Erdős Problem 1167. Let $r \geq 2$ be finite, $\gamma \geq 2$, and $\lambda$ be an infinite cardinal. Let $\kappa_\alpha$ be cardinals for all $\alpha < \gamma$. Is it true that $$2^\lambda \to (\kappa_\alpha + 1)_{\alpha < \gamma}^{r+1}$$ implies $$\lambda \to (\kappa_\alpha)_{\alpha < \gamma}^r?$$ Here $+$ means cardinal addition, so that $\kappa_\alpha + 1 = \kappa_\alpha$ if $\kappa_\alpha$ is infinite.

Infinite-target case. When all $\kappa_\alpha \geq \aleph_0$ are infinite and bounded by $\lambda$, $\kappa_\alpha + 1 = \kappa_\alpha$, so the hypothesis simplifies to a "pure" stepping-down lemma: $$2^\lambda \to (\kappa_\alpha)_{\alpha<\gamma}^{r+1} \implies \lambda \to (\kappa_\alpha)_{\alpha<\gamma}^r.$$ The condition $\kappa_\alpha \leq \lambda$ is needed to avoid a size obstruction: without it, the conclusion would require a subset of $\lambda$ of size $\kappa_\alpha > \lambda$, which is impossible (see infinite_targets_needs_bound).