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Callisto

  • one year ago

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

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  1. modphysnoob
    • one year ago
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    Try DeMorgan

  2. Callisto
    • one year ago
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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!?

  3. UnkleRhaukus
    • one year ago
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    what is bar ?

  4. ParthKohli
    • one year ago
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    Does \(+\) mean \(\wedge\) and \(\bar{a}\) =\(\neg{a}\)

  5. Callisto
    • one year ago
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    \[\overline{a}\] = complement of a

  6. ParthKohli
    • one year ago
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    Okay, so \(\neg a\) indeed.

  7. ParthKohli
    • one year ago
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    \[a + b + \bar{a} \iff a \wedge b \wedge \neg a\]It's zero . . .

  8. Callisto
    • one year ago
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    The answer is 1 :|

  9. ParthKohli
    • one year ago
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    So does \(+\) stand for the OR operator?

  10. Callisto
    • one year ago
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    Yes.

  11. ParthKohli
    • one year ago
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    then it is 1 indeed since either a or neg(a) is 1 :-)

  12. ParthKohli
    • one year ago
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    What does \(ab\) mean? And?

  13. Callisto
    • one year ago
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    Yes :| Don't you write in this way?

  14. ParthKohli
    • one year ago
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    You can write this both ways :-)

  15. UnkleRhaukus
    • one year ago
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    \[a\lor b\lor \neg a\iff a\lor b\lor\neg a\land\neg b\] ?

  16. ParthKohli
    • one year ago
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    Yes, they're pretty much equivalent using truth tables.

  17. Callisto
    • one year ago
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    I can't read (don't understand) all those symbols!! :(

  18. ParthKohli
    • one year ago
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    Eh, it's easy once you are used to it.

  19. ParthKohli
    • one year ago
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    Apparently we're not even caring about \(c\) in that expression.

  20. UnkleRhaukus
    • one year ago
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    truth table ?

  21. Callisto
    • one year ago
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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 :|

  22. ParthKohli
    • one year ago
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    Well, manipulation. I thought you wanted a proof.

  23. ParthKohli
    • one year ago
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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.

  24. UnkleRhaukus
    • one year ago
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    \[\begin{array}{|c|c|c|c|}\hline a&\bar a&b&\bar b&\bar a\bar b&a+b+\bar a& a+b+\bar a \equiv \bar a\bar b \\\hline\ \end{array}\]

  25. ParthKohli
    • one year ago
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    Dangy, that's just a truth table. I was waiting to see the manipulations @UnkleRhaukus does.

  26. Callisto
    • one year ago
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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

  27. ParthKohli
    • one year ago
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    Because they have the same boolean values for each case.

  28. Callisto
    • one year ago
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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 :( )?

  29. UnkleRhaukus
    • one year ago
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    is \(a\)= 1 , \(\bar a\)= 0 ?

  30. ParthKohli
    • one year ago
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    Yes.

  31. ParthKohli
    • one year ago
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    @UnkleRhaukus \(\neg\) lol

  32. Callisto
    • one year ago
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    If a = 1, then \(\bar{a}\)=0.

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