Proof.
Assume $IN \not= N$, $IN' \not= N'$, and $IN'' \not= N''$. Then we can use the characterization of depth using the Ext groups $\mathop{\mathrm{Ext}}\nolimits ^ i(A/I, N)$, see Lemma 47.11.1, and use the long exact cohomology sequence
\[ \begin{matrix} 0 \to \mathop{\mathrm{Hom}}\nolimits _ A(A/I, N') \to \mathop{\mathrm{Hom}}\nolimits _ A(A/I, N) \to \mathop{\mathrm{Hom}}\nolimits _ A(A/I, N'')
\\ \phantom{0\ } \to \mathop{\mathrm{Ext}}\nolimits ^1_ A(A/I, N') \to \mathop{\mathrm{Ext}}\nolimits ^1_ A(A/I, N) \to \mathop{\mathrm{Ext}}\nolimits ^1_ A(A/I, N'') \to \ldots
\end{matrix} \]
from Algebra, Lemma 10.71.6. This argument also works if $IN = N$ because in this case $\mathop{\mathrm{Ext}}\nolimits ^ i_ A(A/I, N) = 0$ for all $i$. Similarly in case $IN' \not= N'$ and/or $IN'' \not= N''$.
$\square$
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