Is it possible to prove distributive law using De Morgan's law?

Question

Is it possible to verify $ A \cap (B \cup C) = (A \cap B) \cup (A \cap C) $ without using distributive law but De Morgan's law?

Answer 1

No, De Morgan's Law only tells us that

$$\neg\Big({A\cap (B\cup C)}\Big)=\neg A\cup\neg(B\cup C)=\neg A\cup(\neg B\cap \neg C)$$

and

$$\neg\Big((A\cap B)\cup(B\cap C)\Big)=\neg(A\cap B)\cap\neg(B\cap C)=(\neg A\cup \neg B)\cap(\neg B\cup \neg C)$$

To use this information to prove distributivity, we ironically need distributivity.

However, (unless you are working with a very unusual set theory), if $\cup$ and $\cap$ are defined, then those definitions should be enough to prove distributivity with first order logic alone.