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In Tourlakis' Mathematical Logic, he claims that $\models A $ if and only if $\emptyset \models A$. This question is on page 36. The first statement implies the second is correct but the converse is incorrect.

A counterexample would be any contradiction say $A \land \neg A$.

Am I missing something ?

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  • $\begingroup$ What's the definition of $\vDash A$ given in the book? I'm only familiar with the symbol as a relation. $\endgroup$
    – jgon
    Dec 1, 2018 at 17:47
  • $\begingroup$ 1st Def) If for every state $A$ is true then $\models A$($A$ is a tautology). 2nd Def) If the state is true for every element in $\Gamma$ then $A$ is also true($\Gamma$ tautologically implies $A$). $\endgroup$
    – Nameless
    Dec 1, 2018 at 17:52
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    $\begingroup$ I don't understand the first definition you wrote. Can you reword it? To me, $\models A$ is simply an abbreviation of $\varnothing \models A$, so they mean the same by definition. $\endgroup$
    – Git Gud
    Dec 1, 2018 at 18:44
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    $\begingroup$ Your counter example doesn't work. Let $\varphi$ be a contradiction. By definition $\varnothing\models \varphi$ means that for every valuation $v$, it holds that $\forall \psi\in \varnothing\left(v(\psi)=T\right)\implies v(\varphi)=T$. The antecedent holds vacuously, but the consequent doesn't. $\endgroup$
    – Git Gud
    Dec 1, 2018 at 19:10
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    $\begingroup$ See the post Problems with using validity symbol ⊨ “vacuously” as well as the post The logical consequence of an empty set of premises. $\endgroup$ Dec 1, 2018 at 20:19

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I think I see your error. I'm not quite sure what you mean by state, but let me try to informally translate the definitions you gave to symbols.

$\models A$ means that $$\forall_{states} A,$$ whereas $\Gamma\models A$ for a collection of statements $\Gamma$ means $$\forall_{states}\left(\left(\forall_{\psi\in\Gamma}\psi\right)\implies A\right).$$

Note the careful parentheses in this second definition, since I'm fairly sure the error is one of misinterpreting the grouping of the quantifiers and symbols here. If I now put $\Gamma=\varnothing$, then I have $\forall_{\psi\in\Gamma}\psi$ becomes vacuously true, or in other words $$\varnothing\models A$$ means that $$\forall_{states} \mathrm{True}\implies A,$$ or $$\forall_{states} A,$$ since $$\mathrm{True}\implies A\text{ if and only if }A.$$

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  • $\begingroup$ I believe "state" in the comment was meant to be "statement." $\endgroup$ Dec 1, 2018 at 18:33
  • $\begingroup$ @FabioSomenzi That was my first thought as well, but "If for every statement $A$ is true" doesn't make very much sense either. $\endgroup$
    – jgon
    Dec 1, 2018 at 18:38
  • $\begingroup$ @FabioSomenzi - a "state" is a truth assignment; see page 26 : "Def 1.3.2. A state $v$ is a function that assigns the value $\text f$ or $\text t$ to each Boolean variable, while it assigns necessarily the value $\text f$ to the constant $\bot$ and necessarily the value $\text t$ to the constant $\top$." $\endgroup$ Dec 1, 2018 at 20:17
  • $\begingroup$ @MauroALLEGRANZA Duly noted. Thanks! $\endgroup$ Dec 2, 2018 at 0:04

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