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I shall first post a couple of screenshots to make it clear what I'm talking about.

enter image description here enter image description here

I am reading A New Introduction to Modal Logic by Hughes and Cresswell. They answer the question in title very explicitly here, but I want to know why this is the case:

enter image description here

But how come the rule of necessiation isn't derivable? The necessiation rule states that if a is a theorem, then so is □a. I will offer a proof and would greatly appreciate it if anyone could point out my mistake.

1.  ⊢ a                    Given
2.  ⊢ (p => p) <=> a        PC
3.  ⊢ □((p & q) => p)       AS1.2
4.  ⊢ □(-(p & q) v p)       PC
5.  ⊢ □(-p v -q v p)        PC
6.  ⊢ (-p v -q v p) <=> ((p => p) v -q)     PC
7.  ⊢ ((p => p) v -q) <=> (p => p)  PC
8.  ⊢ □(p=>p)               eq 5,6,7
9.  ⊢ □a                eq 8,2

So given ⊢ a, we reach ⊢ □a, which is the rule I shouldn't be able to get in S1. What am I doing wrong?

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  • You have saved me time and time again @Bumble, so I shall tag you in hopes of borrowing your power yet again
    – Nick Doe
    Jun 27 '20 at 15:30
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You have shown that every theorem of propositional logic holds in S1 by necessity (which is a property of S2 and S3 as well):

enter image description here

but not the rule of necessitation (which S2 and S3 lack as well):

enter image description here

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3
  • Could you show how to derive N if we added □□(p=>p) as an axiom?
    – Nick Doe
    Jun 28 '20 at 16:24
  • @Nick Doe could you put this down as a separate question; it is a topic going beyond follow-up comments.
    – Tankut Beygu
    Jun 29 '20 at 7:28
  • Thank you, I will do so now
    – Nick Doe
    Jun 29 '20 at 15:55

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