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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 ?

Nameless
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1 Answers1

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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 $$\mathrm{True}\implies A\text{ if and only if }A.$$

jgon
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  • I believe "state" in the comment was meant to be "statement." – Fabio Somenzi Dec 01 '18 at 18:33
  • @FabioSomenzi That was my first thought as well, but "If for every statement $A$ is true" doesn't make very much sense either. – jgon Dec 01 '18 at 18:38
  • @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 $\top$." – Mauro ALLEGRANZA Dec 01 '18 at 20:17
  • @MauroALLEGRANZA Duly noted. Thanks! – Fabio Somenzi Dec 02 '18 at 00:04