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For a uniform set family $\mathcal{A}\subseteq [n{]}^{(r)}$, the Local LYM inequality isTo prove it, count the incident pairs $(B,A)$ with $B\in \partial \mathcal{A}$, $A\in \mathcal{A}$, and $B\subset A$. Every $A$ contributes exactly $r$ pairs, whereas every $B$ belongs to at most $n-r+1$ members of $[n{]}^{(r)}$. Thereforewhich is the displayed inequality after using the binomial coefficient identity $r\left(\genfrac{}{}{0ex}{}{n}{r}\right)=(n-r+1)\left(\genfrac{}{}{0ex}{}{n}{r-1}\right)$.

The LYM inequality says that every antichain $\mathcal{A}\subseteq \mathcal{P}([n])$ satisfiesFor the proof from the local inequality, take the lowest occupied level below the middle and replace that level by its upper shadow. The result remains an antichain: any new containment involving a set in the upper shadow would already give a containment involving the member of $\mathcal{A}$ immediately below it. The dual form of the Local LYM inequality says that this replacement cannot decrease the Lubell mass. Repeating upward below the middle, and similarly replacing high levels by their lower shadows, eventually puts the whole family in one middle level. Its final Lubell mass is at most one, so the original mass is also at most one.

For the maximal chain in a Boolean lattice proof, choose a Uniformly random maximal chain in a Boolean lattice. An $h$-element set lies on it with probability $1/\left(\genfrac{}{}{0ex}{}{n}{h}\right)$. Because an antichain meets each chain at most once, the expected value of the number of its members on the chain is at most one. By linearity of expectation, that expected value is precisely the displayed sum.

The Sperner theorem follows because $\left(\genfrac{}{}{0ex}{}{n}{|A|}\right)\le \left(\genfrac{}{}{0ex}{}{n}{\lfloor n/2\rfloor }\right)$ for every $A$. If equality holds in the resulting cardinality bound, every member lies on a largest level. For even $n$ this is the unique middle level. For odd $n$, the two middle levels have equal size; the regular connected inclusion graph between them and equality in Local LYM inequality force a chosen portion of the lower level to be either empty or the whole level. Hence the maximum antichains are exactly the complete middle level, with either middle level allowed when $n$ is odd.

Write $1_A$ for the indicator function and use the normalized Fourier analysis on a finite abelian groupThere are $N^5$ sextuples satisfying the equation, since any five coordinates determine the sixth. The desired probability is consequentlyBy orthogonality of complex exponentials, the indicator of the equation isSubstitution makes all six sums independent and gives the sixth Fourier moment as a three-sum collision count:Because $1_A$ is real, $\hat{A}(-r)=\overline{\hat{A}(r)}$. Hence the probability is

Let $X=X_1\bigsqcup \cdots \bigsqcup X_r$ and $Y=Y_1\bigsqcup \cdots \bigsqcup Y_s$ be set partitions of the two parts of a bipartite graph. The pair of partitions is $\epsilon$-regular whenThus a uniformly random pair in $X\times Y$ lies in an irregular cell pair with probability at most $\epsilon$.