I know that

$$ 1)\space w ⊩\diamond P\iff there\space is\space some\space worlds\space w' \space such\space that\space wRw': w ⊩ P $$ $$ 2)\space w ⊩\square P\iff for\space all\space worlds\space w' \space such\space that\space wRw': w ⊩ P $$

Now suppose I have a world e which doesn't 'see' any other world. Is it then the case that:

$$ w ⊩\diamond P $$ $$ w ⊮\square P $$

The reason I think so is that (1) doesn't hold for e since it doesn't see any worlds. But (2) seems to me to hold because it is a "for all"-statement. So, at least to me, (2) holds, since it is true that for all worlds e sees (there are none), those force P.

However, I have two questions about this. Couldn't the same reasoning be applied to $$\square ¬P $$ Thus, we would have $$e ⊩ \square P\space and\space e ⊩ \square ¬P $$ This, seems to me to be very strange. But, then again, I'm new to Modal Logic.

Secondly, is it not necessary that we know, at least that $$ e ⊩ P $$ if we are going to even start talking about P being necessary at e?.



The arguments you've given for (1) and (2) seem to say that $w⊩□P$ and $w⊮⋄P$, which is the opposite of what you've actually put in the displayed formula. (The argument works, though -- if there is no $w'$ with $wRw'$, then any property at all holds "for all $w'$ with $wRw'$".)

You are also correct that if there is no $w'$ with $wRw'$, then both $w⊩□P$ and $w⊩□¬P$ hold. This is why we might want to assume that, at least, $wRw$ for all $w$. Your second question has the same answer: the properties we want our logical system to have will put constraints on the relations $R$ that we are willing to consider.

  • $\begingroup$ Sorry! Yes, I meant w⊩□P and w ⊮ ⋄P ... I'm tired after a whole day of logic. Thank you! $\endgroup$ – tom tronbone Nov 14 '14 at 19:48

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