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Write $\left(\genfrac{}{}{0ex}{}{[n]}{k}\right)$ for the $k$th uniform layer of the Boolean cube. For $\mathcal{F}\subseteq \left(\genfrac{}{}{0ex}{}{[n]}{k}\right)$, $1\le k\le n$, its lower shadow is $\partial \mathcal{F}=\{B:|B|=k-1,B\subset A\text{ for some }A\in \mathcal{F}\}$. The Local LYM inequality statesTo prove it, count pairs $(B,A)$ with $A\in \mathcal{F}$, $B\subset A$ and $|B|=k-1$. Each $A$ contributes $k$ pairs; each member of the lower shadow contributes at most $n-k+1$. Thus $k|\mathcal{F}|\le (n-k+1)|\partial \mathcal{F}|$, which is exactly the displayed Local LYM inequality. The corresponding upper shadow inequality follows either by taking complements or by counting upward inclusion pairs.

For an antichain $\mathcal{A}\subseteq \mathcal{P}([n])$, put ${\mathcal{A}}_k=\mathcal{A}\cap \left(\genfrac{}{}{0ex}{}{[n]}{k}\right)$. The LYM inequality isFor the first proof, set ${\mathcal{B}}_n={\mathcal{A}}_n$ and recursively ${\mathcal{B}}_k={\mathcal{A}}_k\cup \partial {\mathcal{B}}_{k+1}$. Every member of ${\mathcal{B}}_{k+1}$ is a subset of an original member of $\mathcal{A}$ of size at least $k+1$. Consequently ${\mathcal{A}}_k$ and $\partial {\mathcal{B}}_{k+1}$ are disjoint: otherwise one original member would properly contain another, contradicting the antichain condition. The Local LYM inequality now givesIterating down to $k=0$ bounds the desired sum by $|{\mathcal{B}}_0|\le 1$, proving the LYM inequality.

For the second proof, a permutation $(i_1,\dots ,i_n)$ determines a maximal chain in a Boolean lattice through the successive initial subsets $\varnothing ,\{i_1\},\dots ,[n]$. A member $A$ of size $k$ lies on exactly $k!(n-k)!$ of these $n!$ maximal chains in a Boolean lattice. An antichain meets each such chain in a partially ordered set at most once. Counting its incidences with the maximal chains in a Boolean lattice therefore yields ${\sum }_{A\in \mathcal{A}}|A|!(n-|A|)!\le n!$. Division by $n!$ is the LYM inequality again.

The Sperner theorem says that an antichain has at most $M=\left(\genfrac{}{}{0ex}{}{n}{\lfloor n/2\rfloor }\right)$ members. Indeed $\left(\genfrac{}{}{0ex}{}{n}{k}\right)\le M$ for every $k$, soThe middle uniform layer of the Boolean cube attains the bound. For equality in Sperner theorem, equality throughout forces every member to lie in a layer of maximum size. If $n=2k$, this forces $\mathcal{A}=\left(\genfrac{}{}{0ex}{}{[n]}{k}\right)$.

If $n=2k+1$, only ranks $k$ and $k+1$ are possible. Let $\mathcal{L}={\mathcal{A}}_k$, $\mathcal{U}={\mathcal{A}}_{k+1}$, and $\mathcal{S}=\nabla \mathcal{L}$ be the upper shadow. The antichain condition gives $\mathcal{U}\cap \mathcal{S}=\varnothing$. The Local LYM inequality for the two equally sized middle layers gives $|\mathcal{S}|\ge |\mathcal{L}|$, while $|\mathcal{L}|+|\mathcal{U}|=M$ forces $|\mathcal{S}|=|\mathcal{L}|$ and $\mathcal{U}=\left(\genfrac{}{}{0ex}{}{[n]}{k+1}\right)\setminus \mathcal{S}$.

Consider the inclusion bipartite graph between these two middle layers. It is a $(k+1)$-regular graph. Since $|\mathcal{S}|=|\mathcal{L}|$, all $(k+1)|\mathcal{S}|$ edges incident to $\mathcal{S}$ must come from $\mathcal{L}$; there are no edges from its complement on the lower side into $\mathcal{S}$. This bipartite graph is a connected graph: any two $k$-subsets can be joined by single-element exchanges, and each exchange passes through their common $(k+1)$-superset; every upper vertex has a lower neighbour. For $k=0$ it is just one edge. Hence either $\mathcal{L}=\varnothing$ or $\mathcal{L}$ is the whole lower layer. We have proved the complete classification:The two possibilities coincide for even $n$; for $n=0$ the unique maximum antichain is $\{\varnothing \}$.

Finally enumerate the members of the separating set system as $A_1,\dots ,A_m$ and encode each point $i\in [n]$ by $T_i=\{a\in [m]:i\in A_a\}$. Separation of the ordered pair $(i,j)$ means $T_i\setminus T_j\ne \varnothing$. Applying the hypothesis also to $(j,i)$ shows that the $n$ sets $T_i$ are distinct and pairwise incomparable. They form an antichain in $\mathcal{P}([m])$, so the Sperner theorem immediately gives

For a family $\mathcal{F}\subseteq \left(\genfrac{}{}{0ex}{}{[n]}{r}\right)$, let its lower shadow $\partial \mathcal{F}$ consist of all $(r-1)$-sets contained in members of $\mathcal{F}$. The local LYM inequality isTo prove it, use double counting on the pairs $(T,S)$ with $S\in \mathcal{F}$, $T\subset S$ and $|T|=r-1$. Each $S$ contributes $r$ pairs; each $T\in \partial \mathcal{F}$ contributes at most $n-r+1$. Hence $r|\mathcal{F}|\le (n-r+1)|\partial \mathcal{F}|$, exactly the displayed Local LYM inequality after using $r\left(\genfrac{}{}{0ex}{}{n}{r}\right)=(n-r+1)\left(\genfrac{}{}{0ex}{}{n}{r-1}\right)$.

To deduce the LYM inequality, start with an antichain $\mathcal{A}$ in the Boolean lattice and assign it weight ${\sum }_{S\in \mathcal{A}}{\left(\genfrac{}{}{0ex}{}{n}{|S|}\right)}^{-1}$. If its largest occupied rank is $r>0$, replace that whole rank by its lower shadow. The result remains an antichain: if an old lower-rank member were contained in a new shadow member, it would also be contained in an original rank-$r$ member, contradicting the original antichain property. Nor can a new member be contained in an old one, since every old member has size at most $r-1$ and equality would give the same contradiction. The Local LYM inequality shows that this replacement does not decrease the weight. Repeating eventually leaves an antichain on rank zero, whose weight is at most one. ThereforeFor the empty family the conclusion is immediate; the same argument includes the singleton family containing the empty set.

The antichains of maximum size are exactly the full middle levels. More precisely, for even $n$ only $\left(\genfrac{}{}{0ex}{}{[n]}{n/2}\right)$ attains the bound in Sperner theorem; for odd $n$ either $\left(\genfrac{}{}{0ex}{}{[n]}{(n-1)/2}\right)$ or $\left(\genfrac{}{}{0ex}{}{[n]}{(n+1)/2}\right)$ does. To prove that there are no others, put $M=\left(\genfrac{}{}{0ex}{}{n}{\lfloor n/2\rfloor }\right)$. If $|\mathcal{A}|=M$, the LYM inequality and $\left(\genfrac{}{}{0ex}{}{n}{r}\right)\le M$ imply equality in every term's bound $1/\left(\genfrac{}{}{0ex}{}{n}{r}\right)\ge 1/M$. Thus every member lies on a middle level. For even $n$, there is only one such level, so it must be full.

For odd $n=2r+1$, let $X$ be the rank-$r$ members and $Y$ the rank-$(r+1)$ members. Consider the bipartite graph of inclusions between those levels. Both parts have size $M$, and the degree of a vertex is $r+1$. If $N(X)$ is the neighbourhood of $X$, double counting gives $|N(X)|\ge |X|$. Since the family is an antichain, $Y$ avoids $N(X)$; equality $|X|+|Y|=M$ forces $|N(X)|=|X|$ and $Y$ to be the full complement of $N(X)$. Equality in the degree count means that every edge into $N(X)$ comes from $X$, so there are no edges between $X\cup N(X)$ and its complement.

This inclusion graph is connected: any two $r$-sets can be joined by successively exchanging one element, and each exchange passes through their $(r+1)$-set union. Every upper vertex has a lower neighbour. Thus $X$ must be empty or the whole lower level, proving the classification. For $n=1$ the graph is just one edge, and for $n=0$ the unique maximum family is $\{\varnothing \}$. This establishes equality in Sperner theorem without assuming that equality in the LYM inequality alone always forces a middle level.

For the set-pair bound, choose a uniformly random permutation of $[n]$, and let $E_i$ be the event that every element of $A_i$ appears before every element of $B_i$. Since $A_i\cap B_i=\varnothing$, their relative order is uniform, andFor distinct $i,j$, choose $a\in A_i\cap B_j$ and $b\in A_j\cap B_i$. These are distinct because $A_i\cap B_i$ is empty. On $E_i$, $a$ precedes $b$; on $E_j$, $b$ precedes $a$. Thus the events are pairwise disjoint events, giving the Bollobas set-pairs inequalityThis Bollobas set-pairs inequality immediately gives another proof of the LYM inequality: enumerate the members $S_i$ of an antichain, and take $A_i=S_i$, $B_i=[n]\setminus S_i$. For $i\ne j$, incomparability means $S_i\setminus S_j\ne \varnothing$, while $A_i\cap B_i=\varnothing$. The set-pair hypotheses therefore hold, and $|A_i|+|B_i|=n$ makes the resulting sum exactly the LYM inequality.