**Proof.**
If (1) is true, then $M_\mathfrak p$ is a reflexive module for all primes of $\mathfrak p$ by Lemma 15.23.4. Thus (1) $\Rightarrow $ (2). Assume (2). Set $N = \mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Hom}}\nolimits _ R(M, R), R)$ so that

\[ N_\mathfrak p = \mathop{\mathrm{Hom}}\nolimits _{R_\mathfrak p}( \mathop{\mathrm{Hom}}\nolimits _{R_\mathfrak p}(M_\mathfrak p, R_\mathfrak p), R_\mathfrak p) \]

for every prime $\mathfrak p$ of $R$. See Algebra, Lemma 10.10.2. We apply Lemma 15.23.14 to the map $j : M \to N$. This is allowed because $M$ is finite and $N$ is torsion free by Lemma 15.22.12. In case (2)(a) the map $M_\mathfrak p \to N_\mathfrak p$ is an isomorphism and in case (2)(b) we have $\text{depth}(M_\mathfrak p) \geq 2$.
$\square$

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