The conditions imply that $B$ is rig-smooth over $(A, I)$, see Lemma \ref{lemma-equivalent-with-artin-smooth}. Write $b' \Delta = b$ in $B$ for some $b' \in B$. Say $I = (a_1, \ldots, a_t)$. Since $V(b) \subset V(IB)$ there exists an integer $c \geq 0$ such that $I^cB \subset bB$. Write $bb_i = a_i^c$ in $B$ for some $b_i \in B$. \medskip\noindent Choose an integer $n \gg 0$ (we will see later how large). Choose polynomials $f'_1, \ldots, f'_m \in A[x_1, \ldots, x_r]$ such that $f_i - f'_i \in I^nA[x_1, \ldots, x_r]^\wedge$. We set $\Delta' = \det_{1 \leq i, j \leq m}(\partial f'_j/\partial x_i)$ and we consider the finite type $A$-algebra $$ C = A[x_1, \ldots, x_r, z_1, \ldots, z_t]/ (f'_1, \ldots, f'_m, z_1\Delta' - a_1^c, \ldots, z_t\Delta' - a_t^c) $$ We will apply Lemma \ref{lemma-presentation-rig-smooth} to $C$. We compute $$ \det\left( \begin{matrix} \text{matrix of partials of} \\ f'_1, \ldots, f'_m, z_1\Delta' - a_1^c, \ldots, z_t\Delta' - a_t^c \\ \text{with respect to the variables} \\ x_1, \ldots, x_m, z_1, \ldots, z_t \end{matrix} \right) = (\Delta')^{t + 1} $$ Hence we see that $\Ext^1_C(\NL_{C/A}, N)$ is annihilated by $(\Delta')^{t + 1}$ for all $C$-modules $N$. Since $a_i^c$ is divisible by $\Delta'$ in $C$ we see that $a_i^{(t + 1)c}$ annihilates these $\Ext^1$'s also. Thus $I^{c_1}$ annihilates $\Ext^1_C(\NL_{C/A}, N)$ for all $C$-modules $N$ where $c_1 = 1 + t((t + 1)c - 1)$. The exact value of $c_1$ doesn't matter for the rest of the argument; what matters is that it is independent of $n$. \medskip\noindent Since $\NL_{C^\wedge/A}^\wedge = \NL_{C/A} \otimes_C C^\wedge$ by Lemma \ref{lemma-NL-is-completion} we conclude that multiplication by $I^{c_1}$ is zero on $\Ext^1_{C^\wedge}(\NL_{C^\wedge/A}^\wedge, N)$ for any $C^\wedge$-module $N$ as well, see More on Algebra, Lemmas \ref{more-algebra-lemma-base-change-property-ext-1-annihilated} and \ref{more-algebra-lemma-two-term-base-change}. In particular $C^\wedge$ is rig-smooth over $(A, I)$. \medskip\noindent Observe that we have a surjective $A$-algebra homomorphism $$ \psi_n : C \longrightarrow B/I^nB $$ sending the class of $x_i$ to the class of $x_i$ and sending the class of $z_i$ to the class of $b_ib'$. This works because of our choices of $b'$ and $b_i$ in the first paragraph of the proof. \medskip\noindent Let $d = d(\text{Gr}_I(B))$ and $q_0 = q(\text{Gr}_I(B))$ be the integers found in  Local Cohomology, Section \ref{local-cohomology-section-uniform}. By Lemma \ref{lemma-get-morphism-general-better} if we take $n \geq \max(q_0 + (d + 1)c_1, 2(d + 1)c_1 + 1)$ we can find a homomorphism $\varphi : C^\wedge \to B$ of $A$-algebras which is congruent to $\psi_n$ modulo $I^{n - (d + 1)c_1}B$. \medskip\noindent Since $\varphi : C^\wedge \to B$ is surjective modulo $I$ we see that it is surjective (for example use Algebra, Lemma \ref{algebra-lemma-completion-generalities}). To finish the proof it suffices to show that $\Ker(\varphi)/\Ker(\varphi)^2$ is annihilated by a power of $I$, see More on Algebra, Lemma \ref{more-algebra-lemma-quotient-by-idempotent}. \medskip\noindent Since $\varphi$ is surjective we see that $\NL_{B/C^{\wedge}}^\wedge$ has cohomology modules $H^0(\NL_{B/C^{\wedge}}^\wedge) = 0$ and $H^{-1}(\NL_{B/C^{\wedge}}^\wedge) = \Ker(\varphi)/\Ker(\varphi)^2$. We have an exact sequence $$ H^{-1}(\NL_{C^\wedge/A}^\wedge \otimes_{C^\wedge} B) \to H^{-1}(\NL_{B/A}^\wedge) \to H^{-1}(\NL_{B/C^{\wedge}}^\wedge) \to H^0(\NL_{C^\wedge/A}^\wedge \otimes_{C^\wedge} B) \to H^0(\NL_{B/A}^\wedge) \to 0 $$ by Lemma \ref{lemma-exact-sequence-NL}. The first two modules are annihilated by a power of $I$ as $B$ and $C^\wedge$ are rig-smooth over $(A, I)$. Hence it suffices to show that the kernel of the surjective map $H^0(\NL_{C^\wedge/A}^\wedge \otimes_{C^\wedge} B) \to H^0(\NL_{B/A}^\wedge)$ is annihilated by a power of $I$. For this it suffices to show that it is annihilated by a power of $b$. In other words, it suffices to show that $$ H^0(\NL_{C^\wedge/A}^\wedge) \otimes_{C^\wedge} B[1/b] \longrightarrow H^0(\NL_{B/A}^\wedge) \otimes_B B[1/b] $$ is an isomorphism. However, both are free $B[1/b]$ modules of rank $r - m$ with basis $\text{d}x_{m + 1}, \ldots, \text{d}x_r$ and we conclude the proof.