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I noticed that in general, the statements $\vdash A\to B$ and "If $\vdash A$ then $\vdash B$" are not equivalent.$^1$

However, this shows that I have a faulty intuition:

  • I thought that I can interpret $\vdash A$ ($A$ is derivable without open assumptions) as "$A$ is true" without causing any harm.

  • Accordingly, I interpret "If $\vdash A$ then $\vdash B$" as "If $A$ is true, then $B$ is true", which is the same as $\vdash A\to B$.

I could just accept that I have to be more careful, but I was hoping that someone could comment on this and give me some additional insight.

$^1$ For example, consider $B=\forall _x A$: $$\text{If }\vdash A\text{, then }\vdash\forall_xA$$ is always true according to the rules of natural deduction, but of course $$\vdash A\to\forall_xA$$ can only be derived if $x$ is not a free variable of $A$ (otherwise we could derive absurd formulas).

Filippo
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    Intuition-wise I would recommend only using "true" when you're working with a specific model or at worst a complete theory. "$\vdash A$" should be interpreted as something like "$A$ is obviously true," and this is a much stricter condition; for example, intuitively everything is either true or false but plenty of things are not "obviously true" or "obviously false," and this corresponds to the behavior of disjunction in this context (see also my answer here). – Noah Schweber Nov 29 '21 at 17:08
  • Regarding your intuition, if $A$ is false, then the meta-conditional "if $A$ is true, then $B$ is true" is true... The same for $\vdash$. – Mauro ALLEGRANZA Nov 30 '21 at 07:07
  • @MauroALLEGRANZA What do you mean by the meta-conditional? – Filippo Nov 30 '21 at 14:57
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    That the proposition is "if..., then ..." in the meta-language and not $\to$ of the object language. But the truth conditions for the two are the same. – Mauro ALLEGRANZA Nov 30 '21 at 14:59
  • @NoahSchweber Thank you for the comment. Unfortunately, I am missing some mathematical background to understand the answer you gave the link to. Do you think that the answer by Bram28 is based on the same idea or were you focusing on another aspect? – Filippo Dec 01 '21 at 07:17

1 Answers1

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I thought that I can interpret $\vdash A$ ($A$ is derivable without open assumptions) as "$A$ is true" without causing any harm.

Sorry, but no. $\vdash A$ is a much stronger statement than simply "$A$ is true". As you point out, it means that $A$ can be derived without any assumptions, and that shows that $A$ isn't just True, but that $A$ is always true: $A$ is a tautology!

Consider the statement "It rains". This statement could be true or false, depending on where and when (basically, in what world) you evaluate it.

This is quite different from a statement like "it rains or it doesn't rain": that statement is true no matter what, and we call it a tautology.

In propositional logic, the first statement would be like $P$, and we cannot prove $P$ from no assumptions whatsovever. But the second statement is of the form $P \lor \neg P$, and that statement we can prove without any further assumptions. That is, where $P$ is an atomic statement, we have $\not \vdash P$, but we do have $\vdash P \lor \neg P$

If for your statement $A$ it is the case that $\vdash A$, then $A$ is like the second statement, not the first.

Bram28
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  • Thank you for your answer! Unfortunately, I don't understand the last sentence. – Filippo Nov 30 '21 at 19:55
  • @Filippo Sorry for not being clear! I meant that if $\vdash A$, then that means that $A$ is a tautology, i.e. it is something like $P \lor \neg P$, or something like $a=a$, rather than something like "It rains" – Bram28 Nov 30 '21 at 20:20
  • I see, thanks! But doesn't it depend on my axioms what a tautology is and what not? For example, I recently learned that $P\lor\lnot P$ is not derivable (without open assumptions) in minimal logic. – Filippo Nov 30 '21 at 21:25
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    @Filippo Ah, sure. I was using classical logic, but yes, in other logics it might not be. Still, you need to distinguish between what is logically true (which may indeed depend on what specific logic you are using, but nevertheless is something that that logic considers always to be true), and what is contingently true (something that, given whatever logic you are using, could be true or false) – Bram28 Nov 30 '21 at 21:45