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For $1\le r\le n$ and a rank-$r$ uniform set family $\mathcal{F}$, let $\partial \mathcal{F}$ be its lower shadow. Count pairs $(B,A)$ with $A\in \mathcal{F}$, $B\subset A$ and $|B|=r-1$. Every $A$ contributes $r$ pairs, whereas each shadow member lies in at most $n-r+1$ rank-$r$ sets. ThusThis is the Local LYM inequality. Complementation gives the upper-shadow form $|\nabla \mathcal{F}|/\left(\genfrac{}{}{0ex}{}{n}{r+1}\right)\ge |\mathcal{F}|/(\genfrac{}{}{0ex}{}{n}{r})$ for $r<n$.

To deduce the LYM inequality, let $\mathcal{A}$ be an antichain, ${\mathcal{A}}_r$ its rank-$r$ part, and $U_r$ all rank-$r$ sets containing some member of $\mathcal{A}$. For $r\ge 1$,The union is disjoint because an antichain member cannot contain a smaller member; every other set of $U_r$ contains a rank-$(r-1)$ superset of its smaller witnessing member. Apply the upper local inequality and put $u_r=|U_r|/(\genfrac{}{}{0ex}{}{n}{r})$:Since $u_0=|{\mathcal{A}}_0|$ and $u_n\le 1$, summing givesThus the deduction uses the local inequality explicitly, rather than quoting a separate maximal-chain argument.

Let $M=\left(\genfrac{}{}{0ex}{}{n}{\lfloor n/2\rfloor }\right)$, the largest rank size. The LYM sum is at least $|\mathcal{A}|/M$, proving the Sperner theorem bound. If $|\mathcal{A}|=M$, equality forces all its members onto ranks with binomial coefficient $M$. For even $n$, there is only the middle rank, so the family is that whole rank.

For $n=2m+1$, only ranks $m,m+1$ are possible, each of size $M$. Let $\mathcal{B}={\mathcal{A}}_{m+1}$. Antichainness gives ${\mathcal{A}}_m\cap \partial \mathcal{B}=\varnothing$, and local LYM gives $|\partial \mathcal{B}|\ge |\mathcal{B}|$. Since the two antichain parts together have size $M$, equality must hold. The inclusion graph between these ranks is $(m+1)$-regular on each side. Its edges from $\mathcal{B}$ already number $(m+1)|\mathcal{B}|$, so equality leaves no edge from $\partial \mathcal{B}$ to the complementary upper vertices.

The graph is connected: exchanging one element between two $m$-sets joins them through their $(m+1)$-element union, and successive exchanges connect all lower vertices; every upper vertex has a lower neighbour. Consequently $\mathcal{B}$ is either empty or the entire upper rank. This connectedness of adjacent-level incidence in a Boolean lattice excludes mixed extremizers. HenceThere is one choice for even $n$ and two for odd $n$. For $n=0$, the sole maximum family is $\{\varnothing \}$, agreeing with this description.