Prove $g(A \cap B) \subseteq g(A) \cap g(B)$ and $g^{-1}(A \cap B) = g^{-1}(A) \cap g^{-1}(B)$

Question

If $g:\mathcal{R} \rightarrow \mathcal{R}$ and $A, B \subseteq \mathcal{R}$, then \begin{align} y \in g(A \cap B) &\rightarrow \exists x \in (A\cap B)~~s.t.~~ y=g(x) \tag{1}\\ & \rightarrow (\exists x \in A~~s.t.~~ y=g(x))~~ \land~~ (\exists x \in B~~s.t.~~ y=g(x)) \tag{2}\\ & \rightarrow y \in g(A) \land y \in g(B) \tag{3}\\ &\rightarrow y \in g(A) \cap g(B) \tag{4} \end{align}

Two questions about the above proof:

1. Is the proof correct?
2. Why can we go from (1) to (2) seems like property that '$s.t.$' can be distributed over $\land$?

if $g^{-1}(B)=\{x \in \mathcal{R} : g(x) \in B\}$ where $B \subseteq \mathcal{R}$,then \begin{align} y \in g^{-1}(A \cap B) &\leftrightarrow \{x \in \mathcal{R} : g(x) \in A\cap B\} \tag{5} \\ &\leftrightarrow \{x \in \mathcal{R} : g(x) \in A\}~~ \land~~ \{x \in \mathcal{R} : g(x) \in B\} \tag{6} \\ &\leftrightarrow y \in g^{-1}(A)~~\land~~ y \in g^{-1}(B)\tag{7}\\ &\leftrightarrow y \in g^{-1}(A) \cap g^{-1}(B) \tag{8} \end{align}

Again my questions are similar to above:

1. Is the proof correct?
2. Why can we go from (5) to (6)? Is there anything in logic that supports this?

Answer 1

Your first proof is correct. Here's how I'd write up the second proof:

Choose $x \in g^{-1}(A \cap B)$. Then $y=g(x) \in A \cap B$. In particular, $y \in A \land y \in B$, so since $g(x)=y$, we have $x \in g^{-1}(A) \land x \in g^{-1}(B)$, which means $x \in g^{-1}(A) \cap g^{-1}(B)$, proving $g^{-1}(A \cap B) \subseteq g^{-1}(A) \cap g^{-1}(B)$.

Conversely, assume $x \in g^{-1}(A) \cap g^{-1}(B)$. Then $x \in g^{-1}(A) \land x \in g^{-1}(B)$. This means $y = g(x) \in A \land y \in B$, so $y \in A \cap B$. Since $g(x) = y$, we then have $x \in g^{-1}(A \cap B)$. This proves $g^{-1}(A) \cap g^{-1}(B) \subseteq g^{-1}(A \cap B)$ and equality follows.