First, I will note that your result above depends on the rounding mode.
Second, IEEE 754 mandates that +0 and -0 are equal (i.e. any equality operator should return True when comparing these two). Therefore both expressions are equal.
NaN has several representations in bits, but they are all "equal" to one another.[1] If an operation gives you NaN, then so will doing it commutatively. It doesn't matter that the underlying bits are the same.
[1] Except for the signaling aspect. But I believe that is preserved in commutative operations.
These are not exceptions.
First, I will note that your result above depends on the rounding mode.
Second, IEEE 754 mandates that +0 and -0 are equal (i.e. any equality operator should return True when comparing these two). Therefore both expressions are equal.
NaN has several representations in bits, but they are all "equal" to one another.[1] If an operation gives you NaN, then so will doing it commutatively. It doesn't matter that the underlying bits are the same.
[1] Except for the signaling aspect. But I believe that is preserved in commutative operations.