The implication ``$\Rightarrow$'' follows from Algebra, Lemmas \ref{algebra-lemma-base-change-finiteness}, \ref{algebra-lemma-flat-base-change}, \ref{algebra-lemma-base-change-smooth}, and \ref{algebra-lemma-etale} because we have $D' \otimes_{B'} B = D$ and $D' \otimes_{B'} A' = C'$ by Lemma \ref{lemma-module-over-fibre-product}. Thus it suffices to prove the implications in the other direction. \medskip\noindent Ad (1). Assume $D$ of finite type over $B$ and $C'$ of finite type over $A'$. We will use the results of Lemma \ref{lemma-module-over-fibre-product} without further mention. Choose generators $x_1, \ldots, x_r$ of $D$ over $B$ and generators $y_1, \ldots, y_s$ of $C'$ over $A'$. Using that $D = D' \otimes_{B'} B$ and $B' \to B$ is surjective we can find $u_1, \ldots, u_r \in D'$ mapping to $x_1, \ldots, x_r$ in $D$. Using that $C' = D' \otimes_{B'} A'$ we can find $v_1, \ldots, v_t \in D'$ such that $y_i = \sum v_j \otimes a'_{ij}$ for some $a'_{ij} \in A'$. In particular, the images of $v_j$ in $C'$ generate $C'$ as an $A'$-algebra. Set $N = r + t$ and consider the cube of rings $$ \xymatrix{ A[x_1, \ldots, x_N] & & A'[x_1, \ldots, x_N] \ar[ll] \\ & A \ar[lu] & & A' \ar[ll] \ar[lu] \\ B[x_1, \ldots, x_N] \ar[uu] & & B'[x_1, \ldots, x_N] \ar[uu] \ar[ll] \\ & B \ar[uu] \ar[lu] & & B' \ar[ll] \ar[uu] \ar[lu] } $$ Observe that the back square is cartesian as well. Consider the ring map $$ B'[x_1, \ldots, x_N] \to D',\quad x_i \mapsto u_i \quad\text{and}\quad x_{r + j} \mapsto v_j. $$ Then we see that the induced maps $B[x_1, \ldots, x_N] \to D$ and $A'[x_1, \ldots, x_N] \to C'$ are surjective, in particular finite. We conclude from Lemma \ref{lemma-relative-finite-module-over-fibre-product} that $B'[x_1, \ldots, x_N] \to D'$ is finite, which implies that $D'$ is of finite type over $B'$ for example by Algebra, Lemma \ref{algebra-lemma-compose-finite-type}. \medskip\noindent Ad (2). The implication ``$\Leftarrow$'' follows from Lemma \ref{lemma-relative-flat-module-over-fibre-product}. Moreover, the final statement follows from the final statement of Lemma \ref{lemma-relative-flat-module-over-fibre-product}. \medskip\noindent Ad (3). Assume $B \to D$ and $A' \to C'$ are flat and of finite presentation. The flatness of $B' \to D'$ we've seen in (2). We know $B' \to D'$ is of finite type by (1). Choose a surjection $B'[x_1, \ldots, x_N] \to D'$. By Algebra, Lemma \ref{algebra-lemma-finite-presentation-independent} the ring $D$ is of finite presentation as a $B[x_1, \ldots, x_N]$-module and the ring $C'$ is of finite presentation as a $A'[x_1, \ldots, x_N]$-module. By Lemma \ref{lemma-relative-finitely-presented-module-over-fibre-product} we see that $D'$ is of finite presentation as a $B'[x_1, \ldots, x_N]$-module, i.e., $B' \to D'$ is of finite presentation. \medskip\noindent Ad (4). Assume $B \to D$ and $A' \to C'$ smooth. By (3) we see that $B' \to D'$ is flat and of finite presentation. By Algebra, Lemma \ref{algebra-lemma-flat-fibre-smooth} it suffices to check that $D' \otimes_{B'} k$ is smooth for any field $k$ over $B'$. If the composition $J \to B' \to k$ is zero, then $B' \to k$ factors as $B' \to B \to k$ and we see that $$ D' \otimes_{B'} k = D' \otimes_{B'} B \otimes_B k = D \otimes_B k $$ is smooth as $B \to D$ is smooth. If the composition $J \to B' \to k$ is nonzero, then there exists an $h \in J$ which does not map to zero in $k$. Then $B' \to k$ factors as $B' \to B'_h \to k$. Observe that $h$ maps to zero in $B$, hence $B_h = 0$. Thus by Lemma \ref{lemma-diagram-localize} we have $B'_h = A'_h$ and we get $$ D' \otimes_{B'} k = D' \otimes_{B'} B'_h \otimes_{B'_h} k = C'_h \otimes_{A'_h} k $$ is smooth as $A' \to C'$ is smooth. \medskip\noindent Ad (5). Assume $B \to D$ and $A' \to C'$ are \'etale. By (4) we see that $B' \to D'$ is smooth. As we can read off whether or not a smooth map is \'etale from the dimension of fibres we see that (5) holds (argue as in the proof of (4) to identify fibres -- some details omitted).