**Proof.**
Let $L_1 = (L \otimes _ K K_1)_{red}$ and $M_1 = (M \otimes _ K K_1)_{red}$ and let $B_1 \subset L_1$ and $C_1 \subset M_1$ be the integral closure of $B$ and $C$. Note that $M_1 = (M \otimes _ L L_1)_{red}$ and that $L_1$ is a (nonempty) finite product of finite extensions of $L$. Hence the ring map $B_1 \to C_1$ is a finite product of ring maps of the form discussed in Remark 15.114.1. In particular, the map $\mathop{\mathrm{Spec}}(C_1) \to \mathop{\mathrm{Spec}}(B_1)$ is surjective. Choose a maximal ideal $\mathfrak m \subset C_1$ and consider the extensions of discrete valuation rings

\[ (A_1)_{A_1 \cap \mathfrak m} \to (B_1)_{B_1 \cap \mathfrak m} \to (C_1)_\mathfrak m \]

If the composition is weakly unramified, so is the map $(A_1)_{A_1 \cap \mathfrak m} \to (B_1)_{B_1 \cap \mathfrak m}$. If the residue field extension $\kappa _{A_1 \cap \mathfrak m} \to \kappa _\mathfrak m$ is separable, so is the subextension $\kappa _{A_1 \cap \mathfrak m} \to \kappa _{B_1 \cap \mathfrak m}$. Taking into account Lemma 15.111.5 this proves (1). A similar argument works for (2).
$\square$

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