Synthetic Basis and Analytic Basis are Compatible

Theorem
$ \def\delimit#1#2{\left#1#2\right} \def\quantify#1#2{\delimit({#1#2}) \;} \def\thereExists{\quantify\exists} \def\forEach{\quantify\forall} \def\topology{\vartheta} \def\space{\delimit({A, \topology})} \def\basis{\mathcal B} \def\powerSet{\mathcal P \delimit(A)} \def\generated{\delimit\{ {\bigcup S \middle| S \subseteq \basis}\} } $ $\basis$ is an Analytic Basis for Topological Space $\space$

$\basis$ is a Synthetic Basis for $A$ that generates $\topology$.

Proof
Assume $\basis$ is an Analytic Basis for Topological Space $\space$.
 * Then $\basis \subseteq \topology \subseteq \powerSet$.
 * Moreover, $A \in \topology = \generated$.
 * Thus,
 * $\displaystyle \thereExists{S \subseteq \basis} A = \bigcup S$
 * Assume $

\def\antecedent{B_1, B_2 \in \basis} \antecedent $.
 * $B_1, B_2 \in \topology$, so $B_1 \cap B_2 \in \topology = \generated$.
 * That means $

\def\consequent{\thereExists{S \subseteq \basis} B_1 \cap B_2 = \bigcup S} \consequent $.
 * $\displaystyle \therefore \forEach{\antecedent} \consequent$.
 * By definition, $\basis$ is a Synthetic Basis for $A$.

$\therefore$ If $\basis$ is an Analytic Basis for Topological Space $\space$, then $\basis$ is a Synthetic Basis for $A$ that generates $\topology$.

Assume $\basis$ is a Synthetic Basis for $A$ that generates $\topology$.
 * Then $\basis = \delimit\{ {\bigcup \delimit\{ {B} \} \middle| \delimit\{ {B} \} \subseteq \basis } \} \subseteq \generated = \topology$.
 * Moreover, $\space$ is a topological space.
 * By definition, $\basis$ is an Analytic Basis for Topological Space $\space$.

$\therefore$ If $\basis$ is a Synthetic Basis for $A$ that generates $\topology$, then $\basis$ is an Analytic Basis for Topological Space $\space$.

The statements are equivalent.