Question

Boolean algebra

How can I get \(a+b+\bar{a}\) from \(a+b+\bar{a}\bar{b}\)?

Answer 1

Try DeMorgan

Answer 2

\[a+b+\bar{a}\bar{b}\]\[=\overline{\bar{a}\bar{b}} + \bar{a}\bar{b}\]\[=1\]*cough* I just get the answer directly!?

Answer 3

what is bar ?

Answer 4

Does \(+\) mean \(\wedge\) and \(\bar{a}\) =\(\neg{a}\)

Answer 5

\[\overline{a}\] = complement of a

Answer 6

Okay, so \(\neg a\) indeed.

Answer 7

\[a + b + \bar{a} \iff a \wedge b \wedge \neg a\]It's zero . . .

Answer 8

The answer is 1 :|

Answer 9

So does \(+\) stand for the OR operator?

Answer 10

Yes.

Answer 11

then it is 1 indeed since either a or neg(a) is 1 :-)

Answer 12

What does \(ab\) mean? And?

Answer 13

Yes :|
Don't you write in this way?

Answer 14

You can write this both ways :-)

Answer 15

\[a\lor b\lor \neg a\iff a\lor b\lor\neg a\land\neg b\]
?

Answer 16

Yes, they're pretty much equivalent using truth tables.

Answer 17

I can't read (don't understand) all those symbols!! :(

Answer 18

Eh, it's easy once you are used to it.

Answer 19

Apparently we're not even caring about \(c\) in that expression.

Answer 20

truth table ?

Answer 21

If ∨ = OR
¬ = NOT
∧ = AND
Then, the way to get a∨b∨¬a⟺a∨b∨¬a∧¬b is what I want know :|

Answer 22

Well, manipulation. I thought you wanted a proof.

Answer 23

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.

Answer 24

\[\begin{array}{|c|c|c|c|}\hline a&\bar a&b&\bar b&\bar a\bar b&a+b+\bar a& a+b+\bar a \equiv \bar a\bar b \\\hline\ \end{array}\]

Answer 25

Dangy, that's just a truth table. I was waiting to see the manipulations @UnkleRhaukus does.

Answer 26

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

Answer 27

Because they have the same boolean values for each case.

Answer 28

...... 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 :( )?

Answer 29

is \(a\)= 1 , \(\bar a\)= 0 ?

Answer 30

Yes.

Answer 31

@UnkleRhaukus \(\neg\) lol

Answer 32

If a = 1, then \(\bar{a}\)=0.