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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