Probability Function Proof: Since $(A_1 \cup A_2)^c = A_1^c \cap A_2^c$, then $P(A_1 \cap A_2) = 1 - P((A_1 \cap A_2)^c) = 1 - P(A_1^c \cup A_2^c)$ 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:
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*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.
EDIT:
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.
 A: Because, as stated, $(A_1\cup A_2)^\complement=A_1^\complement\cap A_2^\complement$, therefore they will have the same probability measure:$$\color{red}{\mathsf P((A_1\cup A_2)^\complement)}=\color{blue}{\mathsf P(A_1^\complement\cap A_2^\complement)}\tag 1$$
Now, by definition, for any event, $A$, then: $\mathsf P(A)=1-\mathsf P(A^\complement)$, and because $A_1\cup A_2$ is an event, therefore by instantiation: 
$$\mathsf P(A_1\cup A_2)=1-\color{red}{\mathsf P((A_1\cup A_2)^\complement)}\tag 2$$
Substitution will give the required result.
$$\therefore\quad\mathsf P(A_1\cup A_2)=1-\color{blue}{\mathsf P(A_1^\complement\cap A_2^\complement)}\tag 3$$
That is all.
A: Assume that $(A_{1}\cup A_{2})^{c}=A_{1}^{c}\cap A_{2}^{c}.$ We want to find an expression for $P(A_{1}\cap A_{2})$. Since the probability of an event is equal to one minus the probability of the complement of the event, we have
$$P(A_{1}\cap A_{2})=1-P((A_{1}\cap A_{2})^{c}).\tag{1}\label{1}$$
From the assumption,
$$(A_{1}^{c}\cup A_{2}^{c})^{c}=(A_{1}^{c})^{c}\cap (A_{2}^{c})^{c}.$$
Since the double complement of a set is equal to the original set, this is
$$(A_{1}^{c}\cup A_{2}^{c})^{c}=A_{1}\cap A_{2}.$$
Plugging this expression into $\eqref{1},$ we get
$$P(A_{1}\cap A_{2})=1-P(((A_{1}^{c}\cup A_{2}^{c})^{c})^{c}).$$
Finally, to simplify the expression on the right-hand side, we once again use the fact that the double complement of a set is equal to the original set. We obtain
$$P(A_{1}\cap A_{2})=1-P(A_{1}^{c}\cup A_{2}^{c}),$$
as desired.
