You are not logged in.
Pages: 1
Hi. I've tried to do this question but i couldn't since there are two conditional logic propositions:
Last edited by JaneFairfax (2007-09-26 10:32:38)
Offline
~(~p -> q) = ~(~p v ~q) = p & q
No?
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
Offline
Offline
In post #2 by Jane, it is in fact perfectly correct!!
And the "tautology" she is referring to means that
for any of the four combinations of p and q, the
result will always be a TRUE value.
So in electronics, we would say, the entire equation
boils down to LOGIC ONE, or True.
igloo myrtilles fourmis
Offline
oh cool thanks so much
Pages: 1