Originally posted on [MathStackExchange](https://math.stackexchange.com/q/4776370/573047) but without answers.

Consider partitions $\mathcal{F}=\{\mathcal{A_1},\ldots,\mathcal{A_n} \}$ of the powerset without the empty set element $Q = \mathcal{P}([n]) \setminus \{\emptyset\}$.

Let $\mathcal{F}_a = \{\mathcal{A} \in \mathcal{F} : \exists B \in \mathcal{A} : a \in B\}$.

We know that for a given $m \le n$, $\vert \mathcal{F}_x \vert \le m, 1 \le x \le n$.

Let $G = \{\{B,C\} : \exists \mathcal{A} \in \mathcal{F} : (B \in \mathcal{A}) \land (C \in \mathcal{A}) \land (B \cap C = \emptyset)\}$.

I would like to find a lower bound for $\vert G \vert$ over all possible partitions as a function of $m$:

$$\vert G \vert \ge f(m)$$

I think it is easier to start with higher values of $m$. For example for $m = n$, $f(m) = 0$ because we can choose $\mathcal{A_1} = \{A \in Q: 1 \in A \}, \mathcal{A_2} = \{A \in Q, A \not\in \mathcal{A_1}: 2 \in A \}, \mathcal{A_3} = \{A \in Q, A \not\in \mathcal{A_1},\mathcal{A_2}: 3 \in A \},\ldots,\mathcal{A_{n-1}} = \{A \in Q, A \not\in \mathcal{A_1},\ldots,\mathcal{A_{n-2}}: n-1 \in A \}=\{\{n-1\},\{n-1,n\}\}, \mathcal{A_n} = \{\{n\}\}$.

I believe $f(n-1) = 1$ although up to now I don't have a proof. Can we prove it and compute $f(m)$ for $m \le n-2$?