# in definition of assigment, what's means 'except possibly a'?

in frist-order logic, part of assignments practice represent like this "if πis βπΌπ, where πΌ is a variable, then β¨vβ³ π iff for every assignment π£' that agrees with π£ on the values of every variable except possibly πΌ, β¨v'β³'π " here I don't understand except possibly πΌ, because I understand 'for all x...' means that 'every variable satisfies ...' why this case postulate except possibly πΌ??

• It means that all the values are the same except at most alpha. Commented Jul 11 at 6:14
• Why is Ξ± excluded? For all variables, why is it that a formula A holds true in the model even if a different value is assigned to the variable Ξ±? Is it because formula A might not include the variable Ξ± itself?
– 유준상
Commented Jul 11 at 6:31
• The idea is simply: we have to check all possible values assigned to variable $x$; thus, we consider all assignments (every ass is defined for every variable of the language) that are exactly as $v$ but assign to the variable $x$ difefrent values. Commented Jul 11 at 6:41
• See also Variable assignments and many more... Commented Jul 11 at 6:48

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".

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 $$x$$.

Intuitively, $$βx \phi$$ is true in an interpretation $$M$$ iff $$\phi[d/x]$$ is true for every object $$d$$ in the domain of $$M$$.

Formally:

$$M \vDash βx \phi[v]$$ iff for every $$d \in |M|$$, we have $$M \vDash \phi[s(x \mid d)]$$.

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 $$x$$.