An $L$-intersecting family on $[n]$ with $\mid L\mid =s$ has at most ${\sum }_{i=0}^s\left(\genfrac{}{}{0px}{}{n}{i}\right)$ members

If $s>n$, then [F1] makes the claimed right-hand side $2^n=\mathrm{\mid }\mathcal{P}([n])\mathrm{\mid }$, so the bound follows from $\mathcal{F}\subseteq \mathcal{P}([n])$. Hence suppose $s\le n$. Work over $\mathbb{R}$, and for each $i$ define $f_i(x):={\prod }_{\mathrm{\ell }\in L,\mathrm{\ell }<\mid F_i\mid }(⟨x,v_{F_i}⟩-\mathrm{\ell })\mathrm{.}$ This is a polynomial of total degree at most $s$.

Evaluating at $v_{F_i}$, [L2] gives $⟨v_{F_i},v_{F_i}⟩=\mid F_i\mid$, so every factor in $f_i(v_{F_i})$ is a positive integer and therefore $f_i(v_{F_i})\mathrm{\ne }0$.

If $j<i$, then $\mid F_i\cap F_j\mid \in L$ and also $\mid F_i\cap F_j\mid \le \mid F_j\mid \le \mid F_i\mid$. Equality with $\mid F_i\mid$ would force $F_i\subseteq F_j$ and then $F_i=F_j$, impossible. So $\mid F_i\cap F_j\mid$ is an element of $L$ strictly below $\mid F_i\mid$, and [L2] makes the corresponding factor of $f_i(v_{F_j})$ equal to $0$.

Let ${\tilde{f}}_i$ be the multilinear reduction of $f_i$. By the cube-agreement lemma, ${\tilde{f}}_i(v_{F_i})\mathrm{\ne }0$ and ${\tilde{f}}_i(v_{F_j})=0$ for $j<i$. If ${\sum }_i{c}_i{\tilde{f}}_i=0$ and $j$ is the least index with $c_j\mathrm{\ne }0$, evaluation at $v_{F_j}$ kills the terms with index larger than $j$ by the vanishing just proved and kills the earlier ones by minimality, leaving $c_j{\tilde{f}}_j(v_{F_j})=0$, a contradiction. Thus the functions are linearly independent.

Each ${\tilde{f}}_i$ is multilinear of total degree at most $s$, so [L1] places all of them in a vector space of dimension ${\sum }_{i=0}^s\left(\genfrac{}{}{0px}{}{n}{i}\right)$. Since they are independent, there can be at most that many of them. Hence $m\le {\sum }_{i=0}^s\left(\genfrac{}{}{0px}{}{n}{i}\right)$.