I'm currently self studying proof and set-theory, and I'm quite new to both of them. As an exercise, I'm practicing proving some basic theorems, so it'll be great if you can give me some feedback on how I'm doing so far. I'm not sure if there are any logical gaps in what I have done here.
Theorem: Suppose $f:A \rightarrow B$ and let $D_1$ and $D_2$ be subsets of $B$. Prove that $$f^{-1}(D_1 \cap D_2) = f^{-1}(D_1) \cap f^{-1}(D_2).$$
Proof: First, let $x \in f^{-1}(D_1 \cap D_2)$. Then, there exists a $y \in D_1 \cap D_2$ that satisfies $f(x) = y.$ As $y \in D_1 \cap D_2$, $y \in D_1$ and $y \in D_2$. Therefore, $x \in f^{-1}(D_1)$ and $x \in f^{-1}(D_2)$, or $x \in f^{-1}(D_1) \cap f^{-1}(D_2)$. Hence, $f^{-1}(D_1 \cap D_2) \subseteq f^{-1}(D_1) \cap f^{-1}(D_2)$.
Conversely, let $x \in f^{-1}(D_1) \cap f^{-1}(D_2)$. Then, $f(x) \in D_1$ and $f(x) \in D_2$ or $f(x) \in D_1 \cap D_2$. Because of this, $x \in f^{-1}(D_1 \cap D_2)$. Therefore, $f^{-1}(D_1) \cap f^{-1}(D_2) \subseteq f^{-1}(D_1 \cap D_2)$.
Thus, ${f^{-1}(D_1 \cap D_2) = f^{-1}(D_1) \cap f^{-1}(D_2)}.$
