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Boolean algebra
How can I get \(a+b+\bar{a}\) from \(a+b+\bar{a}\bar{b}\)?
 one year ago
 one year ago
Boolean algebra How can I get \(a+b+\bar{a}\) from \(a+b+\bar{a}\bar{b}\)?
 one year ago
 one year ago

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CallistoBest 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!?
 one year ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

ParthKohliBest ResponseYou've already chosen the best response.1
It'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.
 one year ago

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

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

CallistoBest ResponseYou've already chosen the best response.0
I 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
 one year ago

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

CallistoBest 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 :( )?
 one year ago

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

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

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