Let $A \to S \to B$ be a good factorization. As $B$ is complete we obtain a factorization $A \to S^\wedge \to B$ where $S^\wedge$ is the completion of $S$. Note that this is also a good factorization: The ring map $S \to S^\wedge$ is flat (Algebra, Lemma \ref{algebra-lemma-completion-flat}), hence $A \to S^\wedge$ is flat. The ring $S^\wedge/\mathfrak m_A S^\wedge = (S/\mathfrak m_A S)^\wedge$ is regular since $S/\mathfrak m_A S$ is regular (More on Algebra, Lemma \ref{more-algebra-lemma-completion-regular}). Let $f_1, \ldots, f_r$ be a minimal sequence of generators of $\Ker(S \to B)$. 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 $S$ if and only if $f_1, \ldots, f_r$ is a regular sequence in the completion $S^\wedge$ by Algebra, Lemma \ref{algebra-lemma-flat-increases-depth}. Moreover, we have $$ S^\wedge/(f_1, \ldots, f_r)R^\wedge = (S/(f_1, \ldots, f_n))^\wedge = B^\wedge = B $$ because $B$ is $\mathfrak m_B$-adically complete (first equality by Algebra, Lemma \ref{algebra-lemma-completion-tensor}). Thus the kernel of $S \to B$ is generated by a regular sequence if and only if the kernel of $S^\wedge \to B$ is generated by a regular sequence. Hence it suffices to consider good factorizations where $S$ is complete. \medskip\noindent Assume we have two factorizations $A \to S \to B$ and $A \to S' \to B$ with $S$ and $S'$ complete. By More on Algebra, Lemma \ref{more-algebra-lemma-dominate-two-surjections} the ring $S \times_B S'$ is a Noetherian complete local ring. Hence, using More on Algebra, Lemma \ref{more-algebra-lemma-embed-map-Noetherian-complete-local-rings} we can choose a good factorization $A \to S'' \to S \times_B S'$ with $S''$ complete. Thus it suffices to show: If $A \to S' \to S \to B$ are comparable good factorizations, then $\Ker(S \to B)$ is generated by a regular sequence if and only if $\Ker(S' \to B)$ is generated by a regular sequence. \medskip\noindent Let $A \to S' \to S \to B$ be comparable good factorizations. First, since $S'/\mathfrak m_R S' \to S/\mathfrak m_R S$ is a surjection of regular local rings, the kernel is generated by a regular sequence $\overline{x}_1, \ldots, \overline{x}_c \in \mathfrak m_{S'}/\mathfrak m_R S'$ which can be extended to a regular system of parameters for the regular local ring $S'/\mathfrak m_R S'$, see (Algebra, Lemma \ref{algebra-lemma-regular-quotient-regular}). Set $I = \Ker(S' \to S)$. By flatness of $S$ over $R$ we have $$ I/\mathfrak m_R I = \Ker(S'/\mathfrak m_R S' \to S/\mathfrak m_R S) = (\overline{x}_1, \ldots, \overline{x}_c). $$ Choose lifts $x_1, \ldots, x_c \in I$. These lifts form a regular sequence generating $I$ as $S'$ is flat over $R$, see Algebra, Lemma \ref{algebra-lemma-grothendieck-regular-sequence}. \medskip\noindent We conclude that if also $\Ker(S \to B)$ is generated by a regular sequence, then so is $\Ker(S' \to B)$, see More on Algebra, Lemmas \ref{more-algebra-lemma-join-koszul-regular-sequences} and \ref{more-algebra-lemma-noetherian-finite-all-equivalent}. \medskip\noindent Conversely, assume that $J = \Ker(S' \to B)$ is generated by a regular sequence. Because the generators $x_1, \ldots, x_c$ of $I$ map to linearly independent elements of $\mathfrak m_{S'}/\mathfrak m_{S'}^2$ we see that $I/\mathfrak m_{S'}I \to J/\mathfrak m_{S'}J$ is injective. Hence there exists a minimal system of generators $x_1, \ldots, x_c, y_1, \ldots, y_d$ for $J$. Then $x_1, \ldots, x_c, y_1, \ldots, y_d$ is a regular sequence and it follows that the images of $y_1, \ldots, y_d$ in $S$ form a regular sequence generating $\Ker(S \to B)$. This finishes the proof of the lemma.