Assignment 0 : Problem 7
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Aaron Albers
Sep 6, 2010, 8:27:39 PM9/6/10
to CSC 2511 Fall 2010
!( !( (p or q) and r) or !q) = ((p or q) and r) or q
Is there anything else that can be done?
Natalia Duque
Sep 6, 2010, 8:30:39 PM9/6/10
to csc-2511-...@googlegroups.com
That's as far as I got. I feel that shouldn't be it tho
Anne S
Sep 6, 2010, 9:07:21 PM9/6/10
to CSC 2511 Fall 2010
what about (p && r) || (q && r)
[&& = and, || = or]
On Sep 6, 6:30 pm, Natalia Duque <duque...@gmail.com> wrote:
> That's as far as I got. I feel that shouldn't be it tho
>
[&& = and, || = or]
On Sep 6, 6:30 pm, Natalia Duque <duque...@gmail.com> wrote:
> That's as far as I got. I feel that shouldn't be it tho
>
Aaron Albers
Sep 6, 2010, 9:11:00 PM9/6/10
to csc-2511-...@googlegroups.com
Is that simplifying it or just expanding (p || q) && r?
Anne S
Sep 6, 2010, 9:14:42 PM9/6/10
to CSC 2511 Fall 2010
Yours applies (reverse) distribution to mine, which I think is better.
I think both are simpler than your original solution.
On Sep 6, 7:11 pm, Aaron Albers <aaroncalb...@gmail.com> wrote:
> Is that simplifying it or just expanding (p || q) && r?
>
I think both are simpler than your original solution.
On Sep 6, 7:11 pm, Aaron Albers <aaroncalb...@gmail.com> wrote:
> Is that simplifying it or just expanding (p || q) && r?
>
Anne S
Sep 6, 2010, 9:16:58 PM9/6/10
to CSC 2511 Fall 2010
wait -- I dropped the last q term. Scratch my comment.
Aaron Albers
Sep 6, 2010, 9:17:19 PM9/6/10
to csc-2511-...@googlegroups.com
Lol I was going to say. =)
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