Why does AND become OR in De Morgan's Law proof?

[Isaac12]

I need some help on understanding the proof of De Morgan's Law.

(A intersection B)' = A' U B'
I know that it is proof is along the lines of

Let x be in (A intersection B)'.
Then x is NOT in A intersection B
<=> x not in A AND x not in B (use logical connectors, negations, etc)
<=> x in A' OR x in B'
<=> x in A' U B'

but my question is how does the AND become an OR?

 

[HallsofIvy]

The statement "I am going to work AND I am going to have pizza for lunch" is true only if BOTH are true. If I want to show "It is NOT true that I am going to work and I going to have pizza for lunch" I only need to show ONE of those two if false. That is, either I am NOT going to work or I am NOT going to have pizza for lunch. I don't have to show that they are both false, only that they are not both false. "Not A and B" is the same as "not A or not B".