The implications (a) $\Rightarrow$ (b), (c) $\Rightarrow$ (a), (d) $\Rightarrow$ (c) are straightforward from the definitions. Assume (b) holds and let $J \subset B$ and $I \subset A$ be as in (a). Choose a commutative diagram $$ \xymatrix{ A \ar[r] & \ldots \ar[r] & A_3 \ar[r] & A_2 \ar[r] & A_1 \\ B \ar[r] \ar[u] & \ldots \ar[r] & B/J_3 \ar[r] \ar[u] & B/J_2 \ar[r] \ar[u] & B/J_1 \ar[u] } $$ such that $A_{n + 1}/J_nA_{n + 1} = A_n$ and such that $A = \lim A_n$ as in Lemma \ref{lemma-representable-property-rings}. For every $m$ there exists a $\lambda$ such that $J_\lambda \subset J_m$. Since $B/J_\lambda \to A/I_\lambda$ is of finite type, this implies that $B/J_m \to A/I_m$ is of finite type. Let $\alpha_1, \ldots, \alpha_n \in A_1$ be generators of $A_1$ over $B/J_1$. Since $A$ is a countable limit of a system with surjective transition maps, we can find $a_1, \ldots, a_n \in A$ mapping to $\alpha_1, \ldots, \alpha_n$ in $A_1$. By Remark \ref{remark-universal-property} we find a continuous map $B\{x_1, \ldots, x_n\} \to A$ mapping $x_i$ to $a_i$. This map induces surjections $(B/J_m)[x_1, \ldots, x_n] \to A_m$ by Algebra, Lemma \ref{algebra-lemma-surjective-mod-locally-nilpotent}. For $m \geq 1$ we obtain a short exact sequence $$ 0 \to K_m \to (B/J_m)[x_1, \ldots, x_n] \to A_m \to 0 $$ The induced transition maps $K_{m + 1} \to K_m$ are surjective because $A_{m + 1}/J_mA_{m + 1} = A_m$. Hence the inverse limit of these short exact sequences is exact, see Algebra, Lemma \ref{algebra-lemma-ML-exact-sequence}. Since $B\{x_1, \ldots, x_n\} = \lim (B/J_m)[x_1, \ldots, x_n]$ and $A = \lim A_m$ we conclude that $B\{x_1, \ldots, x_n\} \to A$ is surjective and open. As $A$ is complete the kernel is a closed ideal. In this way we see that (a), (b), (c), and (d) are equivalent. \medskip\noindent Let a diagram (\ref{equation-localize}) as in Situation \ref{situation-local-property} be given. By Example \ref{example-finite-type-from-finite-type-ring-map} the maps $A \to (A')^\wedge$ and $B \to (B')^\wedge$ satisfy (a), (b), (c), and (d). Moreover, by Lemma \ref{lemma-representable-property-rings} in order to prove Axioms (\ref{item-axiom-1}) and (\ref{item-axiom-2}) we may assume both $B \to A$ and $(B')^\wedge \to (A')^\wedge$ are taut. Now pick a weak ideal of definition $J \subset B$. Let $J' \subset (B')^\wedge$, $I \subset A$, $I' \subset (A')^\wedge$ be the closure of $J(B')^\wedge$, $JA$, $J(A')^\wedge$. By what was said above, it suffices to consider the commutative diagram $$ \xymatrix{ A/I \ar[r] & (A')^\wedge/I' \\ B/J \ar[r] \ar[u]^{\overline{\varphi}} & (B')^\wedge/J' \ar[u]_{\overline{\varphi}'} } $$ and to show (1) $\overline{\varphi}$ finite type $\Rightarrow \overline{\varphi}'$ finite type, and (2) if $A \to A'$ is faithfully flat, then $\overline{\varphi}'$ finite type $\Rightarrow \overline{\varphi}$ finite type. Note that $(B')^\wedge/J' = B'/JB'$ and $(A')^\wedge/I' = A'/IA'$ by the construction of the topologies on $(B')^\wedge$ and $(A')^\wedge$. In particular the horizontal maps in the diagram are \'etale. Part (1) now follows from Algebra, Lemma \ref{algebra-lemma-compose-finite-type} and part (2) from Descent, Lemma \ref{descent-lemma-finite-type-local-source-fppf-algebra} as the ring map $A/I \to (A')^\wedge/I' = A'/IA'$ is faithfully flat and \'etale. \medskip\noindent We omit the proof of Axiom (\ref{item-axiom-3}).