Let $f:\mathfrak Y \to \mathfrak X$ be a morphism of formal schemes. We say that $f$ is formally smooth if it satisfies the infinitesimal lifting property, that is if for every affine $\mathfrak X$-scheme $Z$ and for every closed subscheme $T\hookrightarrow Z$ defined by a square zero ideal, the natural map $$\mathrm{Hom}_{\mathfrak X}(Z,\mathfrak Y) \to \mathrm{Hom}_{\mathfrak X}(T,\mathfrak Y)$$ is surjective. The same definition can be stated for morphisms of schemes.
Let $J$ be an ideal of definition of $\mathfrak X$, and let $I$ be an ideal of definition of $\mathfrak Y$ containing $f^{*}(J)\,\mathcal O_{\mathfrak Y}$. This is possible due to the continuity of $f$. For $n\geq 0$, we consider the schemes $\mathfrak X_n := (|\mathfrak X|,\mathcal O_{\mathfrak X}/J^{n+1})$. Then $\mathfrak X$ can be identified with the inductive system $(\mathfrak X_n)$. We can do similarly for $\mathfrak Y$. Then, the morphism $f$ induces a family of compatible morphisms $f_n:\mathfrak X_n \to \mathfrak Y_n$ of schemes.
If $f$ is formally smooth, is it true that all the $f_n$'s are also formally smooth ?
I have done the following. I fix $n$ and I consider $\mathfrak X_n$-schemes $Z$ and $T$ as above. I also consider a morphism $T\to \mathfrak Y_m$ over $\mathfrak X_m$. Using compositions with $\mathfrak Y_m \hookrightarrow \mathfrak Y$ and $\mathfrak X_m \hookrightarrow \mathfrak X$, I can look at $Z$ and $T$ over $\mathfrak X$ and I get a $T$-point of $\mathfrak Y$ over $\mathfrak X$. By hypothesis, it lifts to a $Z$-point of $\mathfrak Y$ over $\mathfrak X$.
To conclude, I would like to justify that it factors through $\mathfrak Y_m$. That is, if I write $g:Z \to \mathfrak Y$, I would have to check whether the map $g^{*}I^m \to \mathcal O_{Z}$ is zero. Because the structure map $f\circ g: Z \to \mathfrak X$ factors through $\mathfrak X_m$, I know that $g^{*}f^{*}J^m\to \mathcal O_Z$ is zero. Thus, it would be over with the additional assumption that $f$ is adic.
But what if $f$ is not adic ? Currently I do not see how I could progress further. However, I have also been unable to come up with a counter example...