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What is the reason for taking the following as the axiom in normal modal logics: □(p → q) → (□p → □q)? What is special about this formula that makes it an axiom? Does it express some kind of tautology? For example, the following is an axiom of propositional logic and is clearly a tautology: (p → (q → s)) → ((p → q) → (p → s)). I wonder if there are similar justifications for the distribution axiom as well.

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    It holds in every Kripke model. – Berci Mar 11 '20 at 01:40
  • How can one prove that it holds in every Kripke model? –  Mar 13 '20 at 14:07
  • It's maybe not the right question to ask. Nevertheless, assume $w\Vdash \box(p\to q)$ and $w\Vdash \box p$. Then in every world $u$ that $w$ sees (in the given model), we have $u\Vdash p\to q$ and also $u\Vdash p$, hence it follows that $u\Vdash q$, and since it applies to all such $u$, we get $w\Vdash \box q$. – Berci Mar 13 '20 at 14:12
  • @Berci, would you please explain your proof to me? I still do not see how this justifies that we should take the formula as an axiom. –  Mar 14 '20 at 11:57
  • There might be other kind of models for (not necessarily normal) modal logics, and it might not hold there. But as long as we choose Kripke semantics, in a complete logic, this formula should be provable, as it is semantically valid. The easiest way to make it provable is to include it (or an equivalent) as an axiom. – Berci Mar 14 '20 at 14:14

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Yes, it is valid in every Kripke model. So, if we want our deduction system to be complete with respect to Kripke semantics, it needs to be derivable, and for that we can simply put it as an axiom.

To prove it holds in an arbitrary Kripke model $(W, R, \Vdash)$, we use that $w\Vdash (a\to b)$ for a world $w\in W$ iff whenever $w\Vdash a$, we also have $w\Vdash b$. (In other words, $w\not\Vdash (a\to b)\ $ iff $\ w\Vdash a$ and $w\not\Vdash b$.)
Now assume $\def\box{\Box} w\Vdash \box(p\to q)$, then we want $w\Vdash \box p\to\box q$, so assume also $w\Vdash \box p$.

Using Kripke semantics for $\box$, these mean that for all worlds $w'\in W$ that is seen from $w$ (i.e. $w\, R\, w'$) we have both $w'\Vdash p\to q$ and $w'\Vdash p$.
But then it follows that $w'\Vdash q$, and as this holds for all $w'$ which satisfy $w\, R\, w'$, we get $w\Vdash \box q$.

Berci
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