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

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

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Other answers:

Does \(+\) mean \(\wedge\) and \(\bar{a}\) =\(\neg{a}\)
\[\overline{a}\] = complement of a
Okay, so \(\neg a\) indeed.
\[a + b + \bar{a} \iff a \wedge b \wedge \neg a\]It's zero . . .
The answer is 1 :|
So does \(+\) stand for the OR operator?
Yes.
then it is 1 indeed since either a or neg(a) is 1 :-)
What does \(ab\) mean? And?
Yes :| Don't you write in this way?
You can write this both ways :-)
\[a\lor b\lor \neg a\iff a\lor b\lor\neg a\land\neg b\] ?
Yes, they're pretty much equivalent using truth tables.
I can't read (don't understand) all those symbols!! :(
Eh, it's easy once you are used to it.
Apparently we're not even caring about \(c\) in that expression.
truth table ?
If ∨ = OR ¬ = NOT ∧ = AND Then, the way to get a∨b∨¬a⟺a∨b∨¬a∧¬b is what I want know :|
Well, manipulation. I thought you wanted a proof.
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.
\[\begin{array}{|c|c|c|c|}\hline a&\bar a&b&\bar b&\bar a\bar b&a+b+\bar a& a+b+\bar a \equiv \bar a\bar b \\\hline\ \end{array}\]
Dangy, that's just a truth table. I was waiting to see the manipulations @UnkleRhaukus does.
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
Because they have the same boolean values for each case.
...... 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 :( )?
is \(a\)= 1 , \(\bar a\)= 0 ?
Yes.
@UnkleRhaukus \(\neg\) lol
If a = 1, then \(\bar{a}\)=0.

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