Can someone let me know if my proof is okay for showing the following two sets are logically equivalent (in propositional logic)? I asked this a day or so ago but the post was very long, disorganized, and messy, so I redid my answer. Hopefully, it looks a bit more cleaned up. I appreciate anyone's help!
What have you done so far is NOT enough to prove that the two infinite sets of formulas $\Phi = \{\varphi_n \mid n \in \mathbb{N}^+ \}$ and $\Phi' = \{(\bigwedge_{i=1}^{n-1} \varphi_i) \to \varphi_n \mid n \in \mathbb{N}^+\}$ (where, given infinitely many propositional variables $p_1, p_2, \dots$, we set $\varphi_n = p_1 \land \ldots \land p_n$ for all $n \in \mathbb{N}^+$) are logically equivalent. There are two errors in your approach.
First, proving that $\Phi$ and $\Phi'$ are logically equivalent amounts to show that for every truth valuation $\sigma$, one has $\sigma \models \Phi$ if and only if $\sigma \models \Phi'$. In your approach you are just trying to show that if $\sigma \models \Phi$ then $\sigma \models \Phi'$, you forgot to show the vice-versa.
Second (and this is subtler), your proof by induction show that your property $S(n)$ is true for all $n \in \mathbb{N}^+$, so it does not prove that if $\sigma \models \Phi$ then $\sigma \models \Phi'$, it proves only that, for every $n \in \mathbb{N}^+$, if $\sigma \models \Phi_n$ then $\sigma \models \Phi_n'$, where $\Phi_n = \{\varphi_i \mid 1\leq i \leq n\}$ and $\Phi_n' = \{(\bigwedge_{j=1}^{i-1} \varphi_j) \to \varphi_i \mid 1 \leq i \leq n\}$. You can see the difference if you note that $\Phi$ and $\Phi'$ are infinite sets, whereas, given $n \in \mathbb{N}^+$, $\Phi_n$ and $\Phi_n'$are finite sets. Induction allows you to prove only something about finite (although arbitrarily large) sets. To move from finite to infinite you need also another ingredient: compactness theorem.