A question about the use of successions in the quantifier clause of the definition of satisfaction

Question

In first order logic, why do we say that: $\exists$ $xi$ $\phi$ is satisfied by succession si iff there is a different succession from s that satisfied $\phi$?

For instance, maybe the only succession that satisfies $\phi$ is actually si, and there is no other succession different from si that satisfies $\phi$. Why there must be a different succession?

Answer 1

The definition is not that there is a different succession. The definition is that there is some succession. This may be different or identical to the original one.

Answer 2

The basic fact is to define what does it mean for a structure (or: interpretation) $\mathfrak A$ to satisfy formula $\varphi$ with variable assignment function $s$. In symbols:

$\mathfrak A,s \vDash \varphi$ (alternatively: $\mathfrak A \vDash \varphi[s]$).

Intuitively, the definition means that the "translation" of $\varphi$ determined by $\mathfrak A$ is true, where the variable $x$ is translated as $s(x)$ wherever it occurs free in the formula.

The "variable assignment function" trick is needed in order to manage free occurrences of variables in formulas, because a variable has no fixed meaning in an interpretation.

Consider the simple example of formula $(x_1=0)$ where $\mathfrak A$ is $\mathbb N$.

What is its truth-value? It depends...

For assignment $s_1$ such that $s_1(x_1)=0$ we have that $(x_1=0)$ is true (i.e. $\mathbb N,s_1 \vDash (x_1=0)$), while for $s_2$ such that $s_2(x_1)=1$ we have that $(x_1=0)$ is false.

If so, what does it mean that the formula $\exists x_1 (x_1=0)$ is true in $\mathbb N$? It means that there is at least one element of $\mathbb N$ - call it $d$ - such that, when variable $x_1$ is "translated" as $d$, the formula $(x_1=0)[x_1 \leftarrow d]$ is true. Obviously, in $\mathbb N$ the element called $d$ will be the number $\text {zero}$.

Consider the formal defintion:

$\mathfrak A,s \vDash \exists x_1 \varphi[s]$ iff for some $d \in \mathfrak A$, we have that $\mathfrak A,s \vDash \varphi [s(x_1 \mid d)]$,

where $s(x \mid d)$ is the variable assignment function which is exactly like $s$ except for the fact that at the variable $x_1$ it assumes the value $d$.

Now, going back to your example: the formula $(x_1=0)$ is satisfied by $s = (0,\ldots)$, because $s(x_1)=0$ and $s(x_1 \mid 0)=s$.

But the formula is satisfied also by $s_1=(1,\ldots)$, because $s_1(x_1 \mid 0)=(0,\ldots)$ and (assuming for simplicity that $x_1$ is the only variable occurring free in $\varphi$) the only relevant fact is that $s(x_1)=0$.