Is it true that year $y$ is a leap year in the Gregorian calendar if and only if $y \in (4\mathbb{Z}) \Delta (100\mathbb{Z}) \Delta (400\mathbb{Z}$)? A leap year in the Gregorian calendar is any year that is divisible by 4, excluding the ones that are divisible by 100, but including the ones that are divisible by 400.
So, now, my question is: Is it true that year $y$ is a leap year in the Gregorian calendar (proleptic if $y \le 1582)$ if and only if $y \in (4\mathbb{Z}) \Delta (100\mathbb{Z}) \Delta (400\mathbb{Z}$), where $\Delta$ denotes the symmetric difference of sets?
Remember that the intersection of three sets is contained in their symmetric difference. Also, BC years are to be converted to the corresponding non-positive years using astronomical year numbering.
Of course, $m$ is divisible by $n$ if and only if $m \in n\mathbb{Z}$, so this statement may in particular be applied for $n \in \{4,100,400\}$. While this statement is about individual factors, considering symmetric differences makes things more complicated.
 A: Yes, an year $y \in \mathbb{Z}^+$ is leap  if and only if $y \in (4\mathbb{Z}) \,\triangle\, (100\mathbb{Z}) \,\triangle\, (400\mathbb{Z})$. Let us see why.
Note that $(4\mathbb{Z}) \,\triangle\, (100\mathbb{Z}) \,\triangle\, (400\mathbb{Z}) = 4\mathbb{Z} \smallsetminus (100\mathbb{Z} \smallsetminus 400\mathbb{Z})$, since $ 400\mathbb{Z} \subseteq 100\mathbb{Z} \subseteq 4\mathbb{Z}$ (see ${}^*$ below).
Clearly, $4\mathbb{Z} \smallsetminus (100\mathbb{Z} \smallsetminus 400\mathbb{Z})$ is the set of integers that is divisible by 4, excluding the ones that are divisible by 100, but including the ones that are divisible by 400.
We restrict the statement to $\mathbb{Z}^+ = \{1, 2, 3, \dots\}$ because in the Gregorian calendar there are no year $0$ and no negative years, as correctly remarked by @celtschk.
Anyway, from the mathematical point of view, the statement holds for every $y \in \mathbb{Z}$.

${}^*$
More in general, given the sets $A, B, C$, if $A \subseteq  B \subseteq C$, then $C \,\triangle\, B \,\triangle\, A = C \smallsetminus (B \smallsetminus A)$. The proof is straightforward and can be easily visualized using Venn diagrams.
