Theorem. Let $I,J$ be sets and let $\alpha \in I$ and $\beta \in J$ be labels such that $A_{\alpha}$ and $B_{\beta}$ are indexed sets. Then $\left( \bigcup\limits_{\alpha \in I} A_{\alpha} \right) \cap \left( \bigcup\limits_{\beta \in J} B_{\beta} \right) = \bigcup\limits_{(\alpha, \beta) \in I \times J} A_{\alpha} \cap B_{\beta}$.
Proof.
Two sets are equal iff every element of one set is element of the other and vice versa. From the definition of a union we have
$$ y \in \bigcup\limits_{(\alpha, \beta) \in I \times J} A_{\alpha} \cap B_{\beta} \leftrightarrow y \in (\exists (\alpha, \beta) \in I \times J) (y \in A_{\alpha} \wedge y \in B_{\beta}). $$
Whereas from the definition of the Cartesian product it follows that
$$ (\alpha, \beta) \in I \times J \leftrightarrow (\exists x \in I)(\exists y \in J) (\alpha, \beta) = (x, y), $$
which then serves as a new quantifier in the identity
$$ ((\exists \alpha \in I) \: y \in A_{\alpha} \wedge (\exists \beta \in J) \: y \in B_{\beta}) \leftrightarrow (\exists \alpha \in I)(\exists \beta \in J) (y \in A_{\alpha} \wedge y \in B_{\beta}). $$
The rhs translates to
$$
(\exists \alpha \in I)(\exists \beta \in J) (y \in A_{\alpha} \wedge y \in B_{\beta})
\leftrightarrow
(y \in A_{\alpha_1} \vee y \in A_{\alpha_2} \vee ...) \wedge (y \in B_{\beta_1} \vee y \in B_{\beta_2} \vee ...)
\leftrightarrow
(\exists \alpha \in I) \: y \in A_{\alpha} \wedge (\exists \beta \in J) \: y \in B_{\beta}
\leftrightarrow
y \in \left( \bigcup\limits_{\alpha \in I} A_{\alpha} \right) \cap \left( \bigcup\limits_{\beta \in J} B_{\beta} \right).
$$ $\Box$
2015 FEB 15 (v1.0)