Let $f \in A$. To prove that $\varphi$ is rig-flat we have to show that $\varphi_{\{f\}} : A_{\{f\}} \to B_{\{f\}}$ is naively rig-flat. Now either by viewing $\varphi_{\{f\}}$ as a base change of $\varphi$ and using Lemma \ref{lemma-base-change-rig-smooth-continuous} or by using the fact that being rig-smooth is a local property (Lemma \ref{lemma-rig-smooth-axioms}) we see that $\varphi_{\{f\}}$ is rig-smooth. Hence it suffices to show that $\varphi$ is naively rig-flat. \medskip\noindent Choose a presentation $B = A\{x_1, \ldots, x_n\}/J$. In order to check the second part of the lemma it suffices to check $J_\mathfrak q \subset A\{x_1, \ldots, x_n\}_\mathfrak q$ is generated by a regular sequence for $J \subset \mathfrak q$ for $\mathfrak q$ maximal with respect to not containing an ideal of definition, see Algebra, Lemma \ref{algebra-lemma-regular-sequence-in-neighbourhood} (which shows that the set of primes in $V(J)$ where there is a regular sequence generating $J$ is open). In other words, we may assume $\mathfrak q$ is rig-closed in $A\{x_1, \ldots, x_n\}$. And to check that $B$ is naively rig-flat, it also suffices to check that the corresponding localizations $B_\mathfrak q$ are flat over $A$. \medskip\noindent Let $\mathfrak q \subset A\{x_1, \ldots, x_n\}$ be rig-closed with $J \subset \mathfrak q$. By Lemma \ref{lemma-rig-closed-point-after-localization} we may choose an $f \in A$ mapping to a unit in $A\{x_1, \ldots, x_n\}/\mathfrak q$ and such that the prime ideal $\mathfrak p'$ in $A_{\{f\}}$ induced is rig-closed. Below we will use that $A_{\{f\}}\{x_1, \ldots, x_n\} = A\{x_1, \ldots, x_n\}_{\{f\}}$; details omitted. Consider the diagram $$ \xymatrix{ A\{x_1, \ldots, x_n\}_{\mathfrak q} / J_\mathfrak q \ar[r] & A_{\{f\}}\{x_1, \ldots, x_n\}_{\mathfrak q'}/ J A_{\{f\}}\{x_1, \ldots, x_n\}_{\mathfrak q'} \\ A\{x_1, \ldots, x_n\}_{\mathfrak q} \ar[r] \ar[u] & A_{\{f\}}\{x_1, \ldots, x_n\}_{\mathfrak q'} \ar[u] \\ A \ar[r] \ar[u] & A_{\{f\}} \ar[u] } $$ The middle horizontal arrow is faithfully flat as it is a local homomorphism of local rings and flat as $A_{\{f\}}\{x_1, \ldots, x_n\}$ is the completion of a localization of the Noetherian ring $A\{x_1, \ldots, x_n\}$. Similarly the bottom horizontal arrow is flat. Hence to show that $J_\mathfrak q$ is generated by a regular sequence and that $A \to A\{x_1, \ldots, x_n\}_{\mathfrak q} / J_\mathfrak q$ is flat, it suffices to prove the same things for $J A_{\{f\}}\{x_1, \ldots, x_n\}_{\mathfrak q'}$ and $A_{\{f\}} \to A_{\{f\}}\{x_1, \ldots, x_n\}_{\mathfrak q'}/ J A_{\{f\}}\{x_1, \ldots, x_n\}_{\mathfrak q'}$. See Algebra, Lemma \ref{algebra-lemma-flat-increases-depth} or More on Algebra, Lemma \ref{more-algebra-lemma-flat-descent-regular-ideal} for the statement on regular sequences. Finally, we have already seen that $A_{\{f\}} \to B_{\{f\}}$ is rig-smooth. This reduces us to the case discussed in the next paragraph. \medskip\noindent Let $\mathfrak q \subset A\{x_1, \ldots, x_n\}$ be rig-closed with $J \subset \mathfrak q$ such that moreover $\mathfrak p = A \cap \mathfrak q$ is rig-closed as well. By the characterization of rig-smooth algebras given in Lemma \ref{lemma-equivalent-with-artin-smooth} after reordering the variables $x_1, \ldots, x_n$ we can find $m \geq 0$ and $f_1, \ldots, f_m \in J$ such that \begin{enumerate} \item $J_\mathfrak q$ is generated by $f_1, \ldots, f_m$, and \item $\det_{1 \leq i, j \leq m}(\partial f_j/ \partial x_i)$ maps to a unit in $A\{x_1, \ldots, x_n\}_\mathfrak q$. \end{enumerate} By Lemma \ref{lemma-fibre-regular} the fibre ring $$ F = A\{x_1, \ldots, x_n\} \otimes_A \kappa(\mathfrak p) $$ is regular. Observe that the $A$-derivations $\partial / \partial x_i$ extend (uniquely) to derivations $D_i : F \to F$. By More on Algebra, Lemma \ref{more-algebra-lemma-quotient-sequence-regular} we see that $f_1, \ldots, f_m$ map to a regular sequence in $F_\mathfrak q$. By flatness of $A \to A\{x_1, \ldots, x_n\}$ and Algebra, Lemma \ref{algebra-lemma-grothendieck-regular-sequence} this shows that $f_1, \ldots, f_m$ map to a regular sequence in $A\{x_1, \ldots, x_m\}_\mathfrak q$ and the quotient by these elements is flat over $A$. This finishes the proof.