User blog comment:LittlePeng9/First order oodle theory/@comment-30167082-20161222182647/@comment-1605058-20161222212452

First let's look an an example. Let's take a formula \((\not\exists x(x\in y))\land(\not y=z)\). In this formula, two variables appear: \(x\) and \(y\), but in essence, this formula depends only on the variables \(y,z\), since \(x\) appears in the formula with a quantifier (we say that \(x\) is bound). Since this formula depends on \(y\) and \(z\), we cannot say whether it's true or false - that depends on \(y\) and \(z\). So in order to make something that can be true or false, we have to assign some value to \(y,z\). We can assign them to be any sets (it could be something else in a more general situation, but in FOST we require them to be sets). For example, if to \(y\) we assign the value \(\{0,1\}\) and \(z\) the value \(\{2,3\}\), the formula will be false, and if we let \(y\) be the empty set and \(z\) be the set \(\{1\}\), it is true.

Now we have to make something clear - assignment is not something we do within a language. Once we have the formula and we want to assign values, we are talking about the truth in the universe of sets, not something like well-formedness of a formula. Let me say again, this assignment can be essentially arbitrary.

Hoping that this might clarify some things, let me recall the definition of Rayo-namability (which I modify a bit to suit the above example). We say the above formula Rayo-names the set \(m\) if there is an assignment of values to the unbound variables (in the above example, \(y,z\)) in which value \(m\) is assigned to \(y\) and the formula is true, and also for any assignment of values which gives a true formula, \(y\) is assigned value \(m\).

Let me argue that the above formula Rayo-names the empty set. Firstly, there is an assignment which makes the formula true and in which \(y\) is assigned the value \(\varnothing\) - indeed, I give an example at the end of the first paragraph. Now let's consider any assignment of values to the variables. Say the value \(A\) has been assigned to \(y\). The first part of our formula says that there is no possible \(x\) which would be an element of \(A\) - hence \(A=\varnothing\). So both conditions for Rayo-namability are satisfied.

I hope this clarifies the misunderstandings.