**Proof.**
By Lemma 53.19.5 we may change our system of coordinates and the choice of generators during the proof.

If (1) holds, then we may change coordinates such that $x_1, \ldots , x_{n - 2}$ map to zero in $A/I$ and $A/I = k[[x_{n - 1}, x_ n]]/(a x_{n - 1}^2 + b x_{n - 1}x_ n + c x_ n^2)$ for some nondegenerate quadric $a x_{n - 1}^2 + b x_{n - 1}x_ n + c x_ n^2$. Then we can explicitly compute to show that both (2) and (3) are true.

Assume the $(n - 1) \times (n - 1)$ minors of the matrix $(\partial f_ j/\partial x_ i)$ generate $I + \mathfrak m_ A$. Suppose that for some $i$ and $j$ the partial derivative $\partial f_ j/\partial x_ i$ is a unit in $A$. Then we may use the system of parameters $f_ j, x_1, \ldots , x_{i - 1}, \hat x_ i, x_{i + 1}, \ldots , x_ n$ and the generators $f_ j, f_1, \ldots , f_{j - 1}, \hat f_ j, f_{j + 1}, \ldots , f_ m$ of $I$. Then we get a regular system of parameters $x_1, \ldots , x_ n$ and generators $x_1, f_2, \ldots , f_ m$ of $I$. Next, we look for an $i \geq 2$ and $j \geq 2$ such that $\partial f_ j/\partial x_ i$ is a unit in $A$. If such a pair exists, then we can make a replacement as above and assume that we have a regular system of parameters $x_1, \ldots , x_ n$ and generators $x_1, x_2, f_3, \ldots , f_ m$ of $I$. Continuing, in finitely many steps we reach the situation where we have a regular system of parameters $x_1, \ldots , x_ n$ and generators $x_1, \ldots , x_ t, f_{t + 1}, \ldots , f_ m$ of $I$ such that $\partial f_ j/\partial x_ i \in \mathfrak m_ A$ for all $i, j \geq t + 1$.

In this case the matrix of partial derivatives has the following block shape

\[ \left( \begin{matrix} I_{t \times t}
& *
\\ 0
& \mathfrak m_ A
\end{matrix} \right) \]

Hence every $(n - 1) \times (n - 1)$-minor is in $\mathfrak m_ A^{n - 1 - t}$. Note that $I \not= \mathfrak m_ A$ otherwise the ideal of minors would contain $1$. It follows that $n - 1 - t \leq 1$ because there is an element of $\mathfrak m_ A \setminus \mathfrak m_ A^2 + I$ (otherwise $I = \mathfrak m_ A$ by Nakayama). Thus $t \geq n - 2$. We have seen that $t \not= n$ above and similarly if $t = n - 1$, then there is an invertible $(n - 1) \times (n - 1)$-minor which is disallowed as well. Hence $t = n - 2$. Then $A/I$ is a quotient of $k[[x_{n - 1}, x_ n]]$ and Lemma 53.19.2 implies in both cases (2) and (3) that $I$ is generated by $x_1, \ldots , x_{n - 2}, f$ for some $f = f(x_{n - 1}, x_ n)$. In this case the condition on the minors exactly says that the quadratic term in $f$ is nondegenerate, i.e., $A/I$ is as in Lemma 53.19.3.
$\square$

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