Write $X = \Spec(A)$, $X^\wedge = \Spec(A^\wedge)$, $Z = V(I) \subset X$, and $Z^\wedge = V(I^\wedge)$. Let $\mathfrak p' \subset \mathfrak q' \subset A^\wedge$ be primes with $\mathfrak p' \not \in Z^\wedge$ and $\mathfrak q' \in Z^\wedge$. Let $\mathfrak p \subset \mathfrak q$ be the corresponding primes of $A$. Then $\mathfrak p \not \in Z$ and $\mathfrak q \in Z$. Picture $$ \xymatrix{ \mathfrak p' \ar[r] & \mathfrak q' \ar[r] & A^\wedge \\ \mathfrak p \ar[r] \ar@{-}[u] & \mathfrak q \ar[r] \ar@{-}[u] & A \ar[u] } $$ Let us write \begin{align*} a & = \dim(A/\mathfrak p) = \dim(A^\wedge/\mathfrak pA^\wedge),\\ b & = \dim(A/\mathfrak q) = \dim(A^\wedge/\mathfrak qA^\wedge),\\ a' & = \dim(A^\wedge/\mathfrak p'),\\ b' & = \dim(A^\wedge/\mathfrak q') \end{align*} Equalities by More on Algebra, Lemma \ref{more-algebra-lemma-completion-dimension}. We also write \begin{align*} p & = \dim(A^\wedge_{\mathfrak p'}/\mathfrak p A^\wedge_{\mathfrak p'}) = \dim((A^\wedge/\mathfrak p A^\wedge)_{\mathfrak p'}) \\ q & = \dim(A^\wedge_{\mathfrak q'}/\mathfrak p A^\wedge_{\mathfrak q'}) = \dim((A^\wedge/\mathfrak q A^\wedge)_{\mathfrak q'}) \end{align*} Since $A$ is universally catenary we see that $A^\wedge/\mathfrak pA^\wedge = (A/\mathfrak p)^\wedge$ is equidimensional of dimension $a$ (More on Algebra, Proposition \ref{more-algebra-proposition-ratliff}). Hence $a = a' + p$. Similarly $b = b' + q$. By Algebra, Lemma \ref{algebra-lemma-apply-grothendieck-module} applied to the flat local ring map $A_\mathfrak p \to A^\wedge_{\mathfrak p'}$ we have $$ \text{depth}(M^\wedge_{\mathfrak p'}) = \text{depth}(M_\mathfrak p) + \text{depth}(A^\wedge_{\mathfrak p'} / \mathfrak p A^\wedge_{\mathfrak p'}) $$ The quantity we are minimizing for $s_{A, I}(M)$ is $$ s(\mathfrak p, \mathfrak q) = \text{depth}(M_\mathfrak p) + \dim((A/\mathfrak p)_\mathfrak q) = \text{depth}(M_\mathfrak p) + a - b $$ (last equality as $A$ is catenary). The quantity we are minimizing for $s_{A^\wedge, I^\wedge}(M^\wedge)$ is $$ s(\mathfrak p', \mathfrak q') = \text{depth}(M^\wedge_{\mathfrak p'}) + \dim((A^\wedge/\mathfrak p')_{\mathfrak q'}) = \text{depth}(M^\wedge_{\mathfrak p'}) + a' - b' $$ (last equality as $A^\wedge$ is catenary). Now we have enough notation in place to start the proof. \medskip\noindent Let $\mathfrak p \subset \mathfrak q \subset A$ be primes with $\mathfrak p \not \in Z$ and $\mathfrak q \in Z$ such that $s_{A, I}(M) = s(\mathfrak p, \mathfrak q)$. Then we can pick $\mathfrak q'$ minimal over $\mathfrak q A^\wedge$ and $\mathfrak p' \subset \mathfrak q'$ minimal over $\mathfrak p A^\wedge$ (using going down for $A \to A^\wedge$). Then we have four primes as above with $p = 0$ and $q = 0$. Moreover, we have $\text{depth}(A^\wedge_{\mathfrak p'} / \mathfrak p A^\wedge_{\mathfrak p'})=0$ also because $p = 0$. This means that $s(\mathfrak p', \mathfrak q') = s(\mathfrak p, \mathfrak q)$. Thus we get the first inequality. \medskip\noindent Assume that the formal fibres of $A$ are $(S_n)$. Then $\text{depth}(A^\wedge_{\mathfrak p'} / \mathfrak p A^\wedge_{\mathfrak p'}) \geq \min(n, p)$. Hence $$ s(\mathfrak p', \mathfrak q') \geq s(\mathfrak p, \mathfrak q) + q + \min(n, p) - p \geq s_{A, I}(M) + q + \min(n, p) - p $$ Thus the only way we can get in trouble is if $p > n$. If this happens then \begin{align*} s(\mathfrak p', \mathfrak q') & = \text{depth}(M^\wedge_{\mathfrak p'}) + \dim((A^\wedge/\mathfrak p')_{\mathfrak q'}) \\ & = \text{depth}(M_\mathfrak p) + \text{depth}(A^\wedge_{\mathfrak p'} / \mathfrak p A^\wedge_{\mathfrak p'}) + \dim((A^\wedge/\mathfrak p')_{\mathfrak q'}) \\ & \geq 0 + n + 1 \end{align*} because $(A^\wedge/\mathfrak p')_{\mathfrak q'}$ has at least two primes. This proves the second inequality.