In first-order Heyting arithmetic $HA$ with the usual language $0,S,+,\times$ and equality, let $P(x)$ be a formula with at most the variable $x$ free. If $HA \vdash \exists x\,P(x)$, then there is $k \in \mathbb N$ such that $HA \vdash P(\bar{k})$, where $\bar{k}$ is the numeral $S^k0$.