| ▲ | anonymousiam 4 days ago | |
Demorgan's theorem says AND and OR are equivalent, and only depend upon the polarity of the bits. So if "state |1>" is a binary zero, AND is the proper logical operator. | ||
| ▲ | SyzygyRhythm 4 days ago | parent [-] | |
Yes, I think that was implied in my original post, that if you define |0> as logical 1 then it works as an AND. It just seems confusing and unnecessary when they could have framed it to be consistent with classical logic. | ||