Gap in Halmos & Givant's "Logic as Algebra": undefined $\models$? I have been hugely enjoying Logic as Algebra by Halmos and Givant (1998, isbn: 0-88385-327-2)1, largely because I appreciate the authors' careful attention to the point of view of someone who is completely unfamiliar with the subject (like me!).  In particular, the authors take the time to explain notational conventions in detail.  (In fact, while reading this book I've often found myself wishing that all those who write about mathematics were this considerate, that is to say, this polite, to the reader.)
Therefore it came like a punch in the stomach when on page 91 the authors for the first time drop the symbol $\models$ without any prior mention, let alone definition.  In fact, here's this very first occurrence of this symbol in the book:

Deduction theorem. ${\mathbf S}\models q$ if and only if ${\mathbf S}$ has a finite subset $\{p_1,\dots,p_n\}$ such that $\vdash (p_1 \wedge \cdots \wedge p_n) \Rightarrow q$.

(Here $p_1,\dots,p_n,$ and $q$ are individual propositions of the propositional calculus, and ${\mathbf S}$ is a set of such propositions.)
And a few lines later:

${\mathbf S}\models q$ if and only if $p_1 \wedge \cdots \wedge p_n \leq q$ for some finite subset $\{p_1,\dots,p_n\}$ of ${\mathbf S}$.

...and

One important special case of the deduction theorem is that $p\models q$ if and only if $\vdash p \Rightarrow q$ (that is, if and only if $p \leq q$).

Eventually I sorted out the matter, and concluded that these were typos.
I figure that I'd post the corrections here, for the next who run into the same unpleasant surprise (see under Answers below).



1 I'm completely new to this subject, so it's impossible for me to gauge how standard the notation, terminology, and concepts that the authors use are.  Therefore, a full understanding of this post may require access to this book.

 A: As it turns out, on the first paragraph of the following page (p. 92), the authors do define $\models$:

The notation that is often employed to express relationship of semantic consequence between ${\mathbf S}$ and $q$ is ${\mathbf S}\models q$.

This is much more in keeping with the authors' handling of notational matters up to this point in the text.
Therefore, I must conclude that all three occurrences of $\models$ on p. 91 are typos.  They should all have been $\vdash$.  Thus:

Deduction theorem. ${\mathbf S}\vdash q$ if and only if ${\mathbf S}$ has a finite subset $\{p_1,\dots,p_n\}$ such that $\vdash (p_1 \wedge \cdots \wedge p_n) \Rightarrow q$.

…

${\mathbf S}\vdash q$ if and only if $p_1 \wedge \cdots \wedge p_n \leq q$ for some finite subset $\{p_1,\dots,p_n\}$ of ${\mathbf S}$.

…

One important special case of the deduction theorem is that $p\vdash q$ if and only if $\vdash p \Rightarrow q$ (that is, if and only if $p \leq q$).

Further confirmation of this explanation is the fact that these three uses of $\models$ on p. 91 happen shortly after a paragraph, also on p. 91, defining and discussing the use of the symbol $\vdash$.
And not least, if the three rogue $\models$'s are replaced with $\vdash$'s, then the resulting text actually makes sense in the context of the foregoing discussion.
EDIT: One further bit of confirmation for the explanation given above is that, on p. 93, the authors write (my emphasis on the first sentence):

There is a surprising and important semantic analogue of the deduction theorem.
Compactness theorem. For any proposition $q$ and any set ${\mathbf S}$ of propositions in $({\mathbf A}, {\mathbf F})$, we have ${\mathbf S} \models q$ if and only if there is a finite subset $\{p_1,\dots,p_n\}$ of ${\mathbf S}$ such that $\{p_1,\dots,p_n\} \models q$.

