The universal generalization rule of predicate logic says that whenever formula M(x) is valid for its free variable x, we can prefix it with the universal quantifier,

M(x) ⊢ ∀x M(x).

But it seems then makes no sense. Why do you introduce a notion that does not mean anything new?

  • $\begingroup$ It doesn't mean the same thing as $M(x).$ $\endgroup$ – Thomas Andrews Jan 8 '14 at 22:23
  • $\begingroup$ Well, we also have $\forall x M(x)\vdash M(x)$ $\endgroup$ – Hagen von Eitzen Jan 8 '14 at 22:27
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    $\begingroup$ If you are teaching yourself predicate logic in order to write and understand mathematical proofs, may I humbly suggest the tutorial accompanying my proof checking software downloadable at dcproof.com The system of logic presented there is much easier to use than that presented in many philosophy textbooks (as you seem to be working through). It is more like the simplified logic implicitly used in most mathematics textbooks. $\endgroup$ – Dan Christensen Jan 9 '14 at 4:27
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    $\begingroup$ This is a matter of internalization. Saying that "$M(x)$ is valid" is a sentence in the metalanguage; whereas "$\forall x M(x)$" is a sentence in the object language. In general, internalization improves the expressiveness of our language. I wish I could offer a concrete example of this at work, but I can't think of one off the top of my head. $\endgroup$ – goblin Jan 9 '14 at 15:30

Basically, we need quantifiers not so much for making inferences but for having enough expressive "capability" in our language.

For simplicity, I assume the context of classical first-order logic, where the existential quantifier $\exists$ can be defined as an abbreviation for $\lnot \forall \lnot$.

If we are working into formalized arithmetic (e.g.first-order logic with Peano's Axioms), we may stay with the convention that free variables are implicitly universally quantified.

So, the Axiom about the successor function :

$x = y \rightarrow S(x) = S(y)$

can be interpreted as $\forall x \forall y ( ...)$, like in algebra when we write : $(x+y)^2 = x^2 + 2xy +y^2$.

The problem is that not all the relevant "facts" can be rightly expressed with universal quantifiers upfront.

Take the formula : $x = 0$; if we stay with the above convention, its meaning must be : "all numbers are equal to $0$", that is false.

If we negate it, we get $\lnot x = 0$, which, following the convention, must be read as : "all number are different from $0$". But this is false also !

We need a sign for "restricting" the context (the scope) of the quantifier (this wonderful discovery was made by Frege) in order to have :

$\forall x (x = 0)$

Now, its negation will be :

$\lnot \forall x (x = 0)$

and this time we have the intended meaning : "not all numbers are equal to $0$", that is true.

This example is taken form S.C.Kleene, Introduction to Metamathematics (1952).


$M(x)$ doesn't strictly have a truth value; $\forall x.M(x)$ does. So, different beasts. Not to mention "$T\vdash M(x)$" is not a sentence in the language, while again, "$\forall x.M(x)$" is.

The universal generalization rule connects a piece of metatheory ("for any variable '$x$', '$M(x)$' is provable", or some such), with a statement in the theory itself. The distinction is a subtle one, and won't often trip one up in everyday logic use, but when studying some truly abstract reaches of logic like model theory, you can end up in some tangled fallacies if you don't keep theory and metatheory apart.

  • $\begingroup$ Wait, which language are you talking about? Both M(x) and ∀xM(x) are predicate formulas? Why one is not? $\endgroup$ – Val Jan 8 '14 at 23:48
  • $\begingroup$ Unfortunately I can't read the other end of that link on this device, so hopefully I'm not entirely off the mark in what follows; but I'm not disputing that, for example, the formula $M(x)$, or the sentence $\forall x.M(x)$ are both part of the language. But mention of things like "validity", "satisfiability", "provability", and any symbols like $\vdash$ or $\vDash$ that you usually see in a statement of the universal generalization rule, are not part of the first order language. They are statements about the language. $\endgroup$ – Malice Vidrine Jan 9 '14 at 0:49
  • $\begingroup$ I do not ask to contrast M(x) or ∀x.M(x) against M(x) ⊢ ∀x.M(x). I am contrasting M(x) against ∀x.M(x). Both are formulas "in the language" and both mean the same thing. I am asking why do you need the ∀x.M(x) if you have M(x) already. $\endgroup$ – Val Jan 9 '14 at 14:44
  • $\begingroup$ Both do not mean the same thing. $M(x)$, as mentioned, is not a sentence and can be satisfied when $\forall x.M(x)$ is not. Notice the turnstile in your original post; $M(x)$ to the left of the turnstile means something different from just $M(x)$. Point is, there is proof theoretic language above that is relevant. $\endgroup$ – Malice Vidrine Jan 9 '14 at 17:06

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