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I have read in some lecture notes that "pure classical modal logic without any additional conditions is nothing more than propositional logic". I cannot really see what this means and unfortunately the context is not helpful. Any guess?

Here is the link to the lecture notes: https://www.sciencedirect.com/topics/computer-science/classical-modal-logic

Mijito
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    I hope someone knows. Not only do I have no idea why the one is nothing more than the other, I have no idea what "pure classical modal logic" might be. There are roughly 1.1 zillion classical systems of modal logic (cf "An essay in classical modal logic") - I wonder which is the "pure" one... – David C. Ullrich Aug 21 '19 at 20:48
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    Can you link to the lecture notes? They might have more context which would help. – Noah Schweber Aug 21 '19 at 20:50
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    Here it is: https://www.sciencedirect.com/topics/computer-science/classical-modal-logic – Mijito Aug 23 '19 at 13:40

2 Answers2

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Barring additional context, I cannot see a sense in which this is true.

The key point here is that in contrast with propositional logic, modal logic has non-truth-functional operations. Any "propositional term" - that is, any function built out of the usual Boolean operations - is truth-functional, in the sense that the truth-value of the output is determined entirely by the truth-value of the inputs. But this is not the case in modal logic, specifically for $\Box$ and $\Diamond$ - and indeed this is kind of the point of modal logic in general!

(There are various ways to make this more precise. One is the following: for any $n$-ary function $f$ which is a composition of $\wedge$ and $\neg$, the sentence $$\neg[p_1\wedge p_2\wedge ...\wedge p_n\wedge q_1\wedge q_2\wedge ... \wedge q_n\wedge f(p_1,..., p_n)\wedge \neg f(q_1,..., q_n)],$$ where $p_1,..., p_n, q_1,..., q_n$ are propositional variables, is a validity, i.e. is provable from the empty theory, or is true in every model.)

This fails for the modalities: the sentence $$\neg (p\wedge q\wedge \Box p\wedge \neg\Box q)$$ is not a validity in most modal systems. This is a fundamental difference, and I don't see any way around this. Now, perhaps the author is using a very general notion of equivalence, or takes "pure classical modal logic" to be a system which collapses the modalities, but barring further evidence I would say that their claim is fundamentally wrong.

Now there is a "modal/classical connection" - in a precise sense, modal logic (or more precisely, all the usual systems of modal logic) can be identified with a fragment of classical first-order logic - see e.g. this old answer of mine for a special case. But that's quite different.

Noah Schweber
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I think a possible explanation is that the accessibility relation between states for Intuitionistic Logic is reflexive and transitive, which corresponds to S4 Modal Logic. If one adds symmetry to the accessibility relation for Intuitionism, then the resulting logic is Classical Logic; in like manner, adding symmetry to S4’s accessibility relation yields S5. So, there is a translation from Classical Logic to S5. Of course, this all assumes Intuitionistic semantics, since Classical Logic works fine without worlds/states or accessibility relations.

PW_246
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