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Q7

Posted by Tazz 
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Q7
May 30, 2011 11:50AM
7(a) I have no idea what this question requires us to do.
(b)
(i) Does not hold - c is also accessible from itself but p is not true in c
(ii) Does not hold - p is not true in c
(iii) Holds
(iv) Holds
avatar Re: Q7
May 30, 2011 01:16PM
"Is not universally valid, and yet turns out to be true in this set of worlds".

That's my stab in the dark at it. Please take with at least a shovel load of salt. I need to go back to the drawing board, it would seem.
Re: Q7
May 30, 2011 01:44PM
7a) [](p or q)
then add what eddy said..
From each world, another world is accissible which either is true for p, true for q, or both. However, this is only true in this particular model. For it to be a valid formula, it would have to be true for each and every possible model. For instance, if there is another world accissible from d, and that world is not true for p or q (has no label), then this formula is no longer valid...
Re: Q7
May 30, 2011 01:50PM
ii) You should be looking for q's
So, it holds I think... From a, there is at least one world, d. All the worlds accessible from d: only c. and c is true for q.
Re: Q7
May 30, 2011 02:38PM
Thanks Hexium, and yep (ii) was my error. Hope silly mistakes like that don't get the better of me in the exam!
avatar Re: Q7
May 30, 2011 10:25PM
Finally I'm able to contribute here, and I'm stuck. i, ii, iii = no problem I'm aware of.

For iv I get this:

Worlds accessible from b = c
Worlds accessible from c =
a ... With Worlds accessible = d ----- where p is T <-------??
d ... With Worlds accessible = c (satisfies ~p )
c ... With Worlds accessible =
... a = ~~p
... d = ~~ p
... c = ~p.

So even if I put it to a vote, I get more ~p than p. More to the point I don't get exclusively either. What am I doing wrong?
Re: Q7
May 31, 2011 09:59AM
Ok, here's my explanation to (iv)
In order for it to hold box box ~p has to be true in all worlds accessible from b, only c is accessible. In order for box box ~p to be true in c, box~p has to be true in all world accessible from c. Which are c itself, and d. Let's examine c, ~p has to be true in all world accessible from c, in d p is true so .. IT DOES NOT HOLD.
I think I made a mistake earlier. Does this make more sense?
Re: Q7
May 31, 2011 10:00AM
Oh, and a is also accessible from c, so, there it DOES NOT HOLD
avatar Re: Q7
May 31, 2011 12:34PM
I think that's right, then. I'm a bit in the dark here, to be honest. I'm basically trying to apply hexium's methods, as I understand them, because that seems to make sense to me.

I think you're right now, then. Taking B = box, we start with

BBBBphi

B's scope is BBBphi

B's scope BBphi

B's Bphi .. at which point we have to satisfy every little unravelled bit of thread for phi for it to hold.

It would then work pictorially as a kind of tree graph, getting pretty frayed at the end if there are a lot of boxes.

I THINK ...
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