Prove that for any two sets A;B , we have (A n B)^c = A^c u B^c

I think im supposed to start expansion with the de morgan laws my teacher said I was doing it wrong I'm pretty sure this is the correct approach but i dont know how to continue

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- Apr 17th 2010, 04:35 PMtreethetaProve Set theroy
Prove that for any two sets A;B , we have (A n B)^c = A^c u B^c

I think im supposed to start expansion with the de morgan laws my teacher said I was doing it wrong I'm pretty sure this is the correct approach but i dont know how to continue - Apr 17th 2010, 05:19 PMDeadstar
Lol usually if your lecturer tells you it's wrong it's wrong...

Consider an element .

Then x is NOT contained in the intersection of A and B.

Hence x is NOT contained in A or B.

Hence x IS contained in and also .

So .

Then to prove the reverse...

Well, you try it. - Apr 17th 2010, 08:22 PMtreetheta
so then

x is an element of A^c u b^c

hence, X is an element of A^c and B^c

so x is not contained in A or B

so x is an element of (AnB)^c since AnB is everything thats in both A and B

hence

(A n B)^c = A^c u B^c

is that the correct approach? - Apr 19th 2010, 09:24 AMDefunkt