Assume (1), (2), (3), and (4) hold. Choose an object $R \in \Ob(\widehat{\mathcal{C}}_\Lambda)$ such that $\underline{R}|_{\mathcal{C}_\Lambda}$ is smooth. See Lemma \ref{lemma-exists-smooth}. Let $r = \dim_k T\mathcal{F}$ and put $S = R[[X_1, \ldots, X_r]]$. \medskip\noindent We are going to inductively construct for $n \geq 2$ pairs $(J_n, f_{n - 1} : \xi_n \to \xi_{n - 1})$ where $J_n \subset S$ is an decreasing sequence of ideals and $f_{n - 1} : \xi_n \to \xi_{n - 1}$ is a morphism of $\mathcal{F}$ lying over the projection $S/J_n \to S/J_{n - 1}$. \medskip\noindent Step 1. Let $J_1 = \mathfrak m_S$. Let $\xi_1$ be the unique (up to unique isomorphism) object of $\mathcal{F}$ over $k = S/J_1 = S/\mathfrak m_S$ \medskip\noindent Step 2. Let $J_2 = \mathfrak m_S^2 + \mathfrak{m}_R S$. Then $S/J_2 = k[V]$ with $V = kX_1 \oplus \ldots \oplus kX_r$ By (S2) for $\overline{\mathcal{F}}$ we get a bijection $$ \overline{\mathcal{F}}(S/J_2) \longrightarrow T\mathcal{F} \otimes_k V, $$ see Lemmas \ref{lemma-S1-S2-associated-functor} and \ref{lemma-tangent-space-vector-space}. Choose a basis $\theta_1, \ldots, \theta_r$ for $T\mathcal{F}$ and set $\xi_2 = \sum \theta_i \otimes X_i \in \Ob(\mathcal{F}(S/J_2))$. The point of this choice is that $$ d\xi_2 : \Mor_{\mathcal{C}_\Lambda}(S/J_2, k[\epsilon]) \longrightarrow T\mathcal{F} $$ is surjective. Let $f_1 : \xi_2 \to \xi_1$ be the unique morphism. \medskip\noindent Induction step. Assume $(J_n, f_{n - 1} : \xi_n \to \xi_{n - 1})$ has been constructed for some $n \geq 2$. There is a minimal element $J_{n + 1}$ of the set of ideals $J \subset S$ satisfying: (a) $\mathfrak m_S J_n \subset J \subset J_n$ and (b) there exists a morphism $\xi_{n + 1} \to \xi_n$ lying over $S/J \to S/J_n$, see Lemma \ref{lemma-largest-closed-where-lift}. Let $f_n : \xi_{n + 1} \to \xi_n$ be any morphism of $\mathcal{F}$ lying over $S/J_{n + 1} \to S/J_n$. \medskip\noindent Set $J = \bigcap J_n$. Set $\overline{S} = S/J$. Set $\overline{J}_n = J_n/J$. By Lemma \ref{lemma-m-adic-topology} the sequence of ideals $(\overline{J}_n)$ induces the $\mathfrak m_{\overline{S}}$-adic topology on $\overline{S}$. Since $(\xi_n, f_n)$ is an object of $\widehat{\mathcal{F}}_\mathcal{I}(\overline{S})$, where $\mathcal{I}$ is the filtration $(\overline{J}_n)$ of $\overline{S}$, we see that $(\xi_n, f_n)$ induces an object $\xi$ of $\widehat{\mathcal{F}}(\overline{S})$. see Lemma \ref{lemma-formal-objects-different-filtration}. \medskip\noindent We prove $\xi$ is versal. For versality it suffices to check conditions (1) and (2) of Lemma \ref{lemma-versal-criterion}. Condition (1) follows from our choice of $\xi_2$ in Step 2 above. Suppose given a diagram in $\widehat{\mathcal{F}}$ $$ \vcenter{ \xymatrix{             &  y \ar[d] \\ \eta \ar[r]  &  x } } \quad\text{lying over}\quad \vcenter{ \xymatrix{          &   B  \ar[d]^{f} \\ \overline{S} \ar[r] &   A } } $$ in $\widehat{\mathcal{C}}_\Lambda$ with $f: B \to A$ a small extension of Artinian rings. We have to show there is a map $\overline{S} \to B$ fitting into the diagram on the right. Choose $n$ such that $\overline{S} \to A$ factors through $\overline{S} \to S/J_n$. This is possible as the sequence $(\overline{J}_n)$ induces the $\mathfrak m_{\overline{S}}$-adic topology as we saw above. The pushforward of $\xi$ along $\overline{S} \to S/J_n$ is $\xi_n$. We may factor $\xi \to x$ as $\xi \to \xi_n \to x$ hence we get a diagram in $\mathcal{F}$ $$ \vcenter{ \xymatrix{             &  y \ar[d] \\ \xi_n \ar[r]  &  x } } \quad\text{lying over}\quad \vcenter{ \xymatrix{          &   B  \ar[d]^{f} \\ S/J_n \ar[r] &   A . } } $$ To check condition (2) of Lemma \ref{lemma-versal-criterion} it suffices to complete the diagram $$ \xymatrix{ S/J_{n + 1} \ar[d] \ar@{-->}[r] & B \ar[d]^{f} \\ S/J_n   \ar[r] & A } $$ or equivalently, to complete the diagram $$ \xymatrix{   &  S/J_n \times_A B \ar[d]^{p_1} \\ S/J_{n + 1} \ar@{-->}[ur] \ar[r] &  S/J_n. } $$ If $p_1$ has a section we are done. If not, by Lemma \ref{lemma-fiber-product-CLambda} (2) $p_1$ is a small extension, so by Lemma \ref{lemma-essential-surjection} (4) $p_1$ is an essential surjection. Recall that $S = R[[X_1, \ldots, X_r]]$ and that we chose $R$ such that $\underline{R}|_{\mathcal{C}_\Lambda}$ is smooth. Hence there exists a map $h : R \to B$ lifting the map $R \to S \to S/J_n \to A$. By the universal property of a power series ring there is an $R$-algebra map $h : S = R[[X_1, \ldots, X_2]] \to B$ lifting the given map $S \to S/J_n \to A$. This induces a map $g: S \to S/J_n \times_A B$ making the solid square in the diagram $$ \xymatrix{ S \ar[d] \ar[r]_-g  &  S/J_n \times_A B \ar[d]^{p_1} \\ S/J_{n + 1} \ar@{-->}[ur] \ar[r] &  S/J_n } $$ commute. Then $g$ is a surjection since $p_1$ is an essential surjection. We claim the ideal $K = \Ker(g)$ of $S$ satisfies conditions (a) and (b) of the construction of $J_{n + 1}$ in the induction step above. Namely, $K \subset J_n$ is clear and $\mathfrak m_SJ_n \subset K$ as $p_1$ is a small extension; this proves (a). By (S1) applied to $$ \xymatrix{             &  y \ar[d] \\ \xi_n \ar[r]  &  x, } $$ there exists a lifting of $\xi_n$ to $S/K \cong S/J_n \times_A B$, so (b) holds. Since $J_{n + 1}$ was the minimal ideal with properties (a) and (b) this implies $J_{n + 1} \subset K$. Thus the desired map $S/J_{n+1} \to S/K \cong S/J_n \times_A B$ exists.