**Proof.**
Write $C_0 = B_0[x_1, \ldots , x_ n]/(f_{1, 0}, \ldots , f_{m, 0})$. Write $B_ i = A_ i \otimes _{A_0} B_0$ and $C_ i = A_ i \otimes _{A_0} C_0$. Note that $C_ i = B_ i[x_1, \ldots , x_ n]/(f_{1, i}, \ldots , f_{m, i})$ where $f_{j, i}$ is the image of $f_{j, 0}$ in the polynomial ring over $B_ i$. Write $B = A \otimes _{A_0} B_0$ and $C = A \otimes _{A_0} C_0$. Note that $C = B[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$ where $f_ j$ is the image of $f_{j, 0}$ in the polynomial ring over $B$. The assumption is that the map

\[ \text{d} : (f_1, \ldots , f_ m)/(f_1, \ldots , f_ m)^2 \longrightarrow \bigoplus C \text{d}x_ k \]

is an isomorphism. Thus for sufficiently large $i$ we can find elements

\[ \xi _{k, i} \in (f_{1, i}, \ldots , f_{m, i})/(f_{1, i}, \ldots , f_{m, i})^2 \]

with $\text{d}\xi _{k, i} = \text{d}x_ k$ in $\bigoplus C_ i \text{d}x_ k$. Moreover, on increasing $i$ if necessary, we see that $\sum (\partial f_{j, i}/\partial x_ k) \xi _{k, i} = f_{j, i} \bmod (f_{1, i}, \ldots , f_{m, i})^2$ since this is true in the limit. Then this $i$ works.
$\square$

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