**Proof.**
Let $k$ be the residue field of $A$. If the characteristic of $k$ is $p > 0$, then we denote $\Lambda $ a Cohen ring (Algebra, Definition 10.154.5) with residue field $k$ (Algebra, Lemma 10.154.6). If the characteristic of $k$ is $0$ we set $\Lambda = k$. Recall that $\Lambda [[x_1, \ldots , x_ n]]$ for any $n$ is formally smooth over $\mathbf{Z}$, resp. $\mathbf{Q}$ in the $\mathfrak m$-adic topology, see More on Algebra, Lemma 15.38.1. Fix a surjection $\Lambda [[x_1, \ldots , x_ n]] \to A$ as in the Cohen structure theorem (Algebra, Theorem 10.154.8).

Let $R \to A$ be a surjection from a regular local ring $R$. Let $f_1, \ldots , f_ r$ be a minimal sequence of generators of $\mathop{\mathrm{Ker}}(R \to A)$. We will use without further mention that an ideal in a Noetherian local ring is generated by a regular sequence if and only if any minimal set of generators is a regular sequence. Observe that $f_1, \ldots , f_ r$ is a regular sequence in $R$ if and only if $f_1, \ldots , f_ r$ is a regular sequence in the completion $R^\wedge $ by Algebra, Lemmas 10.67.5 and 10.96.2. Moreover, we have

\[ R^\wedge /(f_1, \ldots , f_ r)R^\wedge = (R/(f_1, \ldots , f_ n))^\wedge = A^\wedge = A \]

because $A$ is $\mathfrak m_ A$-adically complete (first equality by Algebra, Lemma 10.96.1). Finally, the ring $R^\wedge $ is regular since $R$ is regular (More on Algebra, Lemma 15.42.4). Hence we may assume $R$ is complete.

If $R$ is complete we can choose a map $\Lambda [[x_1, \ldots , x_ n]] \to R$ lifting the given map $\Lambda [[x_1, \ldots , x_ n]] \to A$, see More on Algebra, Lemma 15.36.5. By adding some more variables $y_1, \ldots , y_ m$ mapping to generators of the kernel of $R \to A$ we may assume that $\Lambda [[x_1, \ldots , x_ n, y_1, \ldots , y_ m]] \to R$ is surjective (some details omitted). Then we can consider the commutative diagram

\[ \xymatrix{ \Lambda [[x_1, \ldots , x_ n, y_1, \ldots , y_ m]] \ar[r] \ar[d] & R \ar[d] \\ \Lambda [[x_1, \ldots , x_ n]] \ar[r] & A } \]

By Algebra, Lemma 10.133.6 we see that the condition for $R \to A$ is equivalent to the condition for the fixed chosen map $\Lambda [[x_1, \ldots , x_ n]] \to A$. This finishes the proof of the lemma.
$\square$

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