I am reading Statistical Theory: A Concise Introduction, by Felix Abramovich and Ya'acov Ritov. In the the appendix, the authors provide a primer on basic probability theory. In discussing the probability function, the authors write the following:
The probability function assigns to each event $A \in \mathcal{A}$ a real number $P(A)$ called the probability of $A$ satisfying the following conditions:
...
- Since $(A_1 \cup A_2)^c = A_1^c \cap A_2^c$, then $P(A_1 \cap A_2) = 1 - P((A_1 \cap A_2)^c) = 1 - P(A_1^c \cup A_2^c)$.
Note that $A$ denotes an event.
Intuitively, I can tell that the proposition is true. However, I wanted to prove it in my notes. I unsuccessfully attempted to do this by using De Morgan's laws and the fact that $P(A^c) = 1 - P(A) \ \forall \ A$.
I was wondering if people could please take the time to show me the correct proof for this.
It seems to me that we have an if/then statement of the form, "Since (If) $(A_1 \cup A_2)^c = A_1^c \cap A_2^c$, then $P(A_1 \cap A_2) = 1 - P((A_1 \cap A_2)^c) = 1 - P(A_1^c \cup A_2^c)$.
A (Hypothesis): $(A_1 \cup A_2)^c = A_1^c \cap A_2^c$.
B (Conclusion): $P(A_1 \cap A_2) = 1 - P((A_1 \cap A_2)^c) = 1 - P(A_1^c \cup A_2^c)$.
So for a proof, we need to either work forwards from A to B or backwards from B to A.
