[thaumasiotes]

Granted. g(a) = f(a,a) is indeed the identity function for all f such that f(a,a) = a. That seems pretty pointless as an observation. We have f, so we're observing that some other, unrelated function is the identity function?

[coinomega]

Assuming you actually care about the answer, this just help me drew the exploratory graph, and this property was useful to fill in cells for which operands were equal. Finding obvious properties is a great way to learn a pattern without even computing things. Unlike what you may think, I had no idea what I was about to find - I had never seen this chart drawn before, and so this post is structured around how I thought about the exploration, not around how it may be presented to an audience who already knows way more than I do on bitwise operations. Sounds fair?

[thaumasiotes]

Sure it's fair. Bitwise operations produce pretty strange numeric results, so no, I wouldn't think you'd expect much going in. But I don't see where the identity function comes into it at all. It looks like the term "the identity function" stuck in your mind, and you reached for it when you found something subjectively similar. But shared semantic space isn't a good reason to say that a function which isn't the identity function, is. Your middle point is the observation "f(a,a) = a", which is perfectly valid, but is not the same finding as the finding you report. Using pretty-simple terms to state something true is generally preferable to using even simpler terms to state something false.

[jonsen]

It is fair enough, that you find the law f(a,a) = a useful. Just don't call it the identity function when it is not.