How do I convert CC-NOT gates to OR operators?

[khkwang]

Fredkin Gates are supposed to be universal. So far I've gotten AND, OR and NOT out of them but I can't figure out XOR. Any help?

I know that A XOR B = (A AND NOT B) OR (B AND NOT A), but trying to recreate that with Fredkin Gates is not very elegant... is that the only way?

Edit: I guess I can't change the title of my thread...

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nvm this bottom part of this post, it's been solved already.

According to Feynman Lectures in Computing, by using only C-NOT, CC-NOT and NOT gates we can recreate AND, OR and XOR gates.

I understand how to create AND and XOR, but I can't work out the OR. I spent a good few hours pondering and trying out different truth tables but I just don't get it. Can anyone demonstrate a way?

 

[rcgldr]

If you have AND and NOT, then NOT ( (NOT (INPUT₁)) AND (NOT (INPUT₂)) ) = ((INPUT₁) OR (INPUT₂))

 

[khkwang]

Ahh so simple. Can't believe I forgot something so basic. Thanks, you actually helped me clear up a whole other question as well.