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