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Set equality

DeMorgan's Law: tex2html_wrap_inline31 .

If we are given specific sets tex2html_wrap_inline33 then we could probably calculate the two sets equated in DeMorgan's Law and see that they are in fact equal. To prove this law in general we can use a technique often used to prove the equality of two sets. The idea is simply this: to prove that A=B we could show that tex2html_wrap_inline37 and tex2html_wrap_inline39 This works very nicely on DeMorgan's Law as follows.

First, assume tex2html_wrap_inline41 This says tex2html_wrap_inline43 so there exists a with tex2html_wrap_inline47 . Thus tex2html_wrap_inline49 (for this particular ) and we have . This short argument establishes the fact that tex2html_wrap_inline55 by showing that an arbitrary element of the first set must be an element of the second.

Second, assume that . This says that there exists a tex2html_wrap_inline45 with tex2html_wrap_
inline49 . Thus tex2html_wrap_inline47 (for this particular tex2html_wrap_inline45 ) and we have tex2html_wrap_inline43 so tex2html_wrap_inline41 This establishes the fact that tex2html_wrap_inline71 .

Putting these two results together verifies DeMorgan's Law. Note that the second argument is essentially the first argument in reverse. This is quite often the case when showing two sets are equal by this method and you can see that most of the work is done when the first argument is complete. That's what makes this approach nice to use in many situations.





Dan Rinne
Fri Aug 2 15:38:49 PDT 1996