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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