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Callisto
 one year ago
Best ResponseYou've already chosen the best response.0\[a+b+\bar{a}\bar{b}\]\[=\overline{\bar{a}\bar{b}} + \bar{a}\bar{b}\]\[=1\]*cough* I just get the answer directly!?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Does \(+\) mean \(\wedge\) and \(\bar{a}\) =\(\neg{a}\)

Callisto
 one year ago
Best ResponseYou've already chosen the best response.0\[\overline{a}\] = complement of a

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Okay, so \(\neg a\) indeed.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1\[a + b + \bar{a} \iff a \wedge b \wedge \neg a\]It's zero . . .

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1So does \(+\) stand for the OR operator?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1then it is 1 indeed since either a or neg(a) is 1 :)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1What does \(ab\) mean? And?

Callisto
 one year ago
Best ResponseYou've already chosen the best response.0Yes : Don't you write in this way?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1You can write this both ways :)

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1\[a\lor b\lor \neg a\iff a\lor b\lor\neg a\land\neg b\] ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Yes, they're pretty much equivalent using truth tables.

Callisto
 one year ago
Best ResponseYou've already chosen the best response.0I can't read (don't understand) all those symbols!! :(

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Eh, it's easy once you are used to it.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Apparently we're not even caring about \(c\) in that expression.

Callisto
 one year ago
Best ResponseYou've already chosen the best response.0If ∨ = OR ¬ = NOT ∧ = AND Then, the way to get a∨b∨¬a⟺a∨b∨¬a∧¬b is what I want know :

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Well, manipulation. I thought you wanted a proof.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1It's just that if b is positive then you have the whole thing true, and if it's false then definitely ¬a∧¬b is true. If it's not true, then definitely a is true, so the whole expression will be true. So we have both expressions equivalent.

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1\[\begin{array}{cccc}\hline a&\bar a&b&\bar b&\bar a\bar b&a+b+\bar a& a+b+\bar a \equiv \bar a\bar b \\\hline\ \end{array}\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Dangy, that's just a truth table. I was waiting to see the manipulations @UnkleRhaukus does.

Callisto
 one year ago
Best ResponseYou've already chosen the best response.0I can get a∨b∨¬a = 1 and a∨b∨¬a∧¬b =1 (so they must be equal), but I don't know how to get a∨b∨¬a from a∨b∨¬a∧¬b

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Because they have the same boolean values for each case.

Callisto
 one year ago
Best ResponseYou've already chosen the best response.0...... I know they are equal, but how to get a∨b∨¬a from a∨b∨¬a∧¬b without showing that they are both equal to 1 and without using truth table (I'm sorry :( )?

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1is \(a\)= 1 , \(\bar a\)= 0 ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1@UnkleRhaukus \(\neg\) lol

Callisto
 one year ago
Best ResponseYou've already chosen the best response.0If a = 1, then \(\bar{a}\)=0.
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