I know it is a simple theorem that $$ f(\bigcup_{\gamma\in \Gamma} A_\gamma) = \bigcup_{\gamma \in \Gamma} f(A_\gamma)$$
I know how to prove it by the definition of the above things, but I want to prove it by logical proposition using $\forall, \exists$.
\begin{align} &y \in f(\bigcup_{\gamma\in \Gamma} A_\gamma) \\ &\Leftrightarrow \exists x \in \bigcup_{\gamma\in \Gamma} A_\gamma : y=f(x)\\ &\Leftrightarrow \exists x(x \in \bigcup_{\gamma\in \Gamma} A_\gamma \wedge y=f(x) )\\ &\Leftrightarrow \exists x\{(\exists \gamma \in \Gamma: x \in A_{\gamma})\wedge y=f(x)\} \end{align}
and then I don't know how to deal with it.
Help me to prove the theorem by using predicate logic.
*edit.
I want to prove it by predicate logic changing the word 'for some x, for all x, and, or' to '$\exists, \forall, \wedge, \vee$' and using bracket. I wrote that I know how to prove it by the definition of things.