serre normality criterion

If $R$ is normal, each prime localization is a normal domain. The domain criterion gives the required depth bound there and regularity when its dimension is at most one. Thus $R$ satisfies both conditions. The zero ring satisfies all three conditions vacuously.

Conversely $(R_1)$ and $(S_2)$ imply $(R_0)$ and $(S_1)$, so $R$ is reduced. Both conditions pass to prime localizations, since prime chains below a prime and successive localizations are unchanged. It is enough to prove that every reduced Noetherian local ring $A$ satisfying them is a normal domain. Induct on its finite dimension $d$. For $d\le 1$, $(R_1)$ makes $A$ regular; in dimension zero its maximal ideal is zero by the generator formula, so it is a field; in dimension one it is a DVR, hence an integrally closed domain by the valuation theorem.

Let $d\ge 2$ and $u\in Q(A)$ be integral over $A$. A monic equation shows $B=A[u]$ is finite, generated by finitely many powers of $u$. For a nonmaximal prime $\mathfrak{p}$ of $A$, the dimension of $A_{\mathfrak{p}}$ is less than $d$ (append the maximal ideal to any chain below $\mathfrak{p}$), so it is a normal domain by induction. A nonzerodivisor of $A$ remains a nonzerodivisor after localization: clear denominators in the equation it kills. Thus $Q(A{)}_{\mathfrak{p}}$ embeds into $Q(A_{\mathfrak{p}})$. The image of $u$ is integral and belongs to $A_{\mathfrak{p}}$, giving $B_{\mathfrak{p}}=A_{\mathfrak{p}}$.

Therefore $B\mathrm{/}A$ is supported only at the maximal ideal. Since $(S_2)$ gives depth at least two, finite-extension rigidity implies $B=A$. Every integral element of $Q(A)$ lies in $A$. The total-fraction component theorem now makes $A$ a finite product of normal domains. A nonzero local ring has no idempotents except zero and one: one of $e,1-e$ is a unit, forcing the other to vanish. Thus the product has a single factor and $A$ is a normal domain. This completes the local induction and hence the global converse.