Proof.
We first prove (1). Assume $s, t$ locally of finite type. By Lemma 83.5.6 we may assume that $R$ is a pre-equivalence relation. Let $k$ be an algebraically closed field over $B$. Suppose $\overline{u}, \overline{u}' \in U(k)$ are $R$-equivalent. Then for some extension field $\Omega /k$ there exists a point $\overline{r} \in R(\Omega )$ mapping to $(\overline{u}, \overline{u}') \in (U \times _ B U)(\Omega )$, see Lemma 83.5.5. Hence
\[ Z = R \times _{j, U \times _ B U, (\overline{u}, \overline{u}')} \mathop{\mathrm{Spec}}(k) \]
is nonempty. As $s$ is locally of finite type we see that also $j$ is locally of finite type, see Morphisms of Spaces, Lemma 67.23.6. This implies $Z$ is a nonempty algebraic space locally of finite type over the algebraically closed field $k$ (use Morphisms of Spaces, Lemma 67.23.3). Thus $Z$ has a $k$-valued point, see Morphisms of Spaces, Lemma 67.24.1. Hence we conclude there exists a $\overline{r} \in R(k)$ with $j(\overline{r}) = (\overline{u}, \overline{u}')$, and we conclude that $\overline{u}, \overline{u}'$ are $R$-equivalent as desired.
The proof of part (2) is the same, except that it uses Morphisms of Spaces, Lemma 67.24.2 instead of Morphisms of Spaces, Lemma 67.24.1. This shows that the assertion holds as soon as $|k| > \lambda (R)$ with $\lambda (R)$ as introduced just above Morphisms of Spaces, Lemma 67.24.1.
$\square$
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