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

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

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

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

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

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

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

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

UnkleRhaukus
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1Yes, they're pretty much equivalent using truth tables.

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

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

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

Callisto
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1Well, manipulation. I thought you wanted a proof.

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

Callisto
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1is \(a\)= 1 , \(\bar a\)= 0 ?

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

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