# Question about independent equivalent subset

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.

It is a general fact about $$\models$$ that if $$\Delta$$ and $$\Sigma$$ are sets of wffs such that $$\Delta \subseteq \Sigma$$ and, if $$\Delta \models \phi$$ for some wff $$\phi$$, then $$\Sigma \models \phi$$ (I can say more about this if it doesn't make sense). In the proof you are looking at, you have that $$\Sigma = \Delta \cup \{\alpha\}$$ and that $$\Delta \models \alpha$$, so any assignment that satisfies $$\Delta$$ also satisfies $$\alpha$$ and hence $$\Sigma$$. You don't need to worry about assignments that satisfy $$\alpha$$, but possibly not other wffs in $$\Delta$$: you are only interested in assignments that satisfy all the wffs in $$\Delta$$.
• Your answer seems to make sense, but I hope you can explain more about the fact that $\Sigma$ $\models$ $\phi$ because $\Delta$ $\models$ $\phi$. Maybe some examples or proof? Aug 12, 2021 at 12:33
• If $I$ is an assignment that satisfies $\Delta$, then $I$ also satisfies $\alpha$ (by definition of the given fact that $\Delta \models \alpha$). So such an $I$ satisfies $\Delta \cup \{\alpha\} = \Sigma$. Aug 12, 2021 at 12:55