We choose an $R$-algebra $B$ and a surjection $B \to A$. Set $\mathfrak p_B = (B \to A)^{-1}(\mathfrak p)$ and denote $r$ the relative dimension of $R \to B$ at $\mathfrak p_B$. We choose $B$ such that $R \to B$ is smooth at $\mathfrak p_B$. For example we can take $B$ to be a polynomial algebra in $r$ variables over $R$. Consider the complex $$ I/I^2 \otimes_A \Lambda \longrightarrow \Omega_{B/R} \otimes_B \Lambda $$ of Lemma \ref{lemma-neron-when-smooth}. By the structure of finite modules over $\Lambda$ (More on Algebra, Lemma \ref{more-algebra-lemma-modules-PID}) we see that the cokernel looks like $$ \Lambda^{\oplus d} \oplus \bigoplus\nolimits_{i = 1, \ldots, n} \Lambda/\pi^{e_i} \Lambda $$ for some $d \geq 0$, $n \geq 0$, and $e_i \geq 1$. Observe that $d$ is the relative dimension of $A/R$ at $\mathfrak q$ (Algebra, Lemma \ref{algebra-lemma-characterize-smooth-over-field}). If the defect $e = \sum_{i = 1, \ldots, n} e_i$ is zero, then we are done by Lemma \ref{lemma-neron-when-smooth}. \medskip\noindent Next, we consider what happens when we perform the N\'eron blowup. Recall that $A'$ is the quotient of $B'/IB'$ by its $\pi$-power torsion (Lemma \ref{lemma-neron-functorial}) and that $R \to B'$ is smooth at $\mathfrak p_{B'}$ (Lemma \ref{lemma-neron-blowup-smooth}). Thus after blowup we have exactly the same setup. Picture $$ \xymatrix{ 0 \ar[r] & I' \ar[r] & B' \ar[r] & A' \ar[r] & 0 \\ 0 \ar[r] & I \ar[u] \ar[r] & B \ar[u] \ar[r] & A \ar[r] \ar[u] & 0 } $$ Since $I \subset \mathfrak p_B$, we see that $I \to I'$ factors through $\pi I'$. Looking at the induced map of complexes we get $$ \xymatrix{ I'/(I')^2 \otimes_{A'} \Lambda \ar[r] & \Omega_{B'/R} \otimes_{B'} \Lambda \ar@{=}[r] & M' \\ I/I^2 \otimes_A \Lambda \ar[r] \ar[u] & \Omega_{B/R} \otimes_B \Lambda \ar[u] \ar@{=}[r] & M } $$ Then $M \subset M'$ are finite free $\Lambda$-modules with quotient $M'/M$ annihilated by $\pi$, see Lemma \ref{lemma-neron-blowup-smooth}. Let $N \subset M$ and $N' \subset M'$ be the images of the horizontal maps and denote $Q = M/N$ and $Q' = M'/N'$. We obtain a commutative diagram $$ \xymatrix{ 0 \ar[r] & N' \ar[r] & M' \ar[r] & Q' \ar[r] & 0 \\ 0 \ar[r] & N \ar[r] \ar[u] & M \ar[r] \ar[u] & Q \ar[r] \ar[u] & 0 } $$ Then $N \subset N'$ are free $\Lambda$-modules of rank $r - d$. Since $I$ maps into $\pi I'$ we see that $N \subset \pi N'$. \medskip\noindent Let $K = \Lambda_\pi$ be the fraction field of $\Lambda$. We have a commutative diagram $$ \xymatrix{ 0 \ar[r] & N' \ar[r] & N'_K \cap M' \ar[r] & Q'_{tor} \ar[r] & 0 \\ 0 \ar[r] & N \ar[r] \ar[u] & N_K \cap M \ar[r] \ar[u] & Q_{tor} \ar[r] \ar[u] & 0 } $$ whose rows are short exact sequences. This shows that the change in defect is given by $$ e - e' = \text{length}(Q_{tor}) - \text{length}(Q'_{tor}) = \text{length}(N'/N) - \text{length}(N'_K \cap M' / N_K \cap M) $$ Since $M'/M$ is annihilated by $\pi$, so is $N'_K \cap M' / N_K \cap M$, and its length is at most $\text{dim}_K(N_K)$. Since $N \subset \pi N'$ we get $\text{length}(N'/N) \ge \text{dim}_K(N_K)$, with equality if and only if $N = \pi N'$. \medskip\noindent To finish the proof we have to show that $N$ is strictly smaller than $\pi N'$ when $A$ is not smooth at $\mathfrak p$; this is the key computation one has to do in N\'eron's argument. To do this, we consider the exact sequence $$ I/I^2 \otimes_B \kappa(\mathfrak p_B) \to \Omega_{B/R} \otimes_B \kappa(\mathfrak p_B) \to \Omega_{A/R} \otimes_A \kappa(\mathfrak p) \to 0 $$ (follows from Algebra, Lemma \ref{algebra-lemma-differential-seq}). Since $R \to A$ is not smooth at $\mathfrak p$ we see that the dimension $s$ of $\Omega_{A/R} \otimes_A \kappa(\mathfrak p)$ is bigger than $d$. On the other hand the first arrow factors through the injective map $$ \mathfrak p B_\mathfrak p/\mathfrak p^2 B_\mathfrak p \to \Omega_{B/R} \otimes_B \kappa(\mathfrak p_B) $$ of Algebra, Lemma \ref{algebra-lemma-computation-differential}; note that $\kappa(\mathfrak p)$ is separable over $k$ by our assumption on $R/\pi R \subset \Lambda/\pi \Lambda$. Hence we conclude that we can find generators $g_1, \ldots, g_t \in I$ such that $g_j \in \mathfrak p^2$ for $j > r - s$. Then the images of $g_j$ in $A'$ are in $\pi^2 I'$ for $j > r - s$. Since $r - s < r - d$ we find that at least one of the minimal generators of $N$ becomes divisible by $\pi^2$ in $N'$. Thus we see that $e$ decreases by at least $1$ and we win.