Callisto
  • Callisto
Boolean algebra How can I get \(a+b+\bar{a}\) from \(a+b+\bar{a}\bar{b}\)?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Try DeMorgan
Callisto
  • Callisto
\[a+b+\bar{a}\bar{b}\]\[=\overline{\bar{a}\bar{b}} + \bar{a}\bar{b}\]\[=1\]*cough* I just get the answer directly!?
UnkleRhaukus
  • UnkleRhaukus
what is bar ?

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

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