Q7

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

"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.

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if there is another world accissible from d, and that world is not true for p or q (has no label), then this formula is no longer valid...

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.

Thanks Hexium, and yep (ii) was my error. Hope silly mistakes like that don't get the better of me in the exam!

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?

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?

Oh, and a is also accessible from c, so, there it DOES NOT HOLD

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 ...