Date: Feb 28, 2007 9:47 PM
Author: Ben Brink
Subject: RE: resolution question
Earth,
This is to congratulate your courage and perseverance for putting these
logic problems online. The "chat" environment can be rough.
If I understand your question correctly we want to resolve (x or Y or not
z) and (not y or not z) and (not x or y or z). Notice the "and" connectives
imply that each parenthesis is true. Assuming the first parenthesis is
true, then at least one of x, Y, or not z must be true. Let's look at these
in turn.
If x is true, we see no contradiction in the second parenthesis. But
notice the third parenthesis contains "not x." Since in standard logic we
can't have both "x" and "not x", we can't conclude anything about x.
If Y is true, we see no further comment in the last two parentheses.
If not z is true, then the second parenthesis agrees, but the third
parenthesis has "z." Just like before, we can't say anything about z.
The only variable left is "y." The second parenthesis now says that not y
is true, but the third parenthesis now says that y is true. Again we have a
contradiction.
As I read your problem, the only conclusion is that Y is true.
I appreciate seeing your posts. If you see a solution either better or
shorter than the one I've written, please be sure to post it. Thanks and
stay with your course!
Ben
>From: Earth <discussions@mathforum.org>
>Reply-To: discretemath@mathforum.org
>To: discretemath@mathforum.org
>Subject: resolution question
>Date: Tue, 27 Feb 2007 20:00:54 EST
>
>conclude on that can be conclude from the statement
>(x v Y v -z) ^ (-y v -z) ^ (-x v y v z)
>i can't find the big upside "v" so i use this ^ instead
>then it say
>applying resolution
>can some one tell me what do i need to do by or what they mean by that
>resolution? thank you
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