Why is the Generalization Axiom considered a Pure Axiom?

Question

If $\varphi$ is a formula in a first order language $\mathcal{L}$ and $x$ is a variable that is not free in $\varphi$, then the following is a pure axiom

$$\varphi \to \forall x\varphi$$

The pure axioms are our building blocks for constructing the logic theorems. There are other pure axioms, e.g. the distribution axiom:

$$\forall x(\varphi \to \psi)\to (\forall x \varphi \to \forall x \psi)$$

My question is:

Why is the generalization axiom necessary? I'm sure (I believe it) it can't be proved from the other pure axioms (but I don't see why isn't this axiom a valid formula), however, it's doesn't seem to be so important in the sense that "it doesn't add any information at all", an example of such an axiom is (in informal language):

If $x=2$ then $x=2$ for all $z$.

Why is it so important? I mean, in my opinion, the other axioms (as the distribution axiom, for example) seem to be more necessary since those "talk" about variables that may be free in the considered formulas. I've seen a couple of lemmas that need the generalization axiom to be proved, but I can't get the intuition behind.

Answer 1

This axiom is "useful" in proving the Generalization Theorem :

If $\Gamma \vdash \varphi$ and $x$ does not occur free in any formula in $\Gamma$, then $\Gamma \vdash ∀x \varphi$.

See :

• Herbert Enderton, A Mathematical Introduction to Logic (2nd ed - 2001), page 117.

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There are other axiomatizations of first-order logic that avoid this "unnatural" axiom; see :

• Joseph Shoenfield, Mathematical Logic (1967), page 21

or

• George Tourlakis, Lectures in Logic and Set Theory. Volume 1 : Mathematical Logic (2003), page 34.

Answer 2

This answer is really just an elaboration of Mauro's answer.

To address your question as to why the Generalization axiom is necessary, we need it (or some mechanism like it) in order to prove lots of theorems that contain universal quantifiers.

Your example $(x = 2) \vdash \forall z (x = 2)$ is, as you describe, a correct but vacuous application of the Generalization axiom. But this Wikipedia example shows a nontrivial application of universal generalization. In fact, using the system in the first edition of the Enderton book cited by Mauro, the distribution axiom you cite in your question can be proven with the help of universal generalization.

Finally, you asked for the intuition behind it. Suppose we can prove $P(c)$ for an arbitrary constant $c$. Let's say we notice that we can follow the same strategy for any other constant. This means it's true for any constant we plug in. So it's safe to generalize it by adding the quantifier to get $\forall x P(x)$.