Prove: (A u B)' = A' n B'?

Question

Lets A and B be sets. (Subsets of some universal set U)

Answer 1

A u B = A n B' + A' n B + A n B
U = A n B' + A' n B + A n B + A' n B'
(A u B)' = U - A u B = A' n B'

Answer 2

A u B consists of all the points in either A or B, so any point not in A u B is in neither A nor B and so is in both the complement of A and the complement of B i.e. it is in A' n B'

For the reverse if a point is in A' n B' then it is not in A and it is not in B so it is not in A u B i.e it is in (A u B)'

so (A u B)' = A' n B'

Answer 3

Consider an element x that belongs in the set (A U B)'.

[x in (A U B)'] <-> [x not in (A U B)] <-> [x not in A] AND [x not in B] <-> [x in A'] AND [x in B'] <-> [x in A' n B']. Since we have proved that: an element belongs in (A U B)' iff it belongs in A' n B',

in other words:
* an element belongs in (A U B)' implies it belongs in A' n B'
* an element belongs in A' n B' implies it belongs in (A U B)'

we have proven the two sets are equal: (A U B)' = A' n B'.

QED.

Answer 4

This is a proof itself...it's De Morgan's law. Also, draw a Venn diagram to see visually.

Answer 5

A B AUB (A u B)' = A'∩B'
0. 0 .. 0 . .. . 1 . . . .. . . 1
0. 1 .. 1 . .. . 0 . . . .. . . 0
1. 0 .. 1 . .. . 0 . . . .. . . 0
1. 1 .. 1 . .. . 0 . . . .. . . 0