Distributive laws for arbitrary sets Given a set $A$, Enderton's book defines the set $\{A\cup X \mid X\in B \}:=\{t\mid\exists X\in B(t=A\cup X) \}$. This set is well defined as this description implies that $t\in P(A\cup \bigcup B) $. 
With this notation, he claims the distributive laws: 


*

*$A\cup \bigcap B=\bigcap\{A\cup X\mid X\in B\}$ for $B\neq \emptyset$

*$A\cap \bigcup B=\bigcup\{A\cap X\mid X\in B\}$
My problem is when trying to prove them. For example, in the the first case, 
$\begin{align} \alpha \in A\cup \bigcap B  &\implies \alpha \in A \vee \alpha \in \bigcap B \\ 
&\implies \alpha \in A \vee \forall X\in B(\alpha \in X )\\
&\implies \forall X\in B (\alpha \in A \vee \alpha \in X)\\
& \implies \forall X \in B(\alpha \in A\cup X)\\
& \implies ???
 \end{align} $ 
conversly, 
$\begin{align} \alpha \in \{A\cup X| X\in B\} &\implies \exists X\in B(\alpha= A\cup X) \\ 
& \implies ???
 \end{align} $ 
 A: For the first one:
$\implies$ Let's suppose first that $\alpha\in A \cup \bigcap B$. Then $\alpha\in A \vee \alpha \in\bigcap B$. Lets also suppose that for an arbitrary set $t$, $\exists X\in B(t=A\cup X)$. If $\alpha \in A$ then $\alpha \in t$. If $\alpha \in \bigcap B$ then $\alpha \in X$, by definition and hence $\alpha \in t$. Therefore $\forall t\in\{A\cup X \mid X\in B\}(\alpha \in t)$. This means $\alpha \in \bigcap\{A\cup X \mid X\in B\}$.
$\Longleftarrow$ Let's suppose that $\alpha \in \bigcap\{A\cup X \mid X\in B\}$. Then $\forall t \in \{A\cup X \mid X\in B\}(\alpha \in t)$. This means $\forall t(\exists X\in B(t=A\cup X)\longrightarrow \alpha \in t)$. Since $B\neq\emptyset$, let's suppose $X$ as an arbitrary set such that $X\in B$. Then there's $t'=A\cup X$ and also $\alpha\in t' $. This means $\alpha \in A \vee \forall X\in B(\alpha \in X)$. Therefore $\alpha \in A\cup \bigcap B$. 
The second one is analogous.  
EDIT:
I think my problem was because I haven't noticed that $\alpha \in \bigcap \{X\cup A\mid X\in B\}\iff \forall X\in B(\alpha \in X\cup A)$ and also $\alpha \in \bigcup \{X\cap A\mid X\in  B\}\iff \exists X\in B(\alpha \in X\cap A)$. 
To prove the first one for example, if $\alpha \in \bigcap \{X\cup A\mid X\in B\}$ then $\forall t\in \{X\cup A\mid X\in B\}(\alpha \in t)$ $(*)$. Now, let's take an arbitrary $X\in B$. Then there exists $t'=X\cup B$. But, by $(*)$ then $\alpha \in t'$. Since $X$ was taken arbitrarily we conclude that $\forall X\in B(\alpha \in X\cup B)$. Conversly, if $\forall X\in B(\alpha \in X\cup A)$ $(**)$ then let's take an arbitrary $t\in \{X\cup A\mid X\in B\} $. Since $t=X\cup A$ for some $X\in B$, by  $(**)$ we conclude that $\alpha \in t$. Since $t$ was arbitrarelly taken, then $\forall t \in \{X\cup A\mid X\in B\}(\alpha \in t)$. This means $\alpha \in \bigcap\{X\cup A\mid X\in B\}$. The proof for the other case is analogous. 
With this work everything is simplified. If $B\neq \emptyset$,


*

*$A\cup \bigcap B=\bigcap\{A\cup X\mid X\in B\}$: 


\begin{align} \alpha \in A\cup \bigcap B  &\iff \alpha \in A \vee \alpha \in \bigcap B \\ 
&\iff \alpha \in A \vee \forall X\in B(\alpha \in X )\\
&\iff \forall X\in B (\alpha \in A \vee \alpha \in X)\\
& \iff \forall X \in B(\alpha \in A\cup X)\\
& \iff  \alpha \in \bigcap\{A\cup X\mid X\in B\}
 \end{align}


*

*$A\cap \bigcup B=\bigcup\{A\cap X\mid X\in B\}$: 


\begin{align} \alpha \in A\cap \bigcup B  &\iff \alpha \in A \wedge \alpha \in \bigcup B \\ 
&\iff \alpha \in A \wedge \exists X\in B(\alpha \in X )\\
&\iff \exists X\in B (\alpha \in A \wedge \alpha \in X)\\
& \iff \exists X \in B(\alpha \in A\cap X)\\
& \iff  \alpha \in \bigcup\{A\cap X\mid X\in B\}
 \end{align}
