Proof of $f^{-1}(A) \cap f^{-1}(B)= f^{-1}(A \cap B)$

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First let me say I am aware of the other threads on this result. The reason for me making this thread is to find out whether or not my proof/proof attempt is correct.

The problem stated in full detail is given below.

Let $X,Y$ be sets and $f:X \rightarrow Y$, let $C,D \subseteq X$ and let $A,B \subseteq Y$. Prove that $f^{-1}(A) \cap f^{-1}(B)= f^{-1}(A \cap B)$.

Here is my attempt:

$f^{-1}(A) \cap f^{-1}(B)= { \{x\in X \mid f(x) \in A\}}\cap { \{x\in X \mid f(x) \in B\}}= \{x\in X \mid f(x) \in A\wedge f(x) \in B\}= \{x\in X \mid f(x) \in A\cap B \}=f^{-1}(A \cap B)$

If anyone would be kind enough to explain to me where I've gone wrong or made an unjustified assumption I would be very grateful!

P.S. Sorry about the bad formating

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Your proof seems perfectly OK to me.

If you aren't sure because the set notation is confusing you, maybe rewriting it without so many sets would also work:

$$\begin{align}x\in f^{-1}(A)\cap f^{-1}(B)\iff &(x\in f^{-1}(A)\land x\in f^{-1}(B)\\\iff& f(x)\in A\land f(x)\in B\\\iff& f(x)\in A\cap B\\\iff& x\in f^{-1}(A\cap B)\end{align}$$

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Your proof is fine ! Everything is correct.