Use of quantors/quantifiers and variables in first-order logic.

Question

Let $v_1,v_2,\dots,v_n$ be variables and $\beta$ the variable assignment $\beta(v_n)=2n$ for $n\geq 0$. Of the following, which are true and which false under $\beta$?

• $\forall v_0 \exists v_1 v_0<v_1$
• $\exists v_0(v_0\cdot v_1)\overset{.}{=} v_1$

Since $\beta$ assigns values to the variables, why the need for quantifiers? Especially the $\forall$, since there is only one $v_0$ anyway. By just blindly replacing $v_n$ with $2n$, all the statements (there are a few more) come out false, which made me suspect that the quantors cannot be ignored, but unfortunately it makes no sense to me.

Answer 1

Hint

Consider the second formula:

$∃v_0(v_0⋅v_1)=v_1$;

it has a free variable: $v_1$.

In order to evaluate its truth-value, we have to assign a "reference" to $v_1$.

Thus, $∃v_0(v_0⋅v_1)=v_1[\beta]$ is "evaluated" to:

"there is an $n \in \mathbb N$ such that $2 \times n = 2$",

which is true.

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Regarding:

$∀v_0∃v_1 (v_0 < v_1)$

we have to consider the clause of the definition of: $\mathfrak A$ to satisfy $\varphi$ with $\beta$, for the case with $\forall$:

$\mathfrak A \vDash ∀x \varphi[\beta]$ iff for every $d \in |\mathfrak A|$ (the domain of the structure), we have $\mathfrak A \vDash \varphi[\beta(x | d)]$,

where $\beta(x | d)$ is the variable assignment which is exactly like $\beta$ except for assigning the object $d$ to the variable $x$.

The sentence $∀v_0∃v_1 (v_0 < v_1)$ is obviously true in $\mathbb N$.

According to the semantical specification, $∀v_0∃v_1 (v_0 < v_1)[\beta]$ is true iff $∃v_1 (v_0 < v_1)[\beta(v_0 | k)]$, for every $k \in \mathbb N$.

And this, in turn, means:

"for every $k \in \mathbb N$, there is an $n \in \mathbb N$ such that $k < n$".