Proving Probability Statement via Axioms of Probability and Set Identities I am given the following problem and I am trying to figure out the last step.

Using the axioms of probability and set identities, prove that
if $(B \cap C) \subset A$, then $P(A) \geq P(B) + P(C) - 1$

The axioms:


*

*$P(A) \geq 0 \text{ for any } A$

*$P(S) = 1$

*$P(A \cup B) = P(A) + P(B) \text{ if } A \cap B = \emptyset$


So far I have:
$Z := B\cap C$
$P(A) = P(A\cap Z) + P(A - Z)$
since $Z \cup A \text{, then }A\cap Z = Z$
$P(Z)= P(Z) + P(A-Z)$
$P(A) = P(Z) + P(A \cap Z^c)$
substitue in
$P(A) = P(B \cap C) + P(A \cap (B \cap C)^c)$
$P(A) \geq P(B\cap C)$
by inclusion-exclusion principle
$ P(A) \geq P(B) + P(C) - P(B \cup C)$
Now I am trying to figure out how I get $P(B \cup C)$ to be $1$.
Thanks.

Hint: You don't need $\mathsf P(B\cup C)=1$, just: $$-\mathsf P(B\cup C)\geq -1$$
So $\mathsf P(A)~{\geq \mathsf P(B)+\mathsf P(C)-\mathsf P(B\cup C)\\\geq \mathsf P(B)+\mathsf P(C)-1}$
