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For a nonempty degree-$m$ layer of a down-set in an $N$-element Boolean lattice, the unique real number $x_m\ge m$ with $|{\mathcal{F}}_m|=\left(\genfrac{}{}{0ex}{}{x_m}{m}\right)$ measures an effective number of available coordinates. The Lovász shadow bound and $\partial {\mathcal{F}}_{m+1}\subseteq {\mathcal{F}}_m$ imply $x_{m+1}\le x_m$ whenever the later layer is nonempty. It is a real parameter, not the actual size of the set family's support.

The lower shadow consists of the sets obtained by removing one element from a member:For a nonempty uniform set family, write its size in its unique decreasing binomial coefficient expansionThe Kruskal-Katona theorem states thatA colexicographic initial segment attains equality; thus this is an exact minimum, not just an asymptotic estimate.

Here is a direct proof of the requested Lovász shadow bound, avoiding any unproved numerical interpolation of the Kruskal-Katona theorem. The real binomial coefficient means $\left(\genfrac{}{}{0ex}{}{x}{r}\right)=x(x-1)\cdots (x-r+1)/r!$. It is strictly increasing for $x>r-1$. Since the set family is nonempty and its size is an integer, necessarily $x\ge r$.

Apply the usual left coordinate shifts of a set family until the set family is fixed by every shift. A shift preserves its size and does not increase its lower shadow; the witness argument is supplied in Solution 3. Termination follows because every nontrivial shift reduces ${\sum }_{A\in \mathcal{A}}{\sum }_{i\in A}i$. It suffices to prove the inequality for this shifted set family.

Induct on $r$, and, for each fixed $r$, on $|\mathcal{A}|$. The cases $r=1$ and $|\mathcal{A}|=1$ give, respectively, the single empty shadow member and the $r$ faces of an $r$-set. Split the set family at coordinate one, writing ${\mathcal{A}}_0$ for the members avoiding it and ${\mathcal{A}}_1$ for the sets obtained by deleting it from members containing it. Every $(r-1)$-subset of a member of ${\mathcal{A}}_0$ belongs to ${\mathcal{A}}_1$: replace the one omitted coordinate by coordinate one. ThereforeThe two terms count shadow members avoiding and containing coordinate one.

Put $b=|{\mathcal{A}}_1|$. This is positive. If $b<\left(\genfrac{}{}{0ex}{}{x-1}{r-1}\right)$, the Pascal's identity gives $|{\mathcal{A}}_0|>\left(\genfrac{}{}{0ex}{}{x-1}{r}\right)$. In this case $x>r$ and ${\mathcal{A}}_0$ is a smaller nonempty set family. The mathematical induction hypothesis on family size gives $|\partial {\mathcal{A}}_0|>\left(\genfrac{}{}{0ex}{}{x-1}{r-1}\right)$, contradicting its containment in ${\mathcal{A}}_1$. Consequently $b\ge \left(\genfrac{}{}{0ex}{}{x-1}{r-1}\right)$. Write $b=\left(\genfrac{}{}{0ex}{}{z}{r-1}\right)$, so $z\ge x-1$. The mathematical induction hypothesis on uniformity and the Pascal's identity now giveFor $r=2$ the degree-zero term is the constant one, which causes no difficulty.

For the graph application, identify a triangle-free graph with a subset of the $N$ possible edges. Removing an edge preserves being triangle-free, so $\partial {\mathcal{G}}_{m+1}\subseteq {\mathcal{G}}_m$. If ${\mathcal{G}}_{m+1}$ is nonempty, the Lovász shadow bound givesThus, for $1\le m<N$ with $p_m>0$,If the next set family is empty its probability is zero, so the same inequality holds without introducing $x_{m+1}$. If $p_m=0$, every later set family is empty; the divided expression is undefined and the correct assertion is this zero-probability conclusion.

Iterate the monotonicity of the effective ground-set parameter of a hereditary uniform layer. Whenever $p_{m+k}>0$, all the intermediate parameters are at least their layer sizes, and all factors below are nonnegative. We obtain the stronger estimateThe second inequality follows from $x_m\le N$ and the fact that $(x_m-t)/(N-t)$ decreases with $t$. If the target set family is empty, the requested bound follows directly when its right side is nonnegative. When $k>x_m+1$, the printed product can have signed factors and should instead be stopped at the first empty layer: its unrestricted signed form is not a meaningful probability bound. In the nonzero target range there is no such ambiguity. A uniformly safe extension, for $0\le k\le N-m$, isIt agrees with the printed expression whenever all its factors are nonnegative. In particular $0\le k\le m$ is always safe when $p_m>0$, and this includes the requested doubling case.

Taking $k=m$ in the nonzero range,If the $2m$-edge set family is empty the conclusion is immediate; if $2m>N$, interpreting the event as impossible gives the same conclusion. These qualifications distinguish the probability assertion from products or ratios outside their domains.