Prove that the intersection of ~(A and B) is equal to the union of ~A and ~B

This is the second of De Morgan's Laws.

We use the "~" to represent "not".

So first we assume there is an object x that is a member of the intersection of ~(A and B)

which is true, if and only if(iff) x is a member of ~A and ~B

which is true if x isn't in A and isn't in B

which is the same as x being in ~A and ~B

which is the same as x being in the union of ~A and ~B.

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