How to justify that x∈B ⟺ ∀i∈I:x∈B

Question

Why does ∀i∈I:x∈B∧x∉Ai equal to x∈B ∧ ∀i∈I:x∉Ai. how to drag the x∈B out from inside the universal quantifier. Isn't it equal to (∀i∈I:x∈B) ∧ (∀i∈I:x∉Ai) instead of x∈B ∧ ∀i∈I:x∉Ai?

Answer 1

Long comment

$∀i∈I:x∈B ∧ x∉A_i$ is equivalent to : $∀i (i∈I \to (x∈B ∧ x∉A_i))$ that in turn is equivalent to : $∀i (i∈I \to x∈B) ∧ ∀i (i∈I \to x∉A_i)$.

Thus, the issue is to show that :

$∀i (i∈I \to x∈B) \equiv x∈B$.

Two cases :

(i) $x \in B$ is TRUE. Thus, by truth-table for conditional (i.e. $\to$) $(i∈I \to x∈B)$ holds, for $i$ whatever.

(ii) $x \in B$ is FALSE.

We assume that $I$ is not empty: in this case, $j \in I$, for some $j$.

Thus $(j∈I \to x∈B)$ is FALSE and thus $∀i(i∈I \to x∈B)$ is FALSE.

The only issue is when $I = \emptyset$.

Answer 2

This is the result of distributive property of intersection of $B$ with the complement of $A_i$

$$B\cap \left(\bigcap_{i\in I}\big(\overline{A_i}\big)\right) = \bigcap_{i\in I}\left(B\cap \overline{A_i}\right)$$