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For an intersection-free uniform set family of rank $r$, fix $X\in \mathcal{F}$. The map $A\mapsto X\cap A$ on $\mathcal{F}\setminus \{X\}$ is injective and its image is an antichain in $\mathcal{P}(X)$: a containment $X\cap A\subseteq X\cap B$ would violate intersection-freeness on the distinct members $X,A,B$. The Sperner theorem then proves the displayed bound.

For disjoint pairs $A_i,B_i$ such that $A_i\cap B_j\ne \varnothing$ whenever $i\ne j$, the displayed sum is at most one. In a uniformly random permutation, the events that all of $A_i$ precedes all of $B_i$ are disjoint events, and their probabilities are the summands. Taking $B_i$ to be the complement of the members $A_i$ of an antichain recovers the LYM inequality.

Two set families $\mathcal{A},\mathcal{B}$ are cross-Sperner when no member of either is a subset of a member of the other, including equality. Thus $\mathcal{A}\cap \mathcal{B}=\varnothing$. Neither family need itself be an antichain. On the Boolean lattice of subsets of $[n]$, the Cross-Sperner inequality bounds $\sqrt{|\mathcal{A}|}+\sqrt{|\mathcal{B}|}$ by $2^{n/2}$.

In any cyclic ordering on $n$ points, an antichain contains at most $n$ nonempty proper cyclic intervals. The intervals with one prescribed final position are nested, so at most one belongs to the antichain. Summing over final positions proves the bound. Averaging it over cyclic orderings proves the LYM inequality; the empty set and full set must be handled separately.

The maximum antichains in the Boolean lattice are exactly its full middle levels. For even $n$ there is one such level, and for odd $n$ either middle level works. The LYM inequality first confines a maximum antichain to those levels; connectedness of the regular inclusion bipartite graph rules out a mixture of the two odd middle levels.

An antichain in the Boolean lattice has Lubell mass one if and only if it is a full rank level. Equality means every maximal chain in a Boolean lattice meets the family exactly once. Swapping two adjacent entries in the defining permutation then forces the family to contain every set obtained from one member by exchanging an included and an excluded element, hence every set of that rank.

An intersection-free uniform set family is a uniform set family $\mathcal{F}$ with no three distinct members $A,B,C$ satisfying $A\cap B\subseteq C$. For distinct members of a rank-$r$ family, $|A\cap B|<r$, so using strict containment gives the same restriction. Fixing one member turns the other members' intersections with it into distinct antichain traces, giving the antichain trace bound for intersection-free families.

A uniform cover is irreducible when it is not the disjoint union, as a multiset, of two nonempty uniform covers. There are only finitely many irreducible uniform covers of a fixed finite set: their multiplicity vectors form an antichain in ${\mathbb{N}}^{2^n}$, and Dickson lemma forbids an infinite antichain there.