**Proof.**
Pick $x \in Z$. To prove (1) suffices to find an open neighbourhood $U \subset X$ of $x$ such that $U \to S$ is flat. Hence the lemma reduces to the case that $X = \mathop{\mathrm{Spec}}(B)$ and $S = \mathop{\mathrm{Spec}}(A)$ are affine and that $Z$ is given by an $H_1$-regular sequence $f_1, \ldots , f_ r \in B$. By assumption $B$ is a finitely presented $A$-algebra and $B/(f_1, \ldots , f_ r)B$ is a flat $A$-algebra. We are going to use absolute Noetherian approximation.

Write $B = A[x_1, \ldots , x_ n]/(g_1, \ldots , g_ m)$. Assume $f_ i$ is the image of $f_ i' \in A[x_1, \ldots , x_ n]$. Choose a finite type $\mathbf{Z}$-subalgebra $A_0 \subset A$ such that all the coefficients of the polynomials $f_1', \ldots , f_ r', g_1, \ldots , g_ m$ are in $A_0$. We set $B_0 = A_0[x_1, \ldots , x_ n]/(g_1, \ldots , g_ m)$ and we denote $f_{i, 0}$ the image of $f_ i'$ in $B_0$. Then $B = B_0 \otimes _{A_0} A$ and

\[ B/(f_1, \ldots , f_ r) = B_0/(f_{0, 1}, \ldots , f_{0, r}) \otimes _{A_0} A. \]

By Algebra, Lemma 10.168.1 we may, after enlarging $A_0$, assume that $B_0/(f_{0, 1}, \ldots , f_{0, r})$ is flat over $A_0$. It may not be the case at this point that the Koszul cohomology group $H_1(K_\bullet (B_0, f_{0, 1}, \ldots , f_{0, r}))$ is zero. On the other hand, as $B_0$ is Noetherian, it is a finitely generated $B_0$-module. Let $\xi _1, \ldots , \xi _ n \in H_1(K_\bullet (B_0, f_{0, 1}, \ldots , f_{0, r}))$ be generators. Let $A_0 \subset A_1 \subset A$ be a larger finite type $\mathbf{Z}$-subalgebra of $A$. Denote $f_{1, i}$ the image of $f_{0, i}$ in $B_1 = B_0 \otimes _{A_0} A_1$. By More on Algebra, Lemma 15.31.3 the map

\[ H_1(K_\bullet (B_0, f_{0, 1}, \ldots , f_{0, r})) \otimes _{A_0} A_1 \longrightarrow H_1(K_\bullet (B_1, f_{1, 1}, \ldots , f_{1, r})) \]

is surjective. Furthermore, it is clear that the colimit (over all choices of $A_1$ as above) of the complexes $K_\bullet (B_1, f_{1, 1}, \ldots , f_{1, r})$ is the complex $K_\bullet (B, f_1, \ldots , f_ r)$ which is acyclic in degree $1$. Hence

\[ \mathop{\mathrm{colim}}\nolimits _{A_0 \subset A_1 \subset A} H_1(K_\bullet (B_1, f_{1, 1}, \ldots , f_{1, r})) = 0 \]

by Algebra, Lemma 10.8.8. Thus we can find a choice of $A_1$ such that $\xi _1, \ldots , \xi _ n$ all map to zero in $H_1(K_\bullet (B_1, f_{1, 1}, \ldots , f_{1, r}))$. In other words, the Koszul cohomology group $H_1(K_\bullet (B_1, f_{1, 1}, \ldots , f_{1, r}))$ is zero.

Consider the morphism of affine schemes $X_1 \to S_1$ equal to $\mathop{\mathrm{Spec}}$ of the ring map $A_1 \to B_1$ and $Z_1 = \mathop{\mathrm{Spec}}(B_1/(f_{1, 1}, \ldots , f_{1, r}))$. Since $B = B_1 \otimes _{A_1} A$, i.e., $X = X_1 \times _{S_1} S$, and similarly $Z = Z_1 \times _ S S_1$, it now suffices to prove (1) for $X_1 \to S_1$ and the relative $H_1$-regular immersion $Z_1 \to X_1$, see Morphisms, Lemma 29.25.7. Hence we have reduced to the case where $X \to S$ is a finite type morphism of Noetherian schemes. In this case we know that $X \to S$ is flat at every point of $Z$ by Lemma 31.22.3. Combined with the fact that the flat locus is open in this case, see Algebra, Theorem 10.129.4 we see that (1) holds. Part (2) then follows from an application of Lemma 31.22.4.
$\square$

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