[proofplan] We collect all formulas satisfying the predicate $P$ into a subset $S$ of $\mathsf{Form}$. The hypotheses say exactly that $S$ contains every atomic formula and $\bot$, and that $S$ is closed under each formula-forming connective. Since $\mathsf{Form}$ is defined as the smallest set with these closure properties, every formula belongs to $S$, which is precisely the desired conclusion. [/proofplan] [step:Collect the formulas satisfying the predicate] Define the subset $S \subset \mathsf{Form}$ by \begin{align*} S := \{A \in \mathsf{Form} : P(A)\text{ holds}\}. \end{align*} By the hypothesis on propositional variables, every $p \in \mathsf{Prop}$ belongs to $S$. By the hypothesis $P(\bot)$, the formula $\bot$ belongs to $S$. [/step] [step:Verify that the set of satisfying formulas is closed under all connectives] Let $A,B \in S$. By the definition of $S$, both $P(A)$ and $P(B)$ hold. The closure hypotheses for $P$ therefore give \begin{align*} P(A \land B), \qquad P(A \lor B), \qquad P(A \to B). \end{align*} Again using the definition of $S$, this means \begin{align*} A \land B \in S, \qquad A \lor B \in S, \qquad A \to B \in S. \end{align*} Thus $S$ is closed under the binary connectives $\land$, $\lor$, and $\to$. Now let $A \in S$. By the definition of $S$, $P(A)$ holds. The closure hypothesis for negation gives $P(\neg A)$, hence $\neg A \in S$. Therefore $S$ is also closed under the unary connective $\neg$. [/step] [step:Use the minimality of the set of formulas] The set $S$ contains $\mathsf{Prop}$ and $\bot$, and it is closed under the operations \begin{align*} (A,B) &\mapsto A \land B, \\ (A,B) &\mapsto A \lor B, \\ (A,B) &\mapsto A \to B, \\ A &\mapsto \neg A. \end{align*} The set $\mathsf{Form}$ is, by definition, the smallest set with exactly these closure properties. Since $S \subset \mathsf{Form}$ and $S$ has all the required closure properties, minimality gives \begin{align*} \mathsf{Form} \subset S. \end{align*} Together with $S \subset \mathsf{Form}$, this yields $S = \mathsf{Form}$. [/step] [step:Conclude the predicate holds for every formula] Let $A \in \mathsf{Form}$ be arbitrary. Since $S = \mathsf{Form}$, we have $A \in S$. By the definition of $S$, this means that $P(A)$ holds. Because $A$ was arbitrary, $P(A)$ holds for every $A \in \mathsf{Form}$. [/step]