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Prove: A\(B∪C)=(A\B)∩(A\C)

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  • tux

Prove: A\(B∪C)=(A\B)∩(A\C)

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Indirect proof means assume the opposite and find a contradiction. If a + c > b + c then a > b
assume that a < b Then there exists a positive number such that a+X = b Now, plug this value into the first part... a + c > (a+X) + c So (a + c) > (a + c) + X So, a number PLUS a positive number is less than the number itself? This is a contradiction. So it is not the case that a < b Part 2: assume a = b a + c > a + c No number (a+c) can be greater than itself, so this is a contradiction. So, it is not the case that a = b So, a is not <= b So a > b By the way, it doesn't matter if a, b or c = 0, the simply can't equal each other.
why these answers get more and more random as the day goes on

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\[ \large A\setminus(B\cup C)=A\cap(B\cup C)^c =A\cap(B^c\cap C^c) \] \[ \large A\cap A\cap B^c\cap C^c=(A\cap B^c)\cap(A\cap C^c)= \]

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