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I'm trying to understand a proof that: $$(A \oplus B) \cap C = (A \cap C) \oplus(B\cap C)$$

So in the example, someone starts the proof from: $$(A \cap C) \oplus(B\cap C)$$

and after a few steps got to the following: $$((A \land C)\lor (B \land C )) \land (( \lnot A \lor \lnot C) \lor (\lnot B \lor \lnot C))$$ and then simplified it to:

$$((A\lor B) \land C)\land((\lnot A\lor \lnot B \lor \lnot C)) $$

I can see that to get from $$( \lnot A \lor \lnot C) \lor (\lnot B \lor \lnot C)$$ to $$(\lnot A\lor \lnot B \lor \lnot C)$$ the associative law and idempotent law has been applied to simplify.

However I can't figure out how to get from: $$(A \land C)\lor (B \land C )$$ to $$(A\lor B) \land C)$$

Many thanks!

jimjim
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Jakub
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    So basically, you need explanation on why $( A \cap C) \cup (B \cap C) = (A \cup B) \cap C$ ? You can refer to this: https://math.stackexchange.com/questions/697305/prove-a-cup-b-cap-c-a-cap-c-cup-b-cap-c – CSch of x May 18 '20 at 11:17
  • I should have been more precise. Thanks! – Jakub May 18 '20 at 11:21
  • @Jak Yes. By $\oplus$, did you mean the disjoint union (partially defined) ? In fact, when $\oplus$ is for the direct sum of vector subspaces (which is common), this is not true (take $A,B,C$ distinct lines in the plane). – Duchamp Gérard H. E. May 18 '20 at 18:44
  • Sorry, for ⊕ I meant to describe symmetric difference. – Jakub May 18 '20 at 20:21

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This is distributive law, check this list of Logical equivalences might be helpful $${\displaystyle (\color{blue}q\wedge\color{red}p)\vee (\color{green}r\wedge\color{red}p)}\equiv (\color{blue}q\vee \color{green}r)\wedge\color{red}p\tag*{distributive law}$$ Hence $(\color{blue}A∧\color{red}C)∨(\color{green}B∧\color{red}C)\equiv(\color{blue}A∨\color{green}B)∧\color{red}C$ hold.

Ethan
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