Boolean functions: functional completeness

What about using only 1 Boolean function?

We showed earlier that OR could be implemented using only NAND:

x | y = (x NAND x) NAND (y NAND y)

In the process of doing so, we also showed that NOT could be implemented with NAND:

~x = x NAND x

Since {OR, NOT} is functionally complete, so is {NAND}

Similarly, we can show that OR and NOT can be implemented with NOR:

~x = ~(x | x)

= x NOR x

x | y = ~~(x | y)

= ~(x NOR y)

= (x NOR y) NOR (X NOR y)