1

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?

  • 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
2

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

  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.