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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?

user525966
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    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 '18 at 00:56
  • @CarlMummert Is "contradiction" same as false, $\bot$, etc? What do you mean by unsatisfiable? – user525966 Sep 24 '18 at 01:04
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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 '18 at 01:08
  • @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 '18 at 01:12
  • Yes, in the case of propositional logic, that is all that a model will be, and your characterization of "satisfies" is right. The model could also be called an interpretation. – Carl Mummert Sep 24 '18 at 01:13
  • @CarlMummert So $p=\top, q=\bot$ would be a model/interpretation of $\Gamma = { (p \to (q \to p))}$, and we'd say $\Gamma$ is satisfiable under this model since $\top \to (\bot \to \top)$ evaluates to $\top$? – user525966 Sep 24 '18 at 01:14
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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 '18 at 01:15
  • See definition of (strong) soundness. – Mauro ALLEGRANZA Sep 24 '18 at 06:22
  • "$\varphi$ is a semantic consequence of $\Gamma$, which I believe is saying $\varphi$ is "true"." NO; it means : "$\varphi$ is TRUE in every interpretation where $\Gamma$ is". – Mauro ALLEGRANZA Sep 24 '18 at 09:28
  • See this post for a simple proof of the consistency of the propositional calculus. Being consistent, we cannot have $\vdash \bot$. Thus, if for some set of assumptions $\Gamma$, we have that $\Gamma \vdash \bot$, it follows from soundness that there is no interpretation where all formulas in $\Gamma$ are TRUE. This means that $\Gamma$ is unsatisfiable (or inconsistent). – Mauro ALLEGRANZA Sep 24 '18 at 12:20
  • @MauroALLEGRANZA So is soundness sort of like how $p \to q$ is defined? $q$ is true whenever $p$ is true, and if $p$ is false, we say $p \to q$ is vacuously true? – user525966 Sep 24 '18 at 12:55

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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 $\Gamma$.

spaceisdarkgreen
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  • 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 '18 at 01:01
  • @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 '18 at 01:03
  • 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 '18 at 01:05
  • 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 '18 at 01:09
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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 '18 at 01:10
  • @user525966 Remember $A\models \phi$ is a statement about all interpretations. So something in $A$ being false doesn't really matter (cause false in what interpretation?). However if a sentence in $A$ is false in all interpretations, then, vacuously, $A\models \phi$ for any $\phi.$ This corresponds to the principle of explosion. – spaceisdarkgreen Sep 24 '18 at 01:13
  • @user525966 I think these two issues might be related. "$\phi$ is false" is different from "$\phi$ is $\bot$." The first means that $\phi$ is false (in some interpretation that should be specified or clear from context). The second means $\phi$ is the sentence $\bot,$ which is a special sentence that is false under all interpretations. – spaceisdarkgreen Sep 24 '18 at 01:20
  • For example let's say I have $\Gamma = { \bot, (p \to (q \to p)) }$. I can prove that $\Gamma \vdash p \to (q \to p)$ and that under all interpretations, this proposition is true. But I can also show that $\Gamma \vdash \bot$, which is false under all interpretations. I don't understand then how this vacuously means we can still say $\Gamma \vDash \varphi$. You say it "doesn't really matter" but I don't know what this means. Is semantic consequence something we can show in other ways? – user525966 Sep 24 '18 at 01:26
  • @user525966 Do you mean to use $\models$ rather than $\vdash$ in the above? – spaceisdarkgreen Sep 24 '18 at 01:28
  • Or, is my definition of semantic consequence wrong and more accurately put, $\Gamma \vDash \varphi$ holds if $\varphi$ is true under every interpretation in which $\Gamma$ is satisfied (i.e. when all the propositions in $\Gamma$ are true in some model), and then since $\bot$ is always false, and therefore $\Gamma$ is never satisfied, semantic consequence is vacuously true because $\varphi$ is true in all satisfied interpretations (all $0$ of them) – user525966 Sep 24 '18 at 01:30
  • No I indeed mean $\vdash$ in my earlier comment, i.e. we can prove $\bot$ syntactically, but I didn't understand why this means it must follow semantically. But I guess we say semantic consequence holds when $\varphi$ is true in all satisfied interpretations of $\Gamma$? i.e. whenever the lefthand side is all true, the righthand side is true? – user525966 Sep 24 '18 at 01:30
  • @user525966 What you say after "more accurately put..." is mostly right. $\Gamma \models \phi$ means $\phi$ is true in all interpretations that satisfy $\Gamma.$ An interpretation satisfies $\Gamma$ iff all sentences of $\Gamma$ are true in this interpretation. In this example, since $\bot\in \Gamma$ and $\bot$ is false in all interpretations, there are no interpretations that satisfy $\Gamma.$ $\phi$ is true in all zero interpretations that satisfy $\Gamma,$ so $\Gamma\models \phi.$ – spaceisdarkgreen Sep 24 '18 at 01:34
  • Why would this be a more helpful definition as opposed to requiring that all propositions of $\Gamma$ be true in all models? Why would we want to allow for semantic consequence to hold when there are falsehoods in $\Gamma$? What happens if we decide to disallow this? – user525966 Sep 24 '18 at 01:38
  • Let us continue this discussion in chat. – spaceisdarkgreen Sep 24 '18 at 01:39