# Q7

Posted by Tazz
Announcements Last Post
SoC Curricula 09/30/2017 01:08PM
Demarcation or scoping of examinations and assessment 02/13/2017 07:59AM
School of Computing Short Learning Programmes 11/24/2014 08:37AM
Unisa contact information 07/28/2011 01:28PM
 Q7 May 30, 2011 11:50AM Registered: 8 years ago Posts: 363 Rating: 0
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
 Re: Q7 May 30, 2011 01:16PM Registered: 10 years ago Posts: 3,496 Rating: 1
"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 Registered: 11 years ago Posts: 111 Rating: 0
7a) [](p or q)
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if there is another world accissible from d, and that world is not true for p or q (has no label), then this formula is no longer valid...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 Registered: 11 years ago Posts: 111 Rating: 0
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 Registered: 8 years ago Posts: 363 Rating: 0
Thanks Hexium, and yep (ii) was my error. Hope silly mistakes like that don't get the better of me in the exam!
 Re: Q7 May 30, 2011 10:25PM Registered: 10 years ago Posts: 3,496 Rating: 1
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 Registered: 8 years ago Posts: 363 Rating: 0
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 Registered: 8 years ago Posts: 363 Rating: 0
Oh, and a is also accessible from c, so, there it DOES NOT HOLD
 Re: Q7 May 31, 2011 12:34PM Registered: 10 years ago Posts: 3,496 Rating: 1
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 ...
Sorry, only registered users may post in this forum.