Why is interpretation of an existential formula defined this way in Model Theory?

Question

I am reading through the text $\textit{Fundamentals of Mathematical Logic}$ by Dr. Peter Hinman. On page 106 there is a discussion regarding how to define an interpretation $V_{\mathcal{U},\alpha}(\exists x \phi)$. For context, the universe of $\mathcal{U}$ is $A$, and $\alpha:Vb \longrightarrow A$ is an interpretation of the variables of the language $L$. $V_{\mathcal{U},\alpha}:Fm_L \longrightarrow \{0,1\}$ is a function that will interpret formulas that will be used in the definition of $\mathcal{U}\models_{\alpha} \phi$. So for example, $\mathcal{U}\models_{\alpha} \phi$ if and only if $V_{\mathcal{U},\alpha}(\phi) = 1$. Hence the recursive structure of truth through this convention comes from the definition of $V_{\mathcal{U},\alpha}$. I understand the definition of $V_{\mathcal{U},\alpha}(\phi)$ where $\phi$ is an atomic formula of a Boolean combination of atomic formulas, and I understand the general intuition of wanting the value of $V_{\mathcal{U},\alpha}(\exists x \phi)$ to reflect our natural way of conducting an existence proof. The thing that confuses me from a technical standpoint is the following. Why exactly should this be independent of the value of $\alpha(x)$? Different values of $\alpha(x)$ would yield different truth values. He goes on to say that it just matters that there is some way of assigning $x$ to an element of $A$, but isn't this essentially changing the function $\alpha$? If we allow for such a change of the value of $\alpha(x)$ to be okay enough for our condition for the statement to be true with respect to $\alpha$, well, why even have it be true with respect to $\alpha$ to begin with? Can someone provide some concrete examples for why we would want it to be this way? Part of my confusion comes from a lack of motivating examples. The one provided with $V_{\mathcal{U},\alpha}(\exists x (\dot Sx \dot = \dot 0)) = 0$ isn't helping since there is only occurrence of $x$ and so changing the one value of $x$ here would seem just like changing the $\alpha$ all together, and this statement will clearly be false no matter what interpretation $\alpha$ we use.

Answer 1

The point is that $x$ is a bound variable in the formula $(\exists x\,\phi)$. The truth of the formula $(\exists x\,\phi)$ doesn't depend on choosing some specific value that $x$ represents; instead, $x$ is just a placeholder variable we are using and we are asking whether there exists a choice of value for it which makes it true. So even though $\alpha$ assigns some specific value to $x$, we don't care about that value. The point of an assignment is to assign values to free variables in formulas, so we can determine whether the formulas are true for those values. It doesn't make sense to say whether a statement like "$n+1=0$" is true, since we haven't defined what $n$ is. On the other hand, we don't need to assign values to bound variables: it makes sense to ask whether a statement "there exists $n$ such that $n+1=0$" is true, even if we have not assigned a specific value to $n$.

Another way to think about it is that the quantifier $\exists x$ makes $x$ into a "local variable". Within that quantifier, $x$ always refers to the value bound by that quantifier. Outside of the quantifier, we might have some different definition for $x$ (which is provided by $\alpha$), but inside the quantifier we redefine $x$ with a new meaning.