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I'm searching for an example proof of the above formula as distribution over and in basic proportional modal logic, I.e. in some of the standard systems K, S4 or S5.

blub
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  • See for example the Blackburn, de Rijke, Venama book "Modal logic". There you can find the hard direction as example 1.40 (at least in the version that I have; at any rate it's in chapter one). The other direction is very easy, it's just monotonicity: $p\land q\to p$ is a tautology, so by necessitation $\Box (p\land q\to p)$ which by K yields $\Box (p\land q)\to \Box p$. You can similarly show that $\Box(p\land q)\to\Box q$ and from this you conclude that $\Box(p\land q)\to\Box p\land \Box q$. – Apostolos May 18 '17 at 03:05
  • Sadly, I don't have access to this book but I'm trying to get it. Thanks – blub May 18 '17 at 06:50
  • I would copy the answer here, but now it's closed. I voted to reopen. Have you found the book? – Apostolos May 19 '17 at 14:54
  • Online to buy, yes but I hadn't bought it yet. Would be extremely helpful if you could post it. – blub May 19 '17 at 15:09
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    I will write it here in the comments until it's reopened. $$1.\vdash p\to(q\to p\land q)$$ $$2. \vdash\Box(p\to(q\to p\land q))$$ $$3. \vdash\Box p\to \Box(q\to p\land q)$$ $$4. \vdash\Box p\to (\Box q\to \Box(p\land q))$$ $$5.\vdash\Box p\land \Box q\to\Box(p\land q)$$ The first step is a tautology, the second is necessitation, the third is K, the forth is K and the fifth is Currying https://en.wikipedia.org/wiki/Currying#Logic (so it's basic propositional logic). – Apostolos May 19 '17 at 15:19

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