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Prove $(A∩B)’=A’∪B’$.
Let $x ∈ (A ∩ B)’$
$∴ (x ∈ A ∩ B)’$
$∴ (x ∈ A ∧ x ∈ B)’$
$∴ (x ∈ A)’ ∨ (x ∈ B)’$
$∴ x ∈ A’ ∨ x ∈ B’$
$∴ x ∈ A’ ∪ B’$
$∴ (A ∩ B)’ ⊆ A’ ∪ B’$

The above is the solution provided. Sorry if it seems trivial but I don't understand how
$∴ (x ∈ A ∧ x ∈ B)’$
leads to
$∴ (x ∈ A)’ ∨ (x ∈ B)’$
and not
$∴ (x ∈ A)’ ∧ (x ∈ B)’$

  • 3
    If it is not true that ($2 + 2 = 4$ and I am the pope), then either $2 + 2 \neq 4$, or I am not the pope. But it is not the case that both $2 + 2 \neq 4$ *and* I am not the pope. http://en.wikipedia.org/wiki/De_Morgan%27s_laws2010-10-21
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    All that needs to happen to make the statement (A and B) false is for at least one of the parts to be false. This is equivalent to (not A or not B) being true.2010-10-21
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    For future reference, it would be clearer if you distinguished in your notation between set complement and logical negation. The standard symbol for the latter is prefix ¬, so the first 3 lines of your proof would read let x ∈ (A ∩ B)’ ∴ ¬(x ∈ A ∩ B) ∴ ¬(x ∈ A ∧ x ∈ B)2010-11-20

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This is just one of DeMorgan's Laws. Not (A and B) is equivalent to (Not A) or (Not B), that is that (A and B) is false if either one is false. If you go to (Not A) and (Not B) you need both to be false.

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    Got it thanks. :) Was really confused cause I thought it had something to do with set identities.2010-10-22