SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\operatorname {r a n k} \left(\mathcal {A} _ {t - 1}\right) \leq \operatorname {r a n k} \left(\mathcal {A} _ {t}\right) \leq \operatorname {r a n k} \left(\mathcal {A} _ {t - 1}\right) + 1.
$$
If  $\mathrm{rank}(\mathcal{A}_t) = \mathrm{rank}(\mathcal{A}_{t - 1})$  then we must also have  $p_{t - 1}^{\prime} = p_{t}^{\prime}$  and the inductive hypothesis holds. Otherwise,  $\mathrm{rank}(\mathcal{A}_t) = \mathrm{rank}(\mathcal{A}_{t - 1}) + 1$ . Note that the  $t^{\mathrm{th}}$  row of  $\mathcal{A}$  has only  $s$  nonzero entries, so  $p_t^\prime \leq p_{t - 1}^\prime +s$ . In this case, we have that
$$
\begin{array}{l} p _ {t} ^ {\prime} - s \cdot \operatorname {r a n k} \left(\mathcal {A} _ {t}\right) = p _ {t} ^ {\prime} - s \cdot \left(\operatorname {r a n k} \left(\mathcal {A} _ {t - 1}\right) + 1\right) \\ \leq (p _ {t - 1} ^ {\prime} + s) - s \cdot (\operatorname {r a n k} (\mathcal {A} _ {t - 1}) + 1) \leq 0 \\ \end{array}
$$
where the last inequality holds due to the inductive hypothesis. Since  $\mathrm{rank}(\mathcal{A}) = r$ , we have that  $p' \leq rs$ . This completes the proof.