Let us recall what the statement signifies. First,  $\textit{WAdm}^{Noeth}$ is the category whose objects are adic Noetherian topological rings and whose morphisms are continuous ring homomorphisms. Consider a commutative diagram $$ \xymatrix{ B \ar[r] & (B')^\wedge \\ A \ar[r] \ar[u]^\varphi & (A')^\wedge \ar[u]_{\varphi'} } $$ satisfying the following conditions: $A$ and $B$ are adic Noetherian topological rings, $A \to A'$ and $B \to B'$ are \'etale ring maps, $(A')^\wedge = \lim A'/I^nA'$ for some ideal of definition $I \subset A$, $(B')^\wedge = \lim B'/J^nB'$ for some ideal of definition $J \subset B$, and $\varphi : A \to B$ and $\varphi' : (A')^\wedge \to (B')^\wedge$ are continuous. Note that $(A')^\wedge$ and $(B')^\wedge$ are adic Noetherian topological rings by Formal Spaces, Lemma \ref{formal-spaces-lemma-completion-in-sub}. We have to show \begin{enumerate} \item $\varphi$ is rig-smooth $\Rightarrow \varphi'$ is rig-smooth, \item if $B \to B'$ faithfully flat, then $\varphi'$ is rig-smooth $\Rightarrow \varphi$ is rig-smooth, and \item if $A \to B_i$ is rig-smooth for $i = 1, \ldots, n$, then $A \to \prod_{i = 1, \ldots, n} B_i$ is rig-smooth. \end{enumerate} The equivalent conditions of Lemma \ref{lemma-finite-type} satisfy conditions (1), (2), and (3). Thus in verifying (1), (2), and (3) for the property ``rig-smooth'' we may already assume our ring maps satisfy the equivalent conditions of Lemma \ref{lemma-finite-type} in each case. \medskip\noindent Pick an ideal of definition $I \subset A$. By the remarks above the topology on each ring in the diagram is the $I$-adic topology and $B$, $(A')^\wedge$, and $(B')^\wedge$ are in the category (\ref{equation-C-prime}) for $(A, I)$. Since $A \to A'$ and $B \to B'$ are \'etale the complexes $\NL_{A'/A}$ and $\NL_{B'/B}$ are zero and hence $\NL_{(A')^\wedge/A}^\wedge$ and $\NL_{(B')^\wedge/B}^\wedge$ are zero by Lemma \ref{lemma-NL-is-completion}. Applying Lemma \ref{lemma-exact-sequence-NL} to $A \to (A')^\wedge \to (B')^\wedge$ we get isomorphisms $$ H^i(\NL_{(B')^\wedge/(A')^\wedge}^\wedge) \to H^i(\NL_{(B')^\wedge/A}^\wedge) $$ Thus $\NL_{(B')^\wedge/A}^\wedge \to \NL_{(B')^\wedge/(A')^\wedge}$ is a quasi-isomorphism. The ring maps $B/I^nB \to B'/I^nB'$ are \'etale and hence are local complete intersections (Algebra, Lemma \ref{algebra-lemma-etale-standard-smooth}). Hence we may apply Lemmas \ref{lemma-exact-sequence-NL} and \ref{lemma-transitive-lci-at-end} to $A \to B \to (B')^\wedge$ and we get isomorphisms $$ H^i(\NL_{B/A}^\wedge \otimes_B (B')^\wedge) \to H^i(\NL_{(B')^\wedge/A}^\wedge) $$ We conclude that $\NL_{B/A}^\wedge \otimes_B (B')^\wedge \to \NL_{(B')^\wedge/A}^\wedge$ is a quasi-isomorphism. Combining these two observations we obtain that $$ \NL_{(B')^\wedge/(A')^\wedge}^\wedge \cong \NL_{B/A}^\wedge \otimes_B (B')^\wedge $$ in $D((B')^\wedge)$. With these preparations out of the way we can start the actual proof. \medskip\noindent Proof of (1). Assume $\varphi$ is rig-smooth. Then there exists a $c \geq 0$ such that $\Ext^1_B(\NL_{B/A}^\wedge, N)$ is annihilated by $I^c$ for every $B$-module $N$. By More on Algebra, Lemmas \ref{more-algebra-lemma-two-term-base-change} and \ref{more-algebra-lemma-base-change-property-ext-1-annihilated} this property is preserved under base change by $B \to (B')^\wedge$. Hence $\Ext^1_{(B')^\wedge}(\NL_{(B')^\wedge/(A')^\wedge}^\wedge, N)$ is annihilated by $I^c(A')^\wedge$ for all $(B')^\wedge$-modules $N$ which tells us that $\varphi'$ is rig-smooth. This proves (1). \medskip\noindent To prove (2) assume $B \to B'$ is faithfully flat and that $\varphi'$ is rig-smooth. Then there exists a $c \geq 0$ such that $\Ext^1_{(B')^\wedge}(\NL_{(B')^\wedge/(A')^\wedge}^\wedge, N')$ is annihilated by $I^c(B')^\wedge$ for every $(B')^\wedge$-module $N'$. The composition $B \to B' \to (B')^\wedge$ is flat (Algebra, Lemma \ref{algebra-lemma-completion-flat}) hence for any $B$-module $N$ we have $$ \Ext^1_B(\NL_{B/A}^\wedge, N) \otimes_B (B')^\wedge = \Ext^1_{(B')^\wedge}(\NL_{B/A}^\wedge \otimes_B (B')^\wedge, N \otimes_B (B')^\wedge) $$ by More on Algebra, Lemma \ref{more-algebra-lemma-base-change-RHom} part (3) (minor details omitted). Thus we see that this module is annihilated by $I^c$. However, $B \to (B')^\wedge$ is actually faithfully flat by our assumption that $B \to B'$ is faithfully flat (Formal Spaces, Lemma \ref{formal-spaces-lemma-etale-surjective}). Thus we conclude that $\Ext^1_B(\NL_{B/A}^\wedge, N)$ is annihilated by $I^c$. Hence $\varphi$ is rig-smooth. This proves (2). \medskip\noindent To prove (3), setting $B = \prod_{i = 1, \ldots, n} B_i$ we just observe that $\NL_{B/A}^\wedge$ is the direct sum of the complexes $\NL_{B_i/A}^\wedge$ viewed as complexes of $B$-modules.