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Soundness property, to my knowledge is the property that:

$\Gamma \vdash \varphi \implies \Gamma \vDash \varphi$

If $\varphi$ is provable (a syntactic consequence) from $\Gamma$ then $\varphi$ is also a semantic consequence of $\varphi$, which I believe is saying $\varphi$ is "true".

But what if, for example, $\varphi \in \Gamma$ and $\varphi = \bot$? It appears conceivable that some of the assumptions in $\Gamma$ are false, and then we might be able to prove things from it, but semantically they would be false.

It's possible I just have the definition of soundness wrong but how is this accounted for? We would normally say that the Hilbert system is both complete and sound but is this still the case even if we begin with a $\Gamma$ that contains some false premises? Or is it "sound only in certain cases"? How does this work?

marked as duplicate by Mauro ALLEGRANZA logic Sep 24 at 6:24

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  • 3
    If $\Gamma$ contains a contradiction, then $\Gamma$ is unsatisfiable, so $\Gamma \vDash \psi$ is vacuously true. The interesting and nontrivial case is when $\Gamma$ is consistent. – Carl Mummert Sep 24 at 0:56
  • @CarlMummert Is "contradiction" same as false, $\bot$, etc? What do you mean by unsatisfiable? – user525966 Sep 24 at 1:04
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    I mean that no model satisfies $\bot$, so a theory that contains $\bot$ is not satisfiable. In that case, trivially "every model of the theory will also be a model of $\phi$", because the theory has no models. For example, for any $\Gamma$ we have $\Gamma \vdash \bot \Longrightarrow \Gamma \vDash \bot$. – Carl Mummert Sep 24 at 1:08
  • @CarlMummert Is a "model" a particular set of boolean inputs to the atomic variables of $\Gamma$? Does "satisfies" mean "everything in $\Gamma$ evaluates to true under a specific model"? – user525966 Sep 24 at 1:12
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    Yes. We would say $\Gamma$ is satisfiable because there is at least one model/interpretation that makes every formula in $\Gamma$ true. – Carl Mummert Sep 24 at 1:15
up vote 8 down vote accepted

If $\bot\in \Gamma,$ then we have $\Gamma\vdash \bot$ and $\Gamma \models \bot,$ so this is not in conflict with the soundness theorem. It is clear that $\Gamma \vdash \bot.$ The reason $\Gamma \models \bot$ is that, since $\bot\in \Gamma,$ there are no interpretations in which all the sentences in $\Gamma$ hold, i.e. no interpretations satisfying $\Gamma$. Hence, vacuously, $\bot$ holds in every interpretation satisfying $\Gamma$.

  • I don't understand... I thought semantic consequence was the idea that if $A \vDash \varphi$ then it means under every interpretation of the contents of $A$ (i.e. all combinations of true/false among all the atomic propositional variables of all the propositions in $A$), we have $\varphi = \top$. Is this wrong? – user525966 Sep 24 at 1:01
  • @user525966 $A\models \phi$ means that under every interpretation in which all sentences in $A$ are true, $\phi$ is true. (Also, I would be careful with the notation $\phi=\top.$ This seems to suggest $\phi$ is the sentence $\top,$ not that $\phi$ is true in a given interpretation.) – spaceisdarkgreen Sep 24 at 1:03
  • I meant it the same way as you when you say "$\varphi$ is true" but I guess I have to use English to represent that and not $=$? Is there some other way to say that $\varphi$ "is" true or "evaluates" to true? $v(\varphi) = \top$ for truth functional $v$ or however it's phrased? – user525966 Sep 24 at 1:05
  • Can I think of $A \vDash \varphi$ as $p \to q$ sort of? Whenever everything in $A$ is true, then so is $\varphi$? And whenever something in $A$ is false, $\varphi$ is vacuously true? Only way for $A \vDash \varphi$ to not hold is if everything in $A$ is true but $\varphi$ is somehow false? – user525966 Sep 24 at 1:09
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    @user525966 Yes, I figured that was what you meant. Yes, it would probably be best to just say "$\phi$ is true under the interpretation" in English. (There are multiple ways to represent it in notation. Most common is if $\mathcal M$ denotes the interpretation, we write $\mathcal M \models \phi,$ in what is perhaps an annoying overloading of the "$\models$" symbol.) – spaceisdarkgreen Sep 24 at 1:10

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