2

I was wondering if someone can perhaps help with this proving. I am not sure how to handle a temporal frame that is not a set of only natural numbers. Any suggestions?

Thank you.

  • I think you need to re-enter the logic statement in the text of your question. The attributes in front of "p" are only square blocks and it is unclear what problem you need to prove. – James Jensen Mar 1 at 15:42
  • 1
    @JamesJensen The "square block" operator in modal logic denotes necessity. – Nick R Mar 1 at 16:29
  • @JamesJensen en.wikipedia.org/wiki/Modal_logic – Not_Here Mar 1 at 16:46
  • @NickR, Oops. My mistake. I thought the blocks were the result of the browser's failure to display certain symbols. – James Jensen Apr 1 at 15:14
2

You handle this exactly the same way you would if the frame would consist of natural numbers. What have you tried? You should also be open about this being a homework question, if it is one. Is it one?

If you want to disprove it, find a valuation and a rational number q where your sentence is not true. Otherwise you show that it is true independent of the valuation.

Let's assume we want to prove it. We choose a rational number q, and assume the left side: q⊨◻◻p. Is it true that q⊨◻p? If so, then for any from q accessible world r, i.e. for any r>q we have to show that r⊨p. Can you take it from here? (Hint: The following is the wrong approach: By assumption, q⊨◻◻p. Therefore r⊨◻p. Now you are stuck.)

  • Yes, this is a question regarding homework exercise. However, I was not able to find enough useful literature on this. If I would want to prove this, I would start by assuming the left side is true and taking a state, lets say 'n', that is in the set of rational numbers. Then it would mean that for all 'm' that are in set Q with respect n<m we have that m⊨◻p such that for all 'k' that are in Q, m<k, k⊨p. Then we would take an arb. 'future' of n:n<r that belongs to Q. Then we want to show that ◻p holds so we look for 'future' that makes this true. Any remarks? @JishinNoben – saraherceg Mar 1 at 18:39
  • @saraherceg I expanded a little bit. There is one "idea" that you have to use, but you haven't mentioned yet. (You could disprove it for the natural numbers, but prove it for the rationals. So you will need to invoke a special property of the rationals.) – Jishin Noben Mar 1 at 19:31
  • I gave it a thought but I am not really sure which property you mean, a bit lost here. @JishinNoben – saraherceg Mar 3 at 0:00
  • @saraherceg Take the case for the natural numbers. The assumption was q⊨◻◻p and one has to show that q⊨◻p. This means you have to show that q+1⊨p, q+2⊨p,... But q⊨◻◻p only gives you q+2⊨p, q+3⊨p,... The intuition is: to loose a modal operator, you have to go at least 1 step towards +∞. For rational numbers, you also have to make a step towards infinity to loose one operator, but they don't have to be of size 1. They could be 1/2, 3/16, or what have you. So what can you do to loose two "◻" 's and still arrive at k? – Jishin Noben Mar 3 at 9:48
  • 1
    Ok, understood it fully! Thank you very much for your help! @JishinNoben – saraherceg Mar 4 at 10:13

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.