On the values of variables

For a moment, let’s forget about our logics and consider everyday mathematical practice. Suppose you have a simple equation of one unknown variable as follows: $$(2\times x)+7=57$$ If you ‘solve’ this equation, you get that $x=25$. Now really, all this means is that if you take the value of the variable $x$ to be $25$, then $(2\times x)+7=57$ will come out correct, while if you take the value of the variable $x$ to be anything else, it will come out incorrect. Thus, one way we can talk about variables is to assign them a value and see whether the resulting formula is correct, computing with the value assigned. In such cases, the variables function like temporary constants. Essentially, what we are saying above is that for a moment, you should understand $(2\times x)+7=57$ the same way as you would understand $(2 \times 25) + 7= 57$. Of course, unlike with constants, what the value of $x$ is is temporarily assigned. For example, in $(5 \times x)+7=107$, the formula will now come out true provided $x$ is assigned the value $20$, not $25$. Clearly, the value of $20$ and $25$ did not change from one equation to the other, but the assigned value of $x$ did, provided the aim is to make the formula correct. And of course, we can assign a value to $x$ that makes neither of these equations come out correct. For example, if $x$ is assigned the value $97$, both equations will come out incorrect. As we shall see presently, our semantics for first-order logic will be founded on these temporary assignments of values to the variables of the language $\mathcal{L}_1$.

Ultimately, our goal is to assign meaning to quantified formulas with no free variables, i.e., quantified closed formulas, in a systematic way. This will be based on a generalization of the idea of temporary assignments of values. For example, take the claim:

What this means is that there is a temporary assignment of a value to the variable $x$ such that $(2\times x)+7=57$ comes out correct. This is clearly a use of the existential quantifier. And clearly, this is true if there really is such an assignment. Moreover, as demonstrated, there is such an assignment (assigning $25$ to $x$), so it is true that the equation has a solution.

On the other hand, take the claim:

This is false, and the reason why it is false is because there is no assignment of a value to the variable $x$ such that $(0\times x)+7=57$.

The same type of reasoning holds for the universal quantifier. For example, for the equation $2 \times (x + 2)=(2 \times x) + 4$, every temporary assignment of a value to the variable $x$ makes the equation come out correct. Incidentally, this is equivalent to the fact that there is no assignment of a value to the variable $x$ which would make the equation come out incorrect.