I am having trouble with Problem 1.2.10a, page 28 of Enderton's A Mathematical Introduction to Logic.
Say that a set $\Sigma_1$ of wffs is equivalent to a set $\Sigma_2$ of wffs iff for any wff a, we have $\Sigma_1$ $\models$ a iff $\Sigma_2$ $\models$ a. A set $\Sigma$ is independent iff no member of $\Sigma$ is tautologically implied by the remaining members in $\Sigma$. Show that the following hold: A finite set of wffs has an independent equivalent subset.
After reading around for a while, I have come to understand the method to attack this problem: You basically take out each wff that is tautologically implied by other members of $\Sigma$. Eventually, you get that independent equivalent subset, because the set is finite. However, I have some problems with understanding tiny details in the full proof. Here's one solution I have found online. The part I don't understand is this:
If $\Sigma$ is not independent, let $\alpha$ $\in$ $\Sigma$ be some formula such that $\Sigma$ \ {$\alpha$} $\models$ $\alpha$. Let $\Delta$ = $\Sigma$ \ {$\alpha$}. Then whenever $\Delta$ $\models$ $\beta$, also $\Sigma$ $\models$ $\beta$ since $\Delta$ $\subseteq$ $\Sigma$.
If $\Delta$ $\models$ $\beta$, then every truth assignment for the sentence symbols in $\Delta$ that satisfies every wffs in $\Delta$ must satisfy $\beta$. We know that $\Sigma$ = $\Delta$ $\cup$ {a}. But how can we know that the truth assignment that satisfies $\alpha$ also satisfies $\beta$? Otherwise, I don't see how we can conclude that $\Sigma$ $\models$ $\beta$ simply because $\Delta$ $\subseteq$ $\Sigma$.
That's the question. I hope someone can clear away my confusion.