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I have heard that the symbol ":=" which means (LHS) is defined to be equal to (RHS) is not a symbol of logic. As far as I know, the symbol intuitively means that the symbol we have on the LHS is from now on a reference to the object that the RHS refers to, thus having exactly the same properties. This looks a lot like the an "informal" use of extensions by definitions (https://en.wikipedia.org/wiki/Extension_by_definitions), which are a part of formal logic. If this is correct however, why is it valid to sometimes have symbols which don't contain the necessary variables? More precisely, if one has a set $$\{x \ | \ P(x)\},$$ one sometimes defines this to be $A$ by $A:=\{x \ | \ P(x)\}$. However, definition by extension only allows me to introduce a new symbol that also contains the occuring variables, such as $A(x)$ or $A_x$. Am I missing something or is this just an informal use of extension by definition?

user1578232
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    If you use the formal language of first order logic you have $\leftrightarrow$ for formulas and $=$ for terms (i.e. "names"). You can use $:=$ in the meta-language, specifying e.g. that the symbol $\to$ is treated as an abbreviation for ... ($p \to q$ means $\lnot p \lor q$). In the context of the theory, when we write $A = { x \mid \ldots }$ we are saying that the set specified by the RHS is the same as the set called $A$. – Mauro ALLEGRANZA Jan 28 '22 at 10:02
  • If you are considering extension by definitions, see your previous post. The "additional axiom" is a formal expression of the language: thus, either $\leftrightarrow$ for predicate or $=$ for terms: constants like $\emptyset$ or functions like $\cap$. – Mauro ALLEGRANZA Jan 28 '22 at 10:06
  • Conclusion: in a formal context you cannot use $:=$ in the "additional axiom" of the extension, unless it is already a symbol of the formal language. – Mauro ALLEGRANZA Jan 28 '22 at 10:29

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Whenever there is a unique set $z$ such that you can prove $\forall x. x \in z \leftrightarrow P(z)$, then you can apply extension by definition to the formula $\exists! z. (\forall x. x \in z \leftrightarrow P(x))$. This allows you to extend your theory with a nullary function symbol $A$ so that $\forall q. (\forall x. x \in q \leftrightarrow P(x)) \leftrightarrow q=A$. This is the actual axiom that describes the behavior of $A$ in our extended theory.

Informally, we might just write $A := \{x|P(x)\}$ in this case, but this is just notation intended to explain what you're doing to the reader. Neither $:=$, nor the $\{x|P(x)\}$ notation itself are part of the logic.

Now, as you write, definition by extension only allows you to introduce a new symbol as a function of the free variables that occur in the firmula that you're using in the definition. But the word "free" is there for a reason: notice the variable $x$ does not occur as a free variable in the formula we actually used above to introduce the definitional extension! It only occurs as a bound variable, bound by the universal quantifier. In fact, that formula has no free variables, so the symbol you introeuce will be a nullary function symbol: a nullary function symbol is the same thing as a constant symbol.

Z. A. K.
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  • Thank you very much, I think I see where I went wrong. If I would have a set of the form: $\forall x \exists!z (\forall y(y \in z \iff P(x,y)))$, I would not be able to define $A$ to be this set, right? I would only be allowed to call it $A_y$ or $A(y)$ but $y$ needs to occur somewhere, if I am not mistaken. – user1578232 Jan 28 '22 at 21:55
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    You'd have to call that $A_x$ or $A(x)$. This is because you're creating the definitional extension using a formula of the form $\forall x. \exists! z. \varphi(x,z)$. Similarly, if you were using a formula of the form $\forall m. \forall n. \exists! z. \varphi(m,n,z)$ then you'd have to define $B(m,n)$. – Z. A. K. Jan 29 '22 at 08:25
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    Again, thank you a lot! – user1578232 Jan 29 '22 at 08:29