= Solution
Write $[n]^{(r)}$ for the rank-$r$ <uniform layer of the Boolean cube>. For a <uniform set family> $\mathcal F\subseteq[n]^{(r)}$, with $1\leq r\leq n$, let $\partial\mathcal F$ be its <lower shadow>. The \b[<Local LYM inequality>] is
$$
\boxed{\frac{|\partial\mathcal F|}{\binom n{r-1}}\geq\frac{|\mathcal F|}{\binom nr}}.
$$
For its proof, count incidences $(B,A)$ with $A\in\mathcal F$, $B\subset A$, and $|B|=r-1$. Each member of $\mathcal F$ contributes $r$ incidences, while each member of the <lower shadow> contributes at most $n-r+1$. Thus $r|\mathcal F|\leq(n-r+1)|\partial\mathcal F|$, which is the displayed bound by the ratio of consecutive <binomial coefficients>. Complementing every set gives the corresponding <upper shadow> bound, $|\nabla\mathcal F|/\binom n{r+1}\geq|\mathcal F|/\binom nr$ for $r<n$.
For an <antichain> $\mathcal A$ in the <Boolean lattice>, the \b[<LYM inequality>] is
$$
\boxed{\sum_{r=0}^n\frac{|\mathcal A\cap[n]^{(r)}|}{\binom nr}\leq1}.
$$
Here is the proof using the <Local LYM inequality>. Denote the rank-$r$ part by $\mathcal A_r$, and let $\mathcal D_r$ consist of the $r$-sets containing some member of $\mathcal A$ of size at most $r$. These families satisfy
$$
\mathcal D_r=\mathcal A_r\mathbin{\dot\cup}\nabla\mathcal D_{r-1}\quad(1\leq r\leq n),\qquad \mathcal D_0=\mathcal A_0.
$$
The union is disjoint because a member of the <antichain> cannot contain a strictly smaller member. Apply the <upper shadow> form of the <Local LYM inequality> to $\mathcal D_{r-1}$:
$$
\frac{|\mathcal D_r|}{\binom nr}\geq\frac{|\mathcal A_r|}{\binom nr}+\frac{|\mathcal D_{r-1}|}{\binom n{r-1}}.
$$
Iterating gives a bound for the entire sum by $|\mathcal D_n|/\binom nn\leq1$. This also covers the empty <antichain> and an <antichain> containing the empty set.
For the second proof, each <permutation> of $[n]$ specifies a <maximal chain in a Boolean lattice> by taking its successive prefixes. There are $n!$ such <maximal chains in a Boolean lattice>, and a fixed $r$-set belongs to $r!(n-r)!$ of them. A <maximal chain in a Boolean lattice> meets an <antichain> at most once. <Double counting> these incidences therefore gives $\sum_r|\mathcal A_r|r!(n-r)!\leq n!$, exactly the <LYM inequality>.
The \b[<Sperner theorem>] says that an <antichain> in $\mathcal P([n])$ has at most $\binom n{\lfloor n/2\rfloor}$ members; a full middle rank attains this bound. Indeed, every <binomial coefficient> $\binom nr$ is at most the central one, so the <LYM inequality> gives
$$
\frac{|\mathcal A|}{\binom n{\lfloor n/2\rfloor}}\leq\sum_r\frac{|\mathcal A_r|}{\binom nr}\leq1.
$$
Now let $\mathcal F$ be an <intersection-free uniform set family>. If it is empty, there is nothing to prove. Fix $X\in\mathcal F$ and consider the traces
$$
\mathcal T=\{X\cap A:A\in\mathcal F\setminus\{X\}\}\subseteq\mathcal P(X).
$$
If $A,B\ne X$ are distinct and $X\cap A\subseteq X\cap B$, then $X\cap A\subseteq B$. This contradicts the defining restriction on the three distinct members $X,A,B$. The intersection has size less than $r$, so it is a proper subset of $B$ even if the printed subset symbol is interpreted strictly. In particular, equal traces are impossible, and the traces form an <antichain>. Thus $|\mathcal T|=|\mathcal F|-1$. Applying the <Sperner theorem> to the $r$-element ground set $X$ proves the <antichain trace bound for intersection-free families>
$$
\boxed{|\mathcal F|\leq1+\binom r{\lfloor r/2\rfloor}}.
$$
The argument also handles $r=0$: a <uniform set family> then has at most one member.
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