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Choose from these sentences:

$y \in C$ and $y \in B_\beta$
$x \in C \cap B_\beta$
$x \in C$ and $\forall \alpha, x \in B_\alpha$
42
$\exists \beta , y \in C \cap B_\beta$
$x \in C$ and $\exists \beta , x \in B_\beta$
$y \in C\cap \bigcup_{\alpha \in \Delta} B_{\alpha}$
For the other direction, presume $y \in \bigcup_{\alpha \in \Delta} (C \cap B_{\alpha})$
$y \in C$ and $y \in \bigcup B_\alpha$
$x \in \bigcup_{\alpha \in \Delta} (C \cap B_{\alpha})$
$\forall \alpha, x \in C \cap B_\alpha$
First, let $x \in C \cap \bigcup_{\alpha \in \Delta} B_{\alpha}$ .

Your Proof:

Choose from these sentences:

$y \in C$ and $y \in B_\beta$
$x \in C \cap B_\beta$
$x \in C$ and $\forall \alpha, x \in B_\alpha$
42
$\exists \beta , y \in C \cap B_\beta$
$x \in C$ and $\exists \beta , x \in B_\beta$
$y \in C\cap \bigcup_{\alpha \in \Delta} B_{\alpha}$
For the other direction, presume $y \in \bigcup_{\alpha \in \Delta} (C \cap B_{\alpha})$
$y \in C$ and $y \in \bigcup B_\alpha$
$x \in \bigcup_{\alpha \in \Delta} (C \cap B_{\alpha})$
$\forall \alpha, x \in C \cap B_\alpha$
First, let $x \in C \cap \bigcup_{\alpha \in \Delta} B_{\alpha}$ .

Your Proof: