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Whenever we use the string

$(\forall x)P(x)$

We are using a meta variable, in this case $x$, which stands for any object in the reference set.

However, the semantics of the symbol $\forall$ indicate that we are refering to all objects that satisfy certain condition $P(x)$.

So, the semantics of "for all the objects over which the metavariable varies, $P(x)$" but we know, a priori, that the meta variable already refers to all the objects in the reference set.

This looks redundant to me.

So, is the universal quantifier redundant?

Darvid
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    Are you saying that the sentence "for all differentiable functions $f: [0,1] \to \Bbb R$, $f$ is continuous" is something we know a priori? How would you write this without using the universal quantifier? – Crostul Apr 03 '21 at 20:38
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    The problem is when you combine quantifiers. For example when you write something like $\forall x \exists y \forall z P(x,y,z)$, you know that $x$ is bound before $y$ and $z$ and this is crucial information to correctly interprete the statement. – Jonathan Schilhan Apr 03 '21 at 20:42
  • @Crostul No. Given the example, "for all differentiable functions f:[0,1]→R, f is continuous", here f is a meta variable that refers to all the objects in the reference set, in this case, all differentiable functions f:[0,1]→R. So, you are saying "for all differentiable functions f:[0,1]→R, over the set of differentiable functions f:[0,1]→R, f is continuous. I'm talking about the meta variable f, in your example. Thank you – Darvid Apr 03 '21 at 20:43
  • @Jonathan, thank you; I'll check if it's a matter of the bounded variables when working with multiple quantifiers, though at first sight I don't see how this avoid the redundancy. – Darvid Apr 03 '21 at 20:46
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    @Darvid The point is that "$\exists\forall\exists$" is very different from "$\forall\exists\exists$" and so on. If we simply erase the $\forall$s and treat every now-free variable as ranging over the whole universe, this has the effect of bringing all the universal quantifiers to the front which vastly changes the meaning (and truth value) of the sentence in general. – Noah Schweber Apr 03 '21 at 21:15

1 Answers1

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"Whenever we use the string (∀)() we are using a meta variable, in this case , which stands for any object in the reference set.

Not so. For a start, you are misusing the term "metavariable" (which means a variable added to mathematical English or Spanish or whatever "metalanguage" you are writing in when discussing formal object language expressions): the variable in (∀)() belongs to the object language not the meta language.

But more importantly, the $x$ here doesn't stand for anything. In fact it is helpful to think of $\forall xP(x)$ as built up like this:

  1. Start from an expression $P(n)$ attributing a property to the object which the name $n$ stands for (and yes, names unlike variables do stand for some object in the relevant domain).
  2. Remove the name, to give us the gappy expression $P(\ )$ which expresses the same property as in (1)
  3. Now complete the gappy expression by applying the quantifier operator $(\forall x)\ldots x \ldots$ to give $(\forall x)P(x)$. This says that everything in the domain has the property expressed by $P(\ )$. But -- and this is important observation -- we should think of the operator $(\forall x)\ldots x \ldots$ as a semantic unit: the variable doesn't have independent significance, doesn't stand for anything, but just serves to bind the quantifier to the open slot in $P(\ )$.

As Bourbaki and others have noted, we could mark the binding in other ways by using arrows or whatever. For example instead of $\forall x\exists yR(y,x)$ we could write $\forall\exists R(\ ,\ )$ with an arrow curving from $\forall$ to the second slot, and another arrow from $\exists$ to the first slot. A binding arrow doesn't stand for anything: nor do the variables in $(\forall x)P(x)$ or $\forall x\exists yR(y,x)$ which do the same binding job.

Standard elementary textbooks should explain this. For some reason, I quite like P&t&r Sm&th's account in this intro logic book which you can freely download at https://www.logicmatters.net/ifl -- look at the short chapters 27 and 28.

Peter Smith
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    I think what the questioner is trying to get at is more like this: the claim that $x²≥0$ (with $x$ free) is semantically very similar to the claim $\forall x. x²≥0$. And in fact we often do omit the quantifier when its domain is clear from context. Typical logical systems for predicate calculus will have introduction and elimination rules that allow either formula to be deduced from the other. So, asks OP, why do we need universal quantifiers at all? – MJD Apr 04 '21 at 14:17
  • @MJD Yes OK -- but the OP's confusions about metavariables and about the "x" referring to objects do rather suggest he needed to get basics nailed down first (before we get to the fancier stuff about the pros and cons of using dialects which allow free variables!) – Peter Smith Apr 04 '21 at 15:22
  • Actually, I was in doubt about the point that @MJD makes on the example $\forall x . x^2 >= 0$. Here, $x$ is a metavariable, because we do not have letters in our reference set, but numbers (to illustrate), so, $x$ is a metavariable that 'stands' for any such number. However, when we use the logical form $\forall x P(x)$ there we already are using the metavariable $x$, so why do we need to say $\forall x$ if, again, $x$ already refers to 'for all' objects where $x$ varies. – Darvid Apr 04 '21 at 21:46
  • but thank you @PeterSmithfor the response. – Darvid Apr 04 '21 at 21:46
  • @Darvid -- No, $x$ here is NOT a metavariable. Sometimes Wikipedia is your friend: https://en.wikipedia.org/wiki/Metavariable – Peter Smith Apr 05 '21 at 10:21