Let $J \subset B$ be the kernel of $B \to A/I$ so that $B/J = A/I$. By Algebra, Lemma \ref{algebra-lemma-application-NL-smooth} the sequence $$ 0 \to I/I^2 \to J/J^2 \to \Omega_{B/A} \otimes_B B/J \to 0 $$ is split exact. Thus $\overline{P} = J/(J^2 + IB) = \Omega_{B/A} \otimes_B B/J$ is a finite projective $A/I$-module. Choose an integer $n$ and a direct sum decomposition $A/I^{\oplus n} = \overline{P} \oplus \overline{K}$. By Lemma \ref{lemma-lift-projective-module} we can find an \'etale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ and a finite projective $A$-module $K$ which lifts $\overline{K}$. We may and do replace $A$ by $A'$. Set $B' = B \otimes_A \text{Sym}_A^*(K)$. Since $A \to \text{Sym}_A^*(K)$ is smooth by Lemma \ref{lemma-symmetric-algebra-smooth} we see that $B \to B'$ is smooth which in turn implies that $A \to B'$ is smooth (see Algebra, Lemmas \ref{algebra-lemma-base-change-smooth} and \ref{algebra-lemma-locally-smooth}). Moreover the section $\text{Sym}^*_A(K) \to A$ determines a section $B' \to B$ and we let $B' \to A/I$ be the composition $B' \to B \to A/I$. Let $J' \subset B'$ be the kernel of $B' \to A/I$. We have $JB' \subset J'$ and $B \otimes_A K \subset J'$. These maps combine to give an isomorphism $$ (A/I)^{\oplus n} \cong J/J^2 \oplus \overline{K} \longrightarrow J'/((J')^2 + IB') $$ Thus, after replacing $B$ by $B'$ we may assume that $J/(J^2 + IB) = \Omega_{B/A} \otimes_B B/J$ is a free $A/I$-module of rank $n$. \medskip\noindent In this case, choose $f_1, \ldots, f_n \in J$ which map to a basis of $J/(J^2 + IB)$. Consider the finitely presented $A$-algebra $C = B/(f_1, \ldots, f_n)$. Note that we have an exact sequence $$ 0 \to H_1(L_{C/A}) \to (f_1, \ldots, f_n)/(f_1, \ldots, f_n)^2 \to \Omega_{B/A} \otimes_B C \to \Omega_{C/A} \to 0 $$ see Algebra, Lemma \ref{algebra-lemma-exact-sequence-NL} (note that $H_1(L_{B/A}) = 0$ and that $\Omega_{B/A}$ is finite projective, in particular flat so the Tor group vanishes). For any prime $\mathfrak q \supset J$ of $B$ the module $\Omega_{B/A, \mathfrak q}$ is free of rank $n$ because $\Omega_{B/A}$ is finite projective and because $\Omega_{B/A} \otimes_B B/J$ is free of rank $n$ (see Algebra, Lemma \ref{algebra-lemma-finite-projective}). By our choice of $f_1, \ldots, f_n$ the map $$ \left((f_1, \ldots, f_n)/(f_1, \ldots, f_n)^2\right)_{\mathfrak q} \to \Omega_{B/A, \mathfrak q} $$ is surjective modulo $J$. Hence we see that this map of modules over the local ring $C_{\mathfrak q}$ has to be an isomorphism (this is because by Nakayama's Algebra, Lemma \ref{algebra-lemma-NAK} the map is surjective and then for example by Algebra, Lemma \ref{algebra-lemma-fun} because $((f_1, \ldots, f_n)/(f_1, \ldots, f_n)^2)_{\mathfrak q}$ is generated by $n$ elements the map is injective). Thus $H_1(L_{C/A})_{\mathfrak q} = 0$ and $\Omega_{C/A, \mathfrak q} = 0$. By Algebra, Lemma \ref{algebra-lemma-smooth-at-point} we see that $A \to C$ is smooth at the prime $\overline{\mathfrak q}$ of $C$ corresponding to $\mathfrak q$. Since $\Omega_{C/A, \mathfrak q} = 0$ it is actually \'etale at $\overline{\mathfrak q}$. Thus $A \to C$ is \'etale at all primes of $C$ containing $JC$. By Lemma \ref{lemma-localize-upstairs} we can find an $f \in C$ mapping to an invertible element of $C/JC$ such that $A \to C_f$ is \'etale. By our choice of $f$ it is still true that $C_f/JC_f = A/I$. The map $C_f/IC_f \to A/I$ is surjective and \'etale by Algebra, Lemma \ref{algebra-lemma-map-between-etale}. Hence $A/I$ is isomorphic to the localization of $C_f/IC_f$ at some element $g \in C$, see Algebra, Lemma \ref{algebra-lemma-surjective-flat-finitely-presented}. Set $A' = C_{fg}$ to conclude the proof.