Before we embark on the proof we briefly explain how to formulate condition (\ref{item-topologically-quasi-nilpotent}) for the case of pairs $(M', \nabla')$ over $D'$. Since $A \to P'$ is smooth the $P'$-module $\Omega_{P'/A}$ is finite projective. Thus we can choose a map $\Omega_{P'/A} \to \bigoplus_{i = 1, \ldots, n} P'$ wich identifies $\Omega_{P'/A}$ with a direct summand of the target. This determines a map $\Omega_{D'} \to \bigoplus_{i = 1, \ldots, n} D'$ identifying the source with a direct summand of the target. Thus for $m' \in M'$ we can write $\nabla'(m') = \sum \theta'_i(m')$ with $\theta'_i(m') \in D'$. The topogical quasi-nilpotence of $\nabla'$ means: for any $m' \in M'$ there are only finitely many pairs $(i, k)$ such that $(\theta'_i)^k(m') \not \in pM'$. (Since in this case there are only finitely many $i$ this just means that $\theta_i^k(m') \in pM'$ for all $k \gg 0$.) \medskip\noindent We can pick the map $P = A[x_i] \to C$ such that it factors through a surjection of $A$-algebras $P \to P'$ (we may have to increase the number of variables in $P$ to do this). Hence we obtain a surjective map $a : D \to D'$ by functoriality of divided power envelopes and completion. Pick $e$ large enough so that $D_e$ is a divided power thickening of $C$ over $A$. Then $D_e \to C$ is a surjection whose kernel is locally nilpotent, see Divided Power Algebra, Lemma \ref{dpa-lemma-nil}. Setting $D'_e = D'/p^eD'$ we see that the kernel of $D_e \to D'_e$ is locally nilpotent. Hence by Algebra, Lemma \ref{algebra-lemma-smooth-strong-lift} we can find a lift $\beta_e : P' \to D_e$ of the map $P' \to D'_e$. Note that $D_{e + i + 1} \to D_{e + i} \times_{D'_{e + i}} D'_{e + i + 1}$ is surjective with square zero kernel for any $i \geq 0$ because $p^{e + i}D \to p^{e + i}D'$ is surjective. Applying the usual lifting property (Algebra, Proposition \ref{algebra-proposition-smooth-formally-smooth}) successively to the diagrams $$ \xymatrix{ P' \ar[r] & D_{e + i} \times_{D'_{e + i}} D'_{e + i + 1} \\ A \ar[u] \ar[r] & D_{e + i + 1} \ar[u] } $$ we see that we can find an $A$-algebra map $\beta : P' \to D$ whose composition with $a$ is the given map $P' \to D'$. By the universal property of the divided power envelope we obtain a map $D_{P', \gamma}(J') \to D$. As $D$ is $p$-adically complete we obtain $b : D' \to D$ such that $a \circ b = \text{id}_{D'}$. \medskip\noindent Consider the base change functor $$ (M, \nabla) \longmapsto (M \otimes^\wedge_D D', \nabla') $$ from pairs for $D$ to pairs for $D'$, see Remark \ref{remark-base-change-connection}. Similarly, we have the base change functor corresponding to the divided power homomorphism $D' \to D$. To finish the proof of the lemma we have to show that the base change for the compositions $b \circ a : D \to D$ and $a \circ b : D' \to D'$ are isomorphic to the identity functor. This is clear for the second as $a \circ b = \text{id}_{D'}$. To prove it for the first, we use the functorial isomorphism $$ c_{\text{id}_D, b \circ a} : M \otimes_{D, \text{id}_D} D \longrightarrow M \otimes_{D, b \circ a} D $$ of the proof of Proposition \ref{proposition-crystals-on-affine}. The only thing to prove is that these maps are horizontal, which we omit. \medskip\noindent The last statement of the proof now follows.