6

My current understanding:

$\vdash$ is "syntactic consequence", e.g. if we have $a \vdash b$ it means if we've written down $a$ as something we know, we can immediately write down $b$ as something we know. This is saying nothing about whether any of these pieces are true, false, something else, etc, it's just saying that we can say something about $b$ by virtue of being able to say something about $a$.

$\vDash$ is "semantic consequence", e.g. if we have $a \vDash b$, then it (usually) means if $a$ is always true in every interpretation (i.e. no matter how we mess with any of the underlying variables driving $a$'s valuation), then $b$ is true as well. However I don't know if this is necessarily just a "true"-only thing. It seems conceivable that we could make one for falsehoods too, i.e. if $a \vDash b$ might mean if $a$ is always false then $b$ is always false, of if $a$ is always "dog" then $b$ is always "dog", etc. But I am unsure of this one.

$\to$ is material implication, which exists solely within the logic system (as opposed to the previous two metalogical relations) and has a specific semantic meaning / definition in terms of a truth table in something like prepositional logic ($a \to b$ is always true unless $a$ is true and $b$ is false). It's defined this way simply because it is a helpful / useful model to describe things like "whenever $a$ is true, $b$ is true".

Do I have the right idea so far?

user525966
  • 5,631

1 Answers1

4

You've done a good job of explaining the main differences. A couple of comments:

$\vdash$ is "syntactic consequence", e.g. if we have $a \vdash b$ it means if we've written down $a$ as something we know, we can immediately write down $b$ as something we know.

I would clarify further that it's less about knowledge and more about proof. $a \vdash b$ is true if there is a valid proof of $b$ from $a$, so it's relative to whatever we have decided "valid proof" means. For example, some philosophers and logicians have insisted on denying certain types of proofs and have proposed very restrictive logical systems where you are restricted in when you can deduce $b$ from $a$. In some such systems, $b$ may follow semantically from $a$ ($a \vDash b$) but there may be no way to prove it (so $a \not \vdash b$).

Concretely, some logicians deny that $\lnot \lnot a \vdash a$ -- they aren't willing to accept $\lnot \lnot a$ as a proof of $a$. So it may be that in every world where $\lnot \lnot a$ is true, $a$ is true (semantically), that is $\lnot \lnot a \vDash a$. But if you insist on a standard of proof where $\lnot \lnot a$ is not a good proof of $a$, then $\lnot \lnot a \not \vdash a$.

$\vDash$ is "semantic consequence" ... However I don't know if this is necessarily just a "true"-only thing. It seems conceivable that we could make one for falsehoods too, i.e. if $a \vDash b$ might mean if $a$ is always false then $b$ is always false, of if $a$ is always "dog" then $b$ is always "dog", etc. But I am unsure of this one.

Your speculation here is a bit inaccurate. $a \vDash b$ usually means some variant of "in every possible world where $a$ is true, $b$ is also true". The definition doesn't consider worlds where $a$ is false. And there are no words where $a$ is always "dog" -- $a$ is some sentence, so it is only ever true or false. Calling $a$ a dog would be a sort of semantic mismatch. Maybe $a$ is some fact about dogs, but not a dog itself.

Do I have the right idea so far?

Yes, and let me comment on one thing. In logic, we insist on being extremely formal and distinguishing between things that have different definitions, even if they ultimately amount to the same thing. For example, we distinguish between $(\lnot a) \land (\lnot b)$ (not $a$ and not $b$) and $\lnot (a \lor b)$, even though these ultimately end up being equivalent.

So it is with $\vDash$, $\vdash$, and $\to$. We insist on distinguishing between them because they are defined differently -- and you have summarized the different definitions well. However, there are important senses in which they are all equivalent. In particular, you will probably prove some theorems like:

  • (Soundness) If $a \vdash b$, then $a \vDash b$.

  • (Completeness) If $a \vDash b$, then $a \vdash b$.

  • (Provability of implication) If $a \to b$ is provable -- that is, if $\vdash (a \to b)$ -- then $a \vdash b$. Similarly, if $a \vdash b$, then $\vdash (a \to b)$.

So, all three of them end up being equivalent concepts in the end, but it is important to keep them mentally distinct. There are even some (generally faulty) logical systems where the three concepts are not equivalent, so you have to be careful to not assume they are the same until it is proven.

Caleb Stanford
  • 45,895
  • The reason I brought up examples like cat/dog/etc is to refer to arbitrary logic systems where we may not be talking about concepts like "true" or "false" (e.g. 3-valued systems, or possibly others) since our semantics/meanings may be totally different, so I am trying to generalize. Or is $\vDash$ literally about the concept of truth and truth alone? – user525966 Sep 08 '18 at 22:23
  • And yes, I'm trying to keep them separate specifically to be open to the idea that there may be systems in which things don't end up being equivalent (even if they happen to be for "simpler" systems) – user525966 Sep 08 '18 at 22:26
  • @user525966 Ah, I see. In that case I think what you wrote is fine. I still wouldn't call $a$ a "dog", but in a multi-valued system maybe it can be true, false, or in-between (for example). Or maybe it can be one of several truth values. The concept $a \vDash b$ would then need an appropriate definition, such as: in all worlds where $a$ is a certain degree of true, $b$ is at least that degree of true. – Caleb Stanford Sep 08 '18 at 22:26
  • I do find one thing strange -- as a possible example: is it possible to have $a \vDash b$ hold but not $a \vdash b$? Basically saying in every conceivable world where $a$ is true, $b$, is true, but at the same time we cannot write $b$ as something we know if $a$ is something we know. Seems kind of contradictory? If we can say something about $b$ semantically from $a$ semantically, does this not imply it is also true syntactically? – user525966 Sep 08 '18 at 22:28
  • I don't know of any logical system where $a \vDash b$ means "whenever $a$ is false, $b$ is false" as this would semantically not line up with what "false" means and so it would be a poor choice of definitions. $a \vDash b$ is supposed to be where $a$ implies $b$ semantically, not where $a$ being false implies $b$ being false. – Caleb Stanford Sep 08 '18 at 22:28
  • @user525966 Yes, that is a conceivable and even potentially common situation. It happens when your proof rules are not strong/general enough, so that $b$ semantically follows from $a$ but there is no way to deduce it or prove it syntactically. – Caleb Stanford Sep 08 '18 at 22:29
  • What's an example of that? (I can't even picture it) -- where we can say something follows semantically and yet we can't deduce or prove it, somehow – user525966 Sep 08 '18 at 22:29
  • 1
    @user525966: It doesn't happen with ordinary propositional logic or first-order logic -- exactly because there's a completeness theorem that holds for those logics. You can make a proof system that has an example, however, simply by omitting some of the axioms or inference rules in your proof system. – hmakholm left over Monica Sep 08 '18 at 22:33
  • @user525966 I added a couple of paragraphs and an example. Does it help? It's just a matter of how strong your "standard of proof" is. If you want to insist on a really high standard of proof, it may be you can't prove certain things. – Caleb Stanford Sep 08 '18 at 22:37
  • 1
    Alternatively, second-order logic has both a $\vdash$ and a $\vDash$, but it is not complete, so in second-order logic there are examples of $T$ and $\varphi$ where $T\vDash\varphi$ but not $T\vdash\varphi$. Unfortunately all the examples I know are rather too complex to describe in this space. – hmakholm left over Monica Sep 08 '18 at 22:37
  • Ah so if our axiom system weren't quite right, we'd be unable to arrive at the syntax we want, even if we were able to show that based on our arbitrary semantic definitions, it would hold semantically if we could arrive at it syntactically? – user525966 Sep 08 '18 at 22:40
  • 1
    @user525966 Yes :) You know in the meta-system that it's semantically true, but you've insisted on a standard of proof such that there's no valid proof of it. It's important to distinguish between what can be proven in the system, and what you know semantically outside the system. – Caleb Stanford Sep 08 '18 at 22:41
  • 1
    This has been really helpful to me, thanks :) – user525966 Sep 08 '18 at 22:42
  • @6005 How do you define "every conceivable world"? – user525966 Sep 09 '18 at 10:18
  • @user525966 Please ask a new question on the site. The discussion isn't supposed to go on in the comments. You can link to this post. The short answer is that this, too, can depend on the system, but usually the possible worlds are all possible truth assignments to the variables. – Caleb Stanford Sep 09 '18 at 17:34