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So I have this homework question, asking to prove (A∩B)∪C= (A∪C)∩(B∪C) , but that doesn't make sense to me. Shouldn't it be A∩(B∪C)= (A∪C)∩(B∪C) ? When I check in the professors notes, that's also what it says, but I'm not sure (I'm very new to maths). Thank you in advance for any help!

J. W. Tanner
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    Try your idea out with $A$ being the set of all animals, $B$ being the set of all bears and $C$ being the set of all chairs. If you are right then chairs are all animals. – Rob Arthan Oct 18 '20 at 22:54
  • Are you familiar with propositional logic, e.g., "element chasing"? If you know set identities, then you are dealing with the distributive property, in both cases. – amWhy Oct 18 '20 at 22:59
  • Draw Venn diagram to convince yourself. – Berci Oct 18 '20 at 23:13
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    In problems like these it's always instructive to take one set to be the empty set. What happens when $A=\emptyset$? – Randy Marsh Oct 18 '20 at 23:33
  • Draw a ven diagram. But not $A\cap (whatever)$ is entirely contained in $A$. But $(whatever)\cup C$ is a set that entirely contains $C$. So if they were equal then $C$ would have to be entirely contained in $A$ and there's no reason at all that should be true. – fleablood Oct 19 '20 at 05:00

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Absolutely not! There are a lot of small cases which would make this claim false, such as the suggestion by Randy Marsh:

$$ A = \emptyset, C = \{ \text{Your Choice} \}$$ And knowing these facts $$ A \cap \emptyset = \emptyset \\ A \cup \emptyset = A$$

You would get that:

$$ ( \emptyset \cap B) \cup C = \emptyset \cup C = C \\ \text{While:} \\ A \cap (B \cup C) = \emptyset \cap (B \cup C) = \emptyset $$

And those are really different... Here is a link that might help: Prove $ (A \cup B) \cap C$ = $(A \cap C) \cup (B \cap C) $

CSch of x
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    Thank you very much! Yeah I'm pretty sure it's a typo in the paper, but I suddently wasn't sure, so thank you for your answer! –  Oct 19 '20 at 12:27