T tintincute New Member Oct 4, 2008 #1 hi i would like to know what is the value of this !A + !A = !A * !A = is it both A? i'm solving this one, and not really sure if i'm doing the right thing here. first i changed the signs of the 3 equations which is this one: ¯¯¯¯¯¯¯¯¯¯¯ A * !B * C into : !A + B + !C can i just interchange the letters here and collect which can be canceled out? ¯¯¯¯¯¯¯ ¯ ¯ ¯¯¯¯¯¯¯¯¯¯¯ Y = A * B + A + C + A * !B * C = !A * !B + !A + !C + !A + B + !C = !A * !A + !A + !C + !C + 1 thanks & regards
hi i would like to know what is the value of this !A + !A = !A * !A = is it both A? i'm solving this one, and not really sure if i'm doing the right thing here. first i changed the signs of the 3 equations which is this one: ¯¯¯¯¯¯¯¯¯¯¯ A * !B * C into : !A + B + !C can i just interchange the letters here and collect which can be canceled out? ¯¯¯¯¯¯¯ ¯ ¯ ¯¯¯¯¯¯¯¯¯¯¯ Y = A * B + A + C + A * !B * C = !A * !B + !A + !C + !A + B + !C = !A * !A + !A + !C + !C + 1 thanks & regards
M MrAl Well-Known Member Most Helpful Member Oct 5, 2008 #2 Hi there, Check out: DeMorgan's Theorem and keep in mind that if you invert all inputs and outputs that you also have to invert *all* logical connectives, not just some. Your examples: !A + !A = !(!!A*!!A) = !(A*A) = !(A) = !A and !A * !A = !(!!A+!!A) = !(A+A) = !(A) = !A Last edited: Oct 5, 2008
Hi there, Check out: DeMorgan's Theorem and keep in mind that if you invert all inputs and outputs that you also have to invert *all* logical connectives, not just some. Your examples: !A + !A = !(!!A*!!A) = !(A*A) = !(A) = !A and !A * !A = !(!!A+!!A) = !(A+A) = !(A) = !A