Understanding the definition of variable assigment: what's means 'except possibly a'?

Question

in frist-order logic, part of assignments practice represent like this "if is ∀, where is a variable, then ⊨vℳ iff for every assignment ' that agrees with on the values of every variable except possibly , ⊨v'ℳ' " here I don't understand except possibly , because I understand 'for all x...' means that 'every variable satisfies ...' why this case postulate except possibly ??

Answer 1

I understand 'for all x...' means that 'every variable satisfies ...'

No, it does not mean that - in fact the opposite. It means that every assignment $v$ to some domain object $a$ for that one variable $\alpha$ satisfies it. Because we want the statement to be true of every object that we can point $\alpha$ to. So we have to look for all ways of assigning an object to $\alpha$, while keeping the assignment of the other variables fixed. This is when we introduce the alternate assignments (or whatever they are called in the textbook you are using) $v'$. The universally quantified statement is true under an assignment if it is true for every possible assignment to the variable $\alpha$ with other other variables the same. Most of these possible assignments will provide a mapping for $\alpha$ different from the original one, but to look at all assignements we also need to consider the one original assignment $v$ that we started out with, so we also need to look at the case where the value is not different from the original $v$. That's why the formulation has "possibly".

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Edit: As an example:

$\exists y \forall x (y < x)$

Suppose in our evaluation we're at the stage that we're looking at $0$ as a candidate value for $y$ (so we have $v: y \mapsto 0$ and, say, $x \mapsto 0$ as a starting point) and now want to check whether $\forall x (x > y)$ holds for that assignment $v$. In order to do that, we need to verify whether every number value we can assign to $x$ satisfies the statement that it is larger than $0$:

$\models_v \forall x (y < x)$ (where $v := [y \mapsto 0, x \mapsto 0]$) iff
for every assignment $v'$ that agrees with $v$ on the values of every variable except possibly $x$: $\models_{v'} \forall x (y < x)$

Note that we keep the value for the variable $y$ fixed because now that we're checking whether $0$ might be a candidate to satisfy the existential claim, we need to check whether that particular $y$ is strictly smaller than every number $x$. So we need to look at

$v := [y \mapsto 0, x \mapsto 0],\\ v' := [y \mapsto 0, x \mapsto 1],\\ v'' := [y \mapsto 0, x \mapsto 2], \ldots$

Note how we still need to consider the assignment with $x \mapsto 0$ because if the universal claim is to be true, we also need to have that the strictly-smaller-than statement is true of $0$ itself. This is not the case, so $\forall x (y < x)$ comes out false for $v: y \mapsto 0$, and as it turns out, for any such $y$. So it was curcial that we checked all assignments, even where they did not differ from the starting assignment $v$ in the value they assigned to $x$.

Answer 2

Intuitively, the formula $∀x \phi$ is true in an interpretation $\mathfrak M$ - in symbols $\mathfrak M \vDash \forall x \phi$ - iff $\phi[d/x]$ is true for every object $d$ in the domain $M$ of the interpretation, i.e. iff:

for every $d \in |M|$, we have $\mathfrak M \vDash \phi[s(x \mid d)] $.

Here, when we write $s(x \mid d)$ for every $d \in |M|$, we consider also the possibility that $d$ is the same object that the variable assignment function $v$ assign to the variable $x$.

Thus, we are considering every $v'$ that agrees with $v$ on the values of every variable except possibly $x$.