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What do the following two notations mean:

$$A \models$$

$$A \vdash$$

The first one would be saying that A is a contradiction? The second one that A is not a theorem? That doesn't sound right to me.

SlowerPhoton
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2 Answers2

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The notation $ A \models \, $ means that the formula $A$ is not satisfiable, i.e. there is no structure (or assignment in propositional logic) that makes $A$ true. This is a semantical notion.

The notation $A \vdash \, $ means that from the formula $A$ you can derive everything (i.e. any other formula), according to some derivation rules already defined. This is a syntactical notion.

An important theorem in propositional and first-order logic (completeness and soundness) states that the two notions coincide: a formula is unsatisfiable if and only if everything is derivable from it, i.e. $A \models \,$ if and only if $A \vdash\,$.

Because of this equivalence, in the literature you can find some ambiguous terminology. A formula is said contradictory or inconsistent if $A \models \,$ in some textbooks, or if $A \vdash \, $ in other textbooks.

The notation used to express that $A$ is not derivable (i.e. it is not a theorem in the considered derivation system) is $\not\vdash A$. This is consistent with the notation $\vdash A$, which says that the formula $A$ is derivable, i.e. that it is a theorem in the considered derivation system. Note that $\not \vdash A$ does not mean $A \vdash\,$: a formula could be non-derivable but still satisfiable.

For the sake of completeness, the notation $\models A$ means that the formula $A$ is valid (a tautology in propositional logic), i.e. every structure makes $A$ true. Again, the notation $\not \models A$ means that $A$ is not valid, i.e. there are some structures that makes $A$ false. Note that $\not \models A$ does not mean $A \models \,$: a formula could be non-valid but still satisfiable.

According to the aforementioned completeness and soundness theorem (in propositional and first-order logic), the notions of validity and derivability coincide: $ \models A$ if and only if $\vdash A$.

Taroccoesbrocco
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  • Regarding $A\vdash$, I wouldn't say it to mean that "you can derive everything" because I think that would implicitly assume the Principle of Exclusion. –  Apr 21 '20 at 05:12
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    @user170039 - Right, thank you for the comment, but I think that this kind of clarifications are off-topic here. The principle of explosion is accepted in classical logic but also in constructive logics such as intuitionistic logic; only in minimal logic it does not hold. The OP clearly refers to a context where the principle of explosion holds, since it uses the word "contradiction". – Taroccoesbrocco Apr 21 '20 at 06:18
  • There is also Paraconsistent Logic where it doesn't hold cf. this but where the talk of contradiction makes sense nevertheless. –  Apr 21 '20 at 06:56
  • I'm not sure to agree... :_) Can you provide some good textbook with a ref to $A \vdash$ ? (except for textbook about sequent calclus, where $A \to$ is used) ? IMO, we have to start from the "canonical" $\Gamma \vdash A$ where $\Gamma$ is a set of formulas (assumptions, axioms) and $A$ a formula, that means that $A$ is derivable from assumptions in $\Gamma$. Then we have $\vdash A$, that is a shorthand for $\emptyset \vdash A$, meaning that $A$ is provable with no assumptions at all (matching with the semantical notion of valid formula, i.e. an "unconditionally" true formula). – Mauro ALLEGRANZA Apr 21 '20 at 07:30
  • If so, following the above syntax, what $A \vdash$ stand for ? At the right of the $\vdash$ symbol we have to imagine $\emptyset$ ? But to say that from $A$ we cannot derive anything is nosense (at least $A \vdash A$ is reasonable)... The canonical syntax has a formula to the right of $\vdash$: thus, what we mean is the "empty fomula" ? – Mauro ALLEGRANZA Apr 21 '20 at 07:32
  • Conclusion : the correct way to say that from $A$ we can derive a contradiction is: $A \vdash \bot$. – Mauro ALLEGRANZA Apr 21 '20 at 07:34
  • @MauroALLEGRANZA - Thank you for you comment, but I disagree. In proof theory, especially in classical sequent calculus, it is natural to consider sequents of the form $\Gamma \vdash \Delta$ where $\Gamma$ and $\Delta$ are finite sets (or multisets or sequences) of formulas, meaning that there is a derivation from the conjunction of the formulas in $\Gamma$ to the disjunction of the formulas in $\Delta$. So, The notation $A \vdash , $ stands for the particular particular case where $\Gamma ={A}$ and $\Delta = \emptyset$. – Taroccoesbrocco Apr 21 '20 at 08:09
  • @MauroALLEGRANZA - A couple of important references where this notation is used are Troelstra, Schwichtenberg, Basic Proof Theory (Ch. 3) and Negri, von Plato, Structural Proof Theory (Ch. 3), in particular for the rule $\bot_L$ in classical sequent calculus. The fact that both use $\Rightarrow$ for the sequent turnstile is not relevant because then $\Gamma \vdash \Delta$ means that the sequent $\Gamma \Rightarrow \Delta$ is derivable. – Taroccoesbrocco Apr 21 '20 at 08:14
  • See my comment above: except for textbook about sequent calclus... – Mauro ALLEGRANZA Apr 21 '20 at 08:18
  • @MauroALLEGRANZA - Sorry, I don't understand your point. OP doesn't provide information about the context of their question but I guess they met the notation $A \vdash ,$ in some textbook, so in my opinion the point here is to explain the meaning of that notation, and not to change the notation they use (at least this is not the first aim, in my opinion). Note also that many textbooks (unfortunately, in my opinion) do not introduce the symbol $\bot$ in the language. We don't know if this is the case for OP's reference textbook. – Taroccoesbrocco Apr 21 '20 at 08:44
  • @MauroALLEGRANZA - Besides I don't understand when you say "what $A \vdash ,$ stand for ? At the right of the $\vdash$ symbol we have to imagine $\emptyset$? But to say that from $A$ we cannot derive anything is nosense". The disjunction of the empty set of formulas is not nothing, is any unsatisfiable formula, from which you can derive every formula (by the principle of explosion). The notation $A \vdash ,$ is not a peculiarity of sequent calculus, and it is consistent with other more common notations. – Taroccoesbrocco Apr 21 '20 at 09:19
  • @MauroALLEGRANZA - By the way, your remarks are interesting and maybe they deserve a separate question, also to see if there are opinions different from mine and yours. – Taroccoesbrocco Apr 21 '20 at 09:20
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For the first one, yes, $A \vDash$ is often used as shorthand for $A \vDash \bot$, i.e. that $A$ is a contradiction.

I have not seen $A \vdash$ ... though I suppose one could likewise use it for $A \vdash \bot$, i.e. that a contradiction can be syntactically derived from $A$ which, assuming the derivational system under consideration is sound, would imply that $A$ is a contradiction.

Note that if we assume we are dealing with a sound derivational system, a statement being a contradiction would imply that it is not a theorem of the derivational system. But the other way around does not. That is, some statement not being a theorem does not mean it is a contradiciton. For example, for any atomic proposition $A$ we have that $A$ is not a theorem, but obviously $A$ is not a contradiction either, as it is a contingency. So, I would never use $A \vdash$ to mean that $A$ is not a theorem. Indeed, to say that $A$ is not a theorem you'd typically do $\not \vdash A$. So, your comment that it didn't sound right to interpreting $A \vdash$ as $A$ not being a theorem, was spot on.

But again, I suppose one could use $A \vdash$ to indicate that a contradiction can be derived from $A$ (especially if $\bot$ is not a proper symbol of the language you are using). And, if you have a complete derivational system, that would also imply that any statement can be derived from $A$.

Bram28
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