Yes. I was actually thinking about this. However, you have a similar problem with the current model: two NaNs can have the same bit pattern, but still have to be unequal.

This is somewhat simpler--as long as either argument to (==) is NaN the answer is False. However, you still have to figure out that at least one bit pattern corresponds to NaN. If you want them to be equal, you would have to figure out that both are NaN. This is certainly a little more difficult, but I think it isn't much worse than the current scenario.

I think though you currently just subtract them and test equality to zero, so it is all done by the hardware.