x	y	~(x & y)	~x \| ~y	~~x
0	0	1	1	0
0	1	1	1	0
1	0	1	1	1
1	1	0	0	1

This verifies the first version of DeMorgan's Law below and the idempotent law (~~x = x)

          ~(x & y) = ~x | ~y

          ~(x | y) = ~x & ~y

          ~~x = x

DeMorgan's Identities