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Assignment 3 - Q6

Posted by brettc 
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avatar Assignment 3 - Q6
July 22, 2010 02:34PM
Anybody completed this question?

I'm not sure how to start the proof so any help/hints/pointers would be awesome smiling smiley

I know that I need to prove that ~K1~K1(p & q) -> K1(p &q) and then visa versa, but I'm just not sure what to do with the ~K1~K1...
avatar Re: Assignment 3 - Q6
July 26, 2010 09:26AM
Anyone?
Re: Assignment 3 - Q6
July 26, 2010 01:43PM
Fought with it on the weekend. I'm not sure if I did it right, but I landed up in each side of the proof assuming the negation of the consequent and then using the rules to introduce a contradiction. No idea if that's the right approach but didn't know what else to try...

What are you trying?
avatar Re: Assignment 3 - Q6
July 26, 2010 03:39PM
I tried a few things on the weekend, even sent an email to the lecturers asking for some assistance.
I tried to assume ~K1(p & q) but not sure what to do with the ~K1 once it's in the box.
Re: Assignment 3 - Q6
July 26, 2010 04:18PM
I assumed ~K1(p &q) I and then followed with using the K5 rule which is to put K1 in front of that assumption.

I went with the idea that you can add and remove K's from the front as you want to but you can't add and remove ~K's.
avatar Re: Assignment 3 - Q6
July 26, 2010 04:52PM
What kind of rules can you apply to ~K1 once it's in the box?
Re: Assignment 3 - Q6
July 27, 2010 09:43AM
I thought that you could use K5 there? As in

~K1(p&q) : assumption
K1~K1(p&q) : K5


I got that idea from looking at page 331 where there are two formal proofs.
avatar Re: Assignment 3 - Q6
July 27, 2010 10:34AM
rezrovs Wrote:
-------------------------------------------------------
> I got that idea from looking at page 331 where
> there are two formal proofs.

YES! That's what I needed, I can't believe I just couldn't see that. Thanks a lot man!smileys with beer
Re: Assignment 3 - Q6
July 27, 2010 10:50AM
Glad I could help - took me absolutely AGES to see it too. I just hope that it's right...
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