[Y\ge 2]$. That is, $A=B\cup C$ and $B\cap C=\emptyset$. We thus have $P(A)=P(B)+P(C)$. Rearranging this gives $P(B)=P(A)-P(C)$.