For every infinite set of wffs, show that there exists an independent equivalent set Also known as Enderton 1.2.10.c. I'm struggling with this question so much. It has been asked and answered on here. I have also read the solution from other sources, but just cannot grasp the main idea. I know that it cannot be the subset of the original infinite set.
As for now, I am following the solution from University of Pennsylvania. I have encountered two main problems.

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*Why are there two separate cases for $\Delta_n$? Specifically, a case in which $\Delta_n$ $\models$ $\sigma_n$, another which $\Delta_n$ $\not\models$ $\sigma_n$.

*Are $\Sigma$ and $\Sigma_n$ two separate sets? If I understand it right, the first one contains a series of sentence symbol, while the second one contains one symbol only.

 A: First regarding your question "Are $\Sigma$ and $\Sigma_n$ two separate sets? If I understand it right, the first one contains a series of sentence symbol, while the second one contains one symbol only.", clearly they're different and $\Sigma_n$ doesn't necessarily contain only one sentence symbol from the definition in your reference:

Let $\Sigma$ = {$\sigma_0$, $\sigma_1$, . . .}. Let $\Sigma_n$ = {$\sigma_i$| i < n}.

So $\Sigma$ is an infinite set of wffs (sentences), while $\Sigma_n$ contains only $n$ such formulas (indexed from $0$ to $n-1$).
Regarding your question "Why are there two separate cases for $\Delta_n$? Specifically, a case in which $\Delta_n$ $\models$ $\sigma_n$, another which $\Delta_n$ $\not\models$ $\sigma_n$."
Here in the proof $\Delta_n$ appears in the general inductive step which is assumed to be tautologically independent and equivalent to $\Sigma_n$, but this by no means
$\Delta_n$ $\models$ $\sigma_n$ or $\Delta_n$ $\not\models$ $\sigma_n$ (remember from above that $\Sigma_n$ contains only $n$ such formulas indexed from $0$ to $n-1$). So we have to proceed with two cases. The first case ($\Delta_n$ $\models$ $\sigma_n$) simply means $\Sigma_n$ $\models$ $\sigma_n$ by inductive hypothesis and thus $\Delta_{n+1}$ = $\Delta_n$ is equivalent to $\Sigma_{n+1}$. So we just find such a construction using $\Delta_n$ itself. Another case we need to construct $\Delta_{n+1}$ according to some specific Horn clause algo as shown in your reference as $\Delta_{n+1}$ = $\Delta_n$ ∪ {$\delta_n$ → $\sigma_n$} where $\delta_n$ is defined as $\alpha_0$ ∧ · · · ∧ $\alpha_k$ each of which is a member of $\Delta_n$. Your another Math Exchange reference mainly discussed this second case since that reference constructed $\Delta_n$ in a more specific way.
