Is "$\alpha \to \forall x \alpha$, where $x$ does not occur free in $\alpha$" a generalization of $\alpha$, which doesn't make any difference?

Question

Enderton's An Mathematical Introduction to Logic says on p112:

Now at last we give the set of logical axioms. These are arranged in six groups. Say that a wff $\phi$ is a generalization of $\psi$ iff for some $n = 0$ and some variables $x_1 , \dots , x_n $, $$ \phi = \forall x_1 \dots \forall x_n \psi.$$ We include the case $n = 0$; any wff is a generalization of itself. The logical axioms are then all generalizations of wffs of the following forms, where $x$ and $y$ are variables and $\alpha$ and $\beta$ are wffs:

Tautologies;

$\forall x \alpha \to \alpha_t^x$, where $t$ is substitutable for $x$ in $\alpha$ ;

$\forall x( \alpha \to \beta) \to ( \forall x \alpha \to \forall x \beta )$;

$\alpha \to \forall x \alpha$, where $x$ does not occur free in $\alpha$.

In 4, if $x$ does not occur free in $\alpha$, adding $\forall x$ to $\alpha$ doesn't make any difference, correct?

Does 4 exactly describe a generalization of $\alpha$? Hasn't generalization already been defined earlier, and why do we need 4?

Thanks.

Yes, it doesn't make any difference, that's why it's a tautology. The definition of a generalization does not include the condition that $x$ doesn't occur free in $\alpha$, that's what makes them generalizations. — Qiaochu Yuan, Oct 06 '23 at 07:37
The formulas $\alpha$ and $\forall x \alpha$ are not the same are they? So "adding $\forall x$ to $\alpha$" does make a real difference. Whether they mean the same or not is irrelevant. — ancient mathematician, Oct 06 '23 at 07:37
As per comments to your previous post, in Enderton's system "generalization" is formalized by GENERALIZATION THEOREM, page 117. As you can verify reading its proof (by induction) Ax.3 and Ax.4 above are used in steps of the proof. Thus, Enderton's choice of quantifier axioms is justified by the result. — Mauro ALLEGRANZA, Oct 06 '23 at 09:43
You can see here a discussion about several choices in the mix axioms+rules. — Mauro ALLEGRANZA, Oct 06 '23 at 10:04
And see also the post Two forms of Generalization Rule in logic. — Mauro ALLEGRANZA, Oct 06 '23 at 10:05

score 2 · Answer 1 · answered Oct 06 '23 at 07:51

Axiom 4 allows us to take propositional tautologies e.g.

$A \to (B \to (A \land B))$

and generalize them so we can prove further theorems/generalizations e.g.

$(\forall x A \land \forall x B) \to \forall x (A \land B)$.

Ultimately, axiom 4 allows us to prove the Generalization Meta-theorem, which states that if $\Gamma \vdash A$ and $x$ does not occur free in $\Gamma$, then $\Gamma \vdash \forall x A$.

Is "$\alpha \to \forall x \alpha$, where $x$ does not occur free in $\alpha$" a generalization of $\alpha$, which doesn't make any difference?

1 Answers1