Need help proving this equation mathematically?

Question

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

(u = union, n = intersect, c = complement)

I know how to show it through Venn Diagrams but I'm not sure how to show it mathematically. Any help/tips would be appreciated.

Answer 1

There is an element in A u B, call it x. If x is in A u B then x is in A or x is in B. Now the complement of that is to say that:
x is not in A and x is not in B. But if x is not in A then x is in A^c and if x is not in B then x is in B^c, so (A^c) n (B^c).
Tips:

Union implies or
intersection implies and
union^c of intersection and intersection^c of union.

Answer 2

Let me suggest that we use ~A for the complement of A. Here is your proof.

Let x be an element of ~(A U B). Then x is not in A U B. That means x is not in A and x is not in B. Therefore, x is in ~A and x is in ~B. But then x is in (~A) n (~B). Thus,
~(A U B) is a subset of (~A) n (~B).

Now use the same techniques to show that (~A) n (~B) is a subset of ~(A U B).

You are astute in noting that mathematicians do not accept arguments based on Venn diagrams as proof.

Answer 3

This is really a set theory question, and the property you show is the distributive property.
You can prove it by selecting a point x as an element of (AUB)nC. This says that x is in AUB and x is in C.
Now assume x is NOT in (AUC) n (BUC).
This says x is not in (AUC) and x is not in (BUC). This implies that x is not in C, which is a contradiction, thus our initial assumption that x is NOT in (AUC) n (BUC) is incorrect, so we conclude x IS in (AUC) n (BUC) throgh indirect proof (Reductio ad Absurdum). QED!