[rezonant]

It might predate HN but you are looking at the XY problem, not the YX problem. In this case, they are not commutative.

[dotancohen]

It seems that the YX problem is a derivative of the XY problem. In both cases we solve for X.

[Dylan16807]

> seems

If you're actually unsure, then be sure. It's a derivative.

[dotancohen]

It's a play on words.

[Dylan16807]

Can you explain it? I don't get it.

I mean, I see how "derivative" could be wordplay, but I don't see how "seems" is wordplay.

[brudgers]

Yes the YX problem has a YX problem.