We will prove this in the rig-smooth case and indicate the necessary changes to prove the rig-\'etale case at the end of the proof. Consider a commutative diagram $$ \xymatrix{ X \times_Y V \ar[r] \ar[d] & V \ar[d] \ar[r] & W \ar[d] \\ X \ar[r] & Y \ar[r] & Z } $$ with $V$ and $W$ affine formal algebraic spaces, $V \to Y$ and $W \to Z$ representable by algebraic spaces and \'etale. We have to show that $V \to W$ corresponds to a rig-smooth map of adic Noetherian topological rings, see Definition \ref{definition-rig-smooth}. We may write $V = \text{Spf}(B)$ and $W = \text{Spf}(C)$ and that $V \to W$ corresponds to an adic ring map $C \to B$ which is topologically of finite type, see Lemma \ref{lemma-finite-type-morphisms}. \medskip\noindent We will use below without further mention that $X \times_Y V \to V$ is rig-smooth and rig-surjective, see Lemmas \ref{lemma-base-change-rig-smooth} and \ref{lemma-base-change-rig-surjective}. Also, the composition $X \times_Y V \to V \to W$ is rig-smooth since $g \circ f$ is rig-smooth. \medskip\noindent Let $I \subset C$ be an ideal of definition. The module Assume $C \to B$ is not rig-smooth to get a contradiction. This means that there exists a prime ideal $\mathfrak q \subset B$ not containing $IB$ such that either $H^{-1}(\NL_{B/C}^\wedge)_\mathfrak p$ is nonzero or $H^0(\NL_{B/C}^\wedge)_\mathfrak p$ is not a finite free $B_\mathfrak q$-module. See Lemma \ref{lemma-equivalent-with-artin-smooth}; some details omitted. We may choose a maximal ideal $IB + \mathfrak q \subset \mathfrak m$. By Algebra, Lemma \ref{algebra-lemma-exists-dvr} we can find a complete discrete valuation ring $R$ and an injective local ring homomorphism $(B/\mathfrak q)_\mathfrak m \to R$. \medskip\noindent After replacing $R$ by an extension, we may assume given a lift $\text{Spf}(R) \to X \times_Y V$ of the adic morphism $\text{Spf}(R) \to V = \text{Spf}(B)$. Choose an \'etale covering $\{\text{Spf}(A_i) \to X \times_Y V\}$ as in Formal Spaces, Definition \ref{formal-spaces-definition-formal-algebraic-space}. By Lemma \ref{lemma-etale-covering-rig-surjective} we may assume $\text{Spf}(R) \to X \times_Y V$ lifts to a morphism $\text{Spf}(R) \to \text{Spf}(A_i)$ for some $i$ (this might require replacing $R$ by another extension). Set $A = A_i$. Consider the ring maps $$ C \to B \to A \to R $$ Let $\mathfrak p \subset A$ be the kernel of the map $A \to R$ and note that $\mathfrak p$ lies over $\mathfrak q$. We know that $C \to A$ and $B \to A$ are rig-smooth. In particular the ring map $B_\mathfrak q \to A_\mathfrak p$ is flat by Lemma \ref{lemma-rig-smooth-rig-flat}. Consider the associated exact sequence $$ \xymatrix{ & H^0(\NL_{B/C}^\wedge) \otimes_B A_\mathfrak p \ar[r] & H^0(\NL_{A/C}^\wedge)_\mathfrak p \ar[r] & H^0(\NL_{A/B}^\wedge)_\mathfrak p \ar[r] & 0 \\ 0 \ar[r] & H^{-1}(\NL_{B/C}^\wedge \otimes_B A)_\mathfrak p \ar[r] & H^{-1}(\NL_{A/C}^\wedge)_\mathfrak p \ar[r] & H^{-1}(\NL_{A/B}^\wedge)_\mathfrak p \ar[llu] } $$ of Lemmas \ref{lemma-exact-sequence-NL} and \ref{lemma-exact-sequence-NL-rig-smooth}. Given the rig-smoothness of $C \to A$ and $B \to A$ we conclude that $H^{-1}(\NL_{B/C}^\wedge \otimes_B A)_\mathfrak p = 0$ and that $H^0(\NL_{B/C}^\wedge) \otimes_B A_\mathfrak p$ is finite free as a kernel of a surjection of finite free $A_\mathfrak p$-modules. Since $B_\mathfrak q \to A_\mathfrak p$ is flat and hence faithfully flat, this implies that $H^{-1}(\NL_{B/C}^\wedge)_\mathfrak q = 0$ and that $H^0(\NL_{B/C}^\wedge)_\mathfrak q$ is finite free which is the contradiction we were looking for. \medskip\noindent In the rig-\'etale case one argues in exactly the same manner but the conclusion obtained is that both $H^{-1}(\NL_{B/C}^\wedge)_\mathfrak q$ and $H^0(\NL_{B/C}^\wedge)_\mathfrak q$ are zero.