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EDITED to crucially distinguish the book excerpts from the OP's own words.

In Bell-Machover's "A course in Mathematical Logic", on pag. 53 we read: "the prime formula $\forall x (\alpha \rightarrow \alpha)$ is satisfied by any valuation, but not every truth valuation" and '$\forall x (\alpha \rightarrow \alpha)$ is logically true but not tautologically true; and it is logically (but not tautologically) equivalent to $\forall x (\alpha \leftrightarrow \alpha)$'

I don't understand how $\forall x (\alpha \rightarrow \alpha)$ can be "not tautologically true".
It is proven that

10.2 Theorem: If $\Phi \vdash_0 \alpha $ then $\Phi \vDash_0 \alpha $. In particular, if $\vdash_0 \alpha $ then $\vDash_0 \alpha$.

That is, if a proposition is provable (propositionally deducible from the empty set) then it is a tautology;

and that

10.3 Lemma: For any $\alpha$, $\vdash_0 (\alpha \rightarrow \alpha) $

That is, $\alpha \rightarrow \alpha $ is provable.

Therefore I should conclude that $\alpha \rightarrow \alpha $ is a tautology (which is clear from the truth table), which means it should be satisfied by any truth valuation. Any truth valuation, though arbitrary, should be by definition compatible with $\neg$ and $\rightarrow$:
$ (\neg \alpha)^\sigma = \top$ iff $(\alpha)^\sigma = \bot \quad $ and $\quad (\alpha \rightarrow \beta)^\sigma =\top$ iff $\alpha^\sigma = \bot $ or $\beta^\sigma =\top $

How can therefore a truth valuation give $\bot$ for $\forall x (\alpha \rightarrow \alpha)$? I feel like I am missing something important!

ryang
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britz
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  • Consider the peculiar choice of the author the use from the start the language of predicate logic (page 15). The propositional part is defined as a "fragment" of it, through truth valuations (page 20) that consider only $\lnot$ and $\to$: they do not "see" quantifiers. Thus, a prime formula $\forall x \alpha$ (page 21) is considered by truth valuations as a propositional atom $p$ and cannot be a tautology. – Mauro ALLEGRANZA Feb 28 '23 at 06:43
  • @MauroALLEGRANZA This makes everythig clear. I confused propositional logic and first order logic. $\alpha \rightarrow \alpha $ is a tautology, but $\forall x (\alpha \rightarrow \alpha)$ cannot be, because it is considered prime. – britz Feb 28 '23 at 11:12

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It's unclear whether the sentence near the middle of your post is an excerpt or your commentary, but my guess is that Bell-Machover defines a tautology by virtue of its truth-functional form; as such, neither $$\forall x (\alpha \rightarrow \alpha)$$ nor $$\forall x (\alpha \leftrightarrow \alpha)$$ nor $$\forall x (\alpha \rightarrow \alpha)\leftrightarrow \forall x (\alpha \leftrightarrow \alpha)$$ are tautologies, but all are logical truths.

EDIT

OP:

Yes, the parts "That is ..." are my remarks.

Yes, I'd guessed that. So, to elaborate: the three formula that I gave above have truth-functional forms $$P$$ and $$Q$$ and $$P\leftrightarrow Q,$$ respectively, so are not (propositional-logic) tautologies.

ryang
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  • Theorems 10.2 and 10.3 are proven in the cited book, pag. 36. – britz Feb 28 '23 at 11:01
  • @britz I wasn't referring to the meat of that theorem, but to (what may be) your commentary within the excerpt. The Google Books preview looks quite different from your excerpt (probably different editions); this reinforces my suspicion that the author's definition of Tautology is as I gave above. $\quad$ EDIT: Mauro's new comment above is reinforcing everything I've said. $\quad$ If you wish to edit your Question, I suggest adding block quotes to clearly distinguish what you've excerpted and which lines are your commentary, which may not be consistent with the author's meaning. – ryang Feb 28 '23 at 11:07
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    Yes, the parts "That is ..." are my remarks. I'll try to edit the post as you suggested. Sorry, I'm new. – britz Feb 28 '23 at 11:25