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I just want to check my idea for what exactly is meant by ${\vdash}$ and ${\vDash}$ is correct (I am a noob to logic).

${S\vdash t}$ means "$t$ is provably true using finitely many elements of $S$ within that system", whereas ${S\vDash t}$ means "$t$ is true given $S$", but doesn't make any statements about whether we could prove it within that system $S$. Is this essentially correct? Thank you.

Riemann'sPointyNose
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    See wiki/Logical Consequence. What do you mean by system? The symbols read “from $Γ$ we derive $φ$”, meaning $φ$ is provable given $Γ$, and “from $Γ$ we conclude $φ$”, meaning $φ$ is true given $Γ$, for $\vdash$ and $\vDash$ respectively, where $Γ$ is a set of formulae and $φ$ is a single formula. The formal system within which we derive or conclude is denoted as a subscript to either $\vdash$ or $\vDash$ and does not appear to the left of these symbols. – k.stm Jan 29 '21 at 20:10
  • Multiple copies of the same question available. – Mauro ALLEGRANZA Jan 30 '21 at 09:14

1 Answers1

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I don't think you have it quite right.

"$S\models\varphi$" is defined purely semantically: it means "If $\mathcal{M}$ is a structure satisfying each sentence in $S$, then $\mathcal{M}$ satisfies $\varphi$ as well." In particular, this notion is secondary to the notion of a sentence being true in a structure (which uses the same "$\models$" symbol, incidentally).

Meanwhile, "$S\vdash\varphi$" is not actually fully precise! It really only makes sense once we specify a proof system $\mathfrak{P}$. This is a totally different object from $S$ and $\varphi$; basically, it's just a set of rules for deriving sentences from sets of sentences. If we can derive $\varphi$ from $S$ via $\mathfrak{P}$, we write "$S\vdash_\mathfrak{P}\varphi$."

A priori these are two totally different things, and we can whip up silly "proof systems" which are obviously unrelated to the semantics: e.g. "bogus ponens," the proof system which lets you deduce anything from anything.

(Note that I haven't defined "proof system" precisely - nor am I going to. There are various definitions floating around in the literature, with no one particular one being universally accepted. And there are further subtleties which can emerge, see e.g. here.)

So how are these connected, and why do we abuse notation by writing "$\vdash$" instead of "$\vdash_\mathfrak{P}$?"

Well, it turns out that for a wide variety of $\mathfrak{P}$s we in fact have soundness and completeness of the proof system with respect to the semantics, that is, $$S\vdash_\mathfrak{P}\varphi\quad\iff\quad S\models\varphi$$ (the left-to-right direction is soundness, the right-to-left is completeness - and it's the latter which is by far the more interesting direction). Moreover, since we're specifically interested in the semantics above, proof systems which are not sound and complete aren't generally of interest.

This means that we can usually write "$\vdash$" without fear of serious confusion, since all the "reasonable" proof systems we look at will behave identically here. But in other contexts, and for certain questions (like questions about proof length), the specific proof system may matter a lot. So when first learning this stuff it's a good idea to fix a single proof system from the get-go, and only generalize later once the basics are understood.

Noah Schweber
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  • Ohhhhhhh I think I see - you are saying that ${S \vDash t}$ means essentially "for any proof system such that $S$ is all true, then $t$ is true"; whereas ${S \vdash_{\mathfrak{P}} t}$ means "using the proof system ${\mathfrak{P}}$, and given all of that in ${S}$, there is a proof"? – Riemann'sPointyNose Jan 29 '21 at 20:30
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    @Riemann'sPointyNose No, $\models$ has nothing to do with proof systems at all (in terms of its definition, at least; per the completeness theorem they are in fact related, but that happens later) - it's just about (first-order) structures. You can study $\models$ without ever thinking about proof systems at all (not that you should) - for example, while some texts prove compactness for $\models$ as a corollary of the completeness theorem, you can just prove compactness "purely semantically." – Noah Schweber Jan 29 '21 at 20:31
  • Ok, I see. So I should think of ${S\vDash t}$ more like "if some structure satisfies everything in ${S}$, it also satisfies $t$" - that "structure" hasn't got anything necessarily to do with any particular proof system; whereas ${S \vdash_{\mathfrak{P}} t}$ says "Given the proof system ${\mathfrak{P}}$, and given everything in $S$ is satisfied by this proof system - I can prove ${t}$ also"? – Riemann'sPointyNose Jan 29 '21 at 20:48
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    @Riemann'sPointyNose Yes, that's exactly right. FWIW I'd recommend studying $\models$ on its own first, and only turning to $\vdash$ once you have a solid understanding of first-order structures. But ultimately your mileage may vary. – Noah Schweber Jan 29 '21 at 21:00
  • Thank you very much for your help! – Riemann'sPointyNose Jan 29 '21 at 21:04