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While I was reading I faced such example

$${(A\cap B) * (C \cap D) = (A * C) \cap (B * D)}$$

But then I faced another example that looks almost equal to this one, but has another solution

$${(A\cup B) * (C \cup D) = (A * C) \cup (A * D) \cup (B * C) \cup (B * D)}$$

So, question is what is the difference? Or there is just two ways how to look at this and I can write second expression with shape of first one like this

$${(A\cup B) * (C \cup D) = (A * C) \cup (B * D)}$$

or I miss something?

Sirop4ik
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    Could you clarify the context of your problem? From what I can tell, you are working with sets, where $\cap$ is intersection, $\cup$ is union, and $$ is some binary operation that satisfies an exchange relation with $\cap$ and is distributive over $\cup$. Does $$ have some specific meaning? – Isaac Ren May 01 '20 at 18:27
  • You start with an equation that has some $\cap$ operations. Why on Earth would you think you can just change every $\cap$ to $\cup$ and still have a true equation? Do you also think you can take any equation that has some multiplication in it an replace every $\times$ symbol with $+$? – David K May 01 '20 at 18:40
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    Is $$ supposed to be the Cartesian product, so $AC$ is actually $A \times C$? You can write the $\times$ symbol like this: \times. – David K May 01 '20 at 19:54

1 Answers1

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First property is cartesian product of intersections, while second is cartesian product of unions, they are different and duality doesn't work here. The last equation is not true for general case. Ex: $A=\{1,2\}, B=\{2,3\}, C=\{2,5\}, D=\{6,5\}$.

You will see that LHS has more elements than RHS if we use last equation.

Lee
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