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I'm trying to understand if the sentence $\square\bot\land \phi$ is consistent in KD. I don't think it is true because it looks like no serial model where this sentence is satisfiable exists. As I understand it, to prove it is not consistent, I must provide a formal proof of $\neg(\square\bot\land \phi)$ in KD. But I failed to do that. Are there other ways of proving inconsistency? (I don't want to appeal to completeness.)


marked as duplicate by Alex Kruckman, Community Mar 11 at 17:27

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    $\begingroup$ What is the system KD? (If you want a purely syntactic proof, it's not enough to just say that it corresponds to the class of serial models.) Incidentally, I don't see what role the "$\phi$" is playing here - if you can give a KD-proof of $\neg\Box \perp$ surely you can give one of $\neg(\Box\perp\wedge\phi)$ (just because $\neg a$ gives $\neg(a\wedge b)$ always). $\endgroup$ – Noah Schweber Mar 11 at 16:57
  • $\begingroup$ (I've just noticed that you asked the exact same question yesterday, and you received an answer which is essentially the same as mine. Please don't duplicate questions on this site.) $\endgroup$ – Alex Kruckman Mar 11 at 17:23
  • $\begingroup$ @AlexKruckman I received that answer after I posted this question. And the reason I posted this question is to find out whether there is another way to prove inconsistency (and/or to make sure that what I asked in the previous question is indeed what I need to do). $\endgroup$ – user643175 Mar 11 at 17:27
  • $\begingroup$ Ah, I see. I misunderstood the point of you question. I've edited my answer. $\endgroup$ – Alex Kruckman Mar 11 at 17:30

There are essentially two ways to prove that a sentence $\varphi$ is inconsistent: Provide a proof of $\lnot \varphi$, or (assuming you know a completeness theorem for your logic) show that $\varphi$ does not hold in any models. But since you specified in the question that you don't want to appeal to completeness, providing a proof is really your only option.

As Noah points out in the comments, it suffices to prove $\lnot \square \bot$, because $\lnot p$ entails $\lnot (p\land q)$ by propositional logic. But $\lnot \square \bot$ is equivalent to $\lozenge \top$, and this is easy to prove!

Start with $\top$. By necessitation, $\square \top$. By D (which is $\square p\rightarrow \lozenge p$), $\lozenge \top$.

  • $\begingroup$ Regarding the first part of your answer: can I do like this? To show that $\phi$ is inconsistent, we need to show $\vdash \neg \phi$. Assume the converse i.e. that $\vdash\phi$. By soundness, $\models \phi$. If I can show that this cannot be true (as in my last question as of now), doesn't this show that $\phi$ is inconsistent? And in such a proof I didn't provide an explicit proof of $\neg \phi$. $\endgroup$ – user643175 Mar 22 at 1:11
  • $\begingroup$ $\vdash \phi$ is not the negation of $\vdash \lnot\phi$! @user643175 $\endgroup$ – Alex Kruckman Mar 22 at 1:12
  • $\begingroup$ Is $\not \vdash \phi$ the negation of $\vdash \phi$? I thought it is, and combining it with the fact that $\not\vdash \psi$ means $\vdash\neg \psi$, we get that $\not \vdash \neg \phi$ is $\vdash \neg \neg \phi$, which is equivalent by propositional logic to $\vdash \phi$. Where am I wrong? $\endgroup$ – user643175 Mar 22 at 1:16
  • $\begingroup$ $\not\vdash\psi$ does not mean $\vdash \lnot \psi$! Just because something isn't provable doesn't mean you have a proof of its negation! Consider the case when $\psi$ is a proposition letter $p$. $\endgroup$ – Alex Kruckman Mar 22 at 1:21

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