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Source. Published pp. 225–226, Claim 3.6. (canonical PDF).

For the blocks and partitions defined in lemma_3_5, every $A^{(i)}$ is an ordered $(l_0,\ldots,l_k)$-partition of $[n]$, and

$$\left|\bigcap_{i=1}^r A^{(i)}_{t_i}\right|\ge b \qquad\text{for every }(t_1,\ldots,t_r)\in\{0,\ldots,k\}^r.$$

Proof.

For each fixed row $i$, its core parts $B^{(i)}_j$ partition all the $L_t$ blocks. Every added block $C_w$ enters exactly one part in that row, namely part $w_i$. All blocks are mutually disjoint, so the row is a partition of the full $n$-element set.

There are $q^{r-1}$ words with a given entry in position $i$. Hence

$$|A^{(i)}_0|=(s-k)l+bq^{r-1}=l_0, \qquad |A^{(i)}_j|=l+bq^{r-1}=l_j\quad(j\ge1).$$

The block $C_{(t_1,\ldots,t_r)}$ belongs to every one of the indicated parts. It has size $b$, proving the lower bound for every joint cell. In particular, all $l_j$ are positive, including $l_0$ if $s=k$.

Bears on. #174.