Prove that $B \setminus (\bigcup_{i\in I} A_i) = \bigcap_{i\in I} (B \setminus A_i)$

Question

Help prove it by first proving that $\Rightarrow$ and later $\Leftarrow$ . I'm confused, if I let $x$ be arbitrary and later want to prove that $x \in B$ , and I've got $∀i ∈ I\,(x ∈ B\; \text{ and }\;x ∉ A_i)$. If I suppose $i ∈ I$, and obtain $x ∈ B$, still I've only proven $∀i ∈ I\,(x ∈ B)$ instead of $x ∈ B$. Thank you very much!! Yes, I've already seen that there's a question about infinite DeMorgan's law question. But what I'm confused of here if about the second identity of Infinite Demorgan's law. And I'm hoping someone would help me understand, if I let $x$ be arbitrary and later want to prove that $x \in B$.

Answer 1

\begin{align} x\in B\setminus\Big(\bigcup_{i\in I}A_i\Big) &\iff x\in B \wedge \neg \Big(x \in \bigcup_{i\in I} A_i\Big)\iff x\in B \wedge \neg (\exists i\in I: x\in A_i)\\ &\iff x\in B \wedge \forall i\in I: x\notin A_i\iff \forall i\in I: x\in B \wedge x\notin A_i\\ &\iff \forall i \in I: x\in B\setminus A_i\iff x\in \bigcap_{i\in I} \,(B\setminus A_i)\end{align}