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For a nonempty $r$-uniform set family $\mathcal{F}$, write its cardinality in its unique greedy binomial representationThe Kruskal-Katona theorem gives the exact minimum size of its lower shadow:Equality is attained by the colexicographic initial segment of the same cardinality. The empty family has empty shadow. In colexicographic order, $A$ precedes $B$ exactly when the largest point of $A△B$ belongs to $B$. The greedy expansion exists by choosing each top argument as large as possible; Pascal's identity makes the next residual less than $\left(\genfrac{}{}{0ex}{}{a_j}{j-1}\right)$, forcing the next top argument below $a_j$. This also proves uniqueness.

Here is a proof by UV-compression of the theorem, including the step that controls the lower shadow. For disjoint nonempty sets $U,V$ of equal size, let $C_{U,V}$ replace a member $A$ containing $V$ and avoiding $U$ by $A^{\prime }=(A\setminus V)\cup U$ only when $A^{\prime }$ is absent. Other members stay. This UV-compression preserves cardinality. Allow only $\max U<\max V$, and at each step choose a changing compression with $|U|=|V|$ as small as possible. Every smaller allowed UV-compression then fixes the current family.

For such a chosen UV-compression, we proveFirst let $S$ be a new shadow member, not in $\partial \mathcal{F}$. It has the form $A^{\prime }\setminus \{x\}$ for a moved member $A$. If $x\in U$, choose $y\in V$ with $y\ne \max V$ when $|V|>1$. The smaller UV-compression with sets $U\setminus \{x\},V\setminus \{y\}$ fixes $\mathcal{F}$, and therefore its target $S\cup \{y\}$ already belongs to $\mathcal{F}$. This would put $S$ in the old shadow, a contradiction. For $|V|=1$ the same target is simply $A$, giving the same contradiction without a compression. Hence $x\notin U$. Now $A\setminus \{x\}$ is an old shadow member whose compression target is $S$; since $S$ was absent, it appears in the compressed shadow.

It remains to check that a surviving old shadow member is not removed on the right side. Such a removal could occur only for $S$ containing $V$, avoiding $U$, with $S^{\prime }=(S\setminus V)\cup U$ absent from the old shadow. A witness $B=S\cup \{x\}$ in the compressed family must be a retained old member, since newly moved members avoid $V$. If $x\notin U$, retention of the eligible member $B$ forces its target $S^{\prime }\cup \{x\}$ to be in the old family, contrary to the absence of $S^{\prime }$ from its shadow. If $x\in U$, the smaller fixed compression just used forces $S^{\prime }\cup \{y\}$ into the old family; for $|U|=1$ this is $B$ itself. Again there is a contradiction. This proves the containment, and hence the chosen compression cannot increase lower shadow cardinality.

Each change strictly decreases the integer potential ${\sum }_{A\in \mathcal{F}}{\sum }_{a\in A}{2}^a$. The procedure therefore terminates. Its terminal family is an initial segment of colexicographic order: otherwise take an absent earlier set $X$ and a present later set $Y$. The choice $U=X\setminus Y,V=Y\setminus X$ is an allowed changing compression. Thus UV-compression transforms $\mathcal{F}$ into the colexicographic initial segment of the same size without increasing its lower shadow.

Finally compute that shadow. The first block comprises all $r$-subsets of $[a_r]$ and contributes $\left(\genfrac{}{}{0ex}{}{a_r}{r-1}\right)$ shadow members. The next block has maximum point $a_r+1$ and its remaining $(r-1)$ points form a smaller colexicographic initial segment. Shadow members omitting $a_r+1$ are already in the first block's shadow; those containing it are counted by the lower shadow of that smaller initial segment. Recursion through the binomial representation gives precisely the displayed sum. This completes the proof.

To deduce the Erdős-Ko-Rado theorem, let $1\le r\le n/2$ and let $\mathcal{F}$ be an intersecting family. Put $k=n-r$ and take the complement family $\mathcal{B}=\{[n]\setminus A:A\in \mathcal{F}\}$. Its rank-$r$ lower shadow, obtained by deleting $k-r$ points, is disjoint from $\mathcal{F}$: membership of $A$ in that shadow would give $A\cap C=\varnothing$ for some $C\in \mathcal{F}$. If $|\mathcal{F}|>\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)=\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)$, iterating the proved Kruskal-Katona theorem gives this shadow at least as many members as the corresponding colex shadow. The colex family contains every $k$-set of $[n-1]$ and a further $k$-set containing $n$. Its rank-$r$ shadow therefore contains every $r$-set of $[n-1]$ and at least one more $r$-set containing $n$. This applies also when $k=r$, with no shadow operation. The two disjoint families would consequently have total size greater thana contradiction. Thus $\boxed{|\mathcal{F}|\le \left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)}$, attained by all $r$-sets containing a fixed point.

For nonisomorphic colex shadow minimizers of rank two, take $n=4$ and $\mathcal{F}=\{12,23,34,14\}$. Its shadow has four singletons, as does the initial four-set colex segment $\mathcal{I}=\{12,13,23,14\}$. The first family is the edge set of a cycle graph; its four point degrees are all two. The second has point degrees $3,2,2,1$, so no permutation makes them isomorphic.

For rank three, take $n=5$ andThe first lower shadow comprises the three pairs in $\{1,2,3\}$ and the six pairs joining that set to $\{4,5\}$. The second shadow comprises every pair on $[5]$ except $45$. Both therefore have size nine. Their point-degree multisets are respectively $\{4,4,4,3,3\}$ and $\{5,4,4,3,2\}$, so again the two families are not isomorphic. These examples can be embedded in larger ground sets by adding unused points.

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 an $r$-uniform set family $\mathcal{F}$, its lower shadow is the set family of all $(r-1)$-sets contained in a member. Write a positive integer $m$ in its unique binomial representationThe Kruskal-Katona theorem states thatFor $m=0$ the bound is zero. Equality is attained by the first $m$ $r$-sets in colexicographic order. In that order $A$ precedes $B$ when the largest element of their symmetric difference belongs to $B$.

Here is a complete UV-compression proof of the Kruskal-Katona theorem. For disjoint equal-sized sets $U,V$, put $C(A)=(A\setminus V)\cup U$ when $A$ contains every element of $V$ and no element of $U$, and put $C(A)=A$ otherwise. The UV-compression of a set family replaces an eligible $A$ only if its image is absent. It preserves cardinality: a retained collision keeps both the original set and its already present image.

We first prove the required Shadow lemma for UV-compressions. Suppose that, for each $x\in U$, a $y\in V$ can be chosen so the set family is fixed by $C_{U\setminus \{x\},V\setminus \{y\}}$. We claimConsider a lower shadow member obtained by deleting $z$ from a moved set $A^{\prime }=(A\setminus V)\cup U$. If $z\notin U$, it is the compressed image of $A\setminus \{z\}$, an old lower shadow member. If $z=x\in U$, the smaller compression of $A$ produces $A^{\prime }\setminus \{x\}\cup \{y\}$, which is in $\mathcal{F}$ by the assumed invariance. Thus $A^{\prime }\setminus \{x\}$ already belongs to the old lower shadow; it is not eligible to move since it lacks $x$.

It remains to consider a lower shadow member $B$ from a retained original member $A=B\cup \{z\}$. If $B$ is not eligible to move, it stays in the compressed lower shadow. If it is eligible, let $B^{\prime }=(B\setminus V)\cup U$. When $z\notin U$, retaining the eligible $A$ means its image already belongs to $\mathcal{F}$, and that image contains $B^{\prime }$. When $z=x\in U$, the smaller compression of $A$ belongs to $\mathcal{F}$ and equals $B^{\prime }\cup \{y\}$. In either case $B^{\prime }$ is in the old lower shadow too. Consequently the collision rule retains $B$ as well. This proves the claimed containment in every case. Since compression preserves cardinality, it cannot increase lower shadow size.

If $\mathcal{F}$ is not a colexicographic initial segment, an earlier absent set $A$ and a later present set $B$ give $U=A\setminus B$, $V=B\setminus A$ with $\max U<\max V$, and a nontrivial UV-compression. Choose such a compression with $|U|$ smallest. If $|U|>1$, choose $y\in V\setminus \{\max V\}$. For every $x\in U$, the smaller compression still has its largest differing element on the $V$ side and must fix $\mathcal{F}$ by minimality. If $|U|=1$, the smaller compression is the identity. Thus the lower shadow argument applies.

Each nontrivial step strictly decreases the positive integer weightbecause the largest differing coordinate belongs to $V$. Iteration therefore terminates. A terminal set family is a colexicographic initial segment, since an earlier gap and a later member would otherwise provide another nontrivial compression. The final set family has $m$ members and no larger lower shadow.

To compute its lower shadow, take $a_r$ maximal with $\left(\genfrac{}{}{0ex}{}{a_r}{r}\right)\le m$. All $r$-sets of $[a_r]$ occur first. The next block consists of sets containing $a_r+1$ whose remaining $(r-1)$-set runs through the initial segment of length $m-\left(\genfrac{}{}{0ex}{}{a_r}{r}\right)$. By Pascal's identity, the remainder is smaller than $\left(\genfrac{}{}{0ex}{}{a_r}{r-1}\right)$, so its next leading top index is strictly less than $a_r$. Repeating gives the stated unique binomial representation. The first block contributes all $\left(\genfrac{}{}{0ex}{}{a_r}{r-1}\right)$ $(r-1)$-sets in $[a_r]$ to the lower shadow. Deleting $a_r+1$ from the next block contributes only sets already counted; deleting another coordinate contributes $a_r+1$ joined to the lower shadow of the smaller initial segment. This recursive decomposition gives exactly the displayed sum. The base $r=1$ has lower shadow $\{\varnothing \}$ for every nonempty set family. This completes the proof and the sharpness assertion.

Now let $\mathcal{C}$ be the set family of vertex sets of the graph's cliques of order four. Its twice-iterated lower shadow consists of edges of the graph. If there were at least sixteen such copies, choose a sixteen-member subfamily. The relevant binomial representation isThe proved theorem gives at least $\left(\genfrac{}{}{0ex}{}{6}{3}\right)+\left(\genfrac{}{}{0ex}{}{3}{2}\right)=23$ triples in its first lower shadow. Apply it again to any twenty-three of those triples, using $23=\left(\genfrac{}{}{0ex}{}{6}{3}\right)+\left(\genfrac{}{}{0ex}{}{3}{2}\right)$. Their edge lower shadow has at leastmembers, contradicting the fifteen available edges. Conversely, the complete graph $K_6$ has $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$ edges and $\left(\genfrac{}{}{0ex}{}{6}{4}\right)=15$ copies of $K_4$. Therefore the maximum number of copies is exactly 15.