The interpretation of $\nvdash$

Question

I have a question about the use of $\nvdash$. $\nvdash$ is commonly used as a meta-level symbol.

Let $A\vdash\perp$, by the deduction theorem, we reach $\vdash A\rightarrow\perp$, which is equivalent to $\vdash\neg A$ (that $A$ is false).

My question is: what does it mean when we write $A\nvdash\perp$? If I interpret it as saying that $\perp$ does not follow from $A$, it seems to be equivalent to saying that no contradiction follows from $A$. But that suggests that $A$ is true, or at least not false. But if that's the case, there is no difference between $A\nvdash\perp$ and $\vdash\neg(A\rightarrow\perp)$. So, I must be wrong.

Then, by stating $A\nvdash\perp$, it should be possible for $A$ to be undecidable, but that's not trivial to me. Can anyone explain it briefly? Thanks!

Answer 1

You're exactly right - $A \not \vdash \perp$ means that $A$ is consistent. That is, we cannot derive a contradiction using $A$ as hypotheses. But there is a difference between "we can't prove it false" and "it's true"!

Here's an easy example. Let's work with the theory of groups. Then

$$ xy = yx \not \vdash \perp$$

Why is this? Because if $xy = yx \vdash \perp$, that would mean no groups could satisfy $xy = yx$ (as the extra axiom would be inconsistent). Of course, abelian groups exist, and the claim follows.

However, this does not mean that $xy = yx$ is true! We have only shown that $xy=yx$ is not always false.

This is exactly what "undecidable" properties are. It's nothing complicated or scary. We have a theory and we have some models of that theory (in this case groups). "Undecidability" of $\varphi$ just means that, looking only at the theory, you can't decide if $\varphi$ is true in every model. And that happens exactly because the opinion of different models differs. This all comes back to the completeness theorem:

Provability is the same thing as truth in all models

The reason people (myself included) get confused by undecidable properties is because we typically only discuss them in settings where we have a particular model in mind. Imagine if $\mathbb{Z}$ was the only group we ever worked with. Then it might be hard to conceive of "$\mathbb{Z}$s" (by this I mean groups) where $xy=yx$ fails. Or moreover, that there might be facts about "$\mathbb{Z}$" that we cannot prove from the group axioms!

But this is exactly what happens with arithmetic. We have a distinguished model of PA, ZFC, etc. in mind when we talk about the "theory" as a whole. A big part of logic is learning to divorce your opinion about the one "real" model from all the others. Once you do that, though, it gets easier

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I hope this helps ^_^

Answer 2

You are right: There should be a difference between $A \nvdash \bot$ and $\vdash \neg (A \to \bot)$, and there is.

As you note, $A \to \bot$ is logically equivalent to/taken to be the meaning of the abbreviation $\neg A$, so $\neg(A \to \bot)$ is equivalent to $\neg \neg A$, and this is in turn equivalent to (and from which is derivable) $A$. That is, $\vdash \neg(A \to \bot)$ entails (assuming semantic completeness of the system) $\vdash A$.

$A \vdash \bot$ would mean that a contradiction can be derived from $A$; assuming soundness that means that $A$ is contradictory, i.e. false under all interpretations. $A \nvdash \bot$ says that this is not the case and thus means that $A$ is satisfiable, i.e. not false under all interpretations, i.e. true under at least one interpretation.
If $A$ were true under all interpretations, it would be tautological, and assuming completeness, it should be derivable: $\vdash A$, and with the above equivalene, $\vdash \neg(A \to \bot)$.

The quantification over interpretations is important here; it does not make sense to say that A is "true" without specifying under which interpretation, becauase truth is only defined relative to interpretations (interpretation = valuation functions in the case of propositional logic, and structures with domain and interpretation function in predicate logic).

However, $A$ being satisfiable (= "not always false") does not imply that $A$ is tautological (= "always true"): It may be true under some but not all interpretations. Hence, again taking the proof system to be sound, from the non-derivability of a contradiction from $A$ we can not infer derivability of the negation of $A \to \bot$ (which would entail derivability of $A$):
$A \nvdash \bot \ \nRightarrow \ \vdash \neg(A \to \bot)$, although the two don't exclude each other ($A$ may be derivable, in which case also no contradiction can be derived from it).

If both $A \nvdash \bot$ (i.e. $\nvdash \neg A$) and $\nvdash \neg(A \to \bot)$ (i.e. $\nvdash A$), then we do indeed have the case that $A$ is undecidable or independent of the theory, and the proof system is syntactically incomplete.