**Proof.**
Let $\mathfrak q$ be a prime of $S$ lying over a prime $\mathfrak p$ of $R$. By Lemma 10.163.2 we have

\[ \text{depth}(S_{\mathfrak q}) = \text{depth}(S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) + \text{depth}(R_{\mathfrak p}). \]

On the other hand, we have

\[ \dim (R_{\mathfrak p}) + \dim (S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) \geq \dim (S_{\mathfrak q}) \]

by Lemma 10.112.6. (Actually equality holds, by Lemma 10.112.7 but strictly speaking we do not need this.) Finally, as the fibre rings of the map are assumed $(S_ k)$ we see that $\text{depth}(S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) \geq \min (k, \dim (S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}))$. Thus the lemma follows by the following string of inequalities

\begin{eqnarray*} \text{depth}(S_{\mathfrak q}) & = & \text{depth}(S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) + \text{depth}(R_{\mathfrak p}) \\ & \geq & \min (k, \dim (S_{\mathfrak q}/\mathfrak pS_{\mathfrak q})) + \min (k, \dim (R_{\mathfrak p})) \\ & = & \min (2k, \dim (S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) + k, k + \dim (R_\mathfrak p), \dim (S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) + \dim (R_{\mathfrak p})) \\ & \geq & \min (k, \dim (S_{\mathfrak q})) \end{eqnarray*}

as desired.
$\square$

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