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Quick boolean algebra problem..help needed

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  • 22-11-2006 7:54pm
    #1
    techguy
    Registered Users Posts: 2,234 ✭✭✭


    HI,
    I hope this is the right forum..
    Well I am just doing a bit of revising for boolean algebra and de morgans rule and came across this in an equation.. I cant seem to work out how part 1 = part 2 or if it is even correct? COuld someone please shed some light on it? See attachment.

    Thanks,
    Hakko


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    #2
    Hobbes
    Registered Users Posts: 21,264 ✭✭✭✭Hobbes


    http://www.boards.ie/vbulletin/forumdisplay.php?f=380


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    #3
    techguy
    Registered Users Posts: 2,234 ✭✭✭techguy


    thanks hobbes I knew there was a maths forums but it was tos up as to which forum to post in..


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    #4
    Mutant_Fruit
    Closed Accounts Posts: 4,943 ✭✭✭Mutant_Fruit


    Well, you know the rule for boolean algebra? Break the line, change the sign.

    Also, you can add two "not's" on top of a problem without changing it. i.e. stick two more lines on top of the left hand problem so that it's
    not(not(not(NOT(A) + NOT(B) + C))) (works best if you write it down :p)

    Then just break one of the lines, change the signs of everything beneath the line that you broke. You can then remove any excess double-nots over individual letters (not(not(A))) = A, then solve the remainder using normal techniques.


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    #5
    Webmonkey
    Registered Users Posts: 9,579 ✭✭✭Webmonkey


    It isn't a valid logic statement. They are not equal.

    And the breaking line changing sign is DeMorgans Theorm.


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    #6
    Mutant_Fruit
    Closed Accounts Posts: 4,943 ✭✭✭Mutant_Fruit


    Webmonkey wrote:
    It isn't a valid logic statement. They are not equal.
    I never tried actually solving it myself. I was too lazy to work it out :p
    And the breaking line changing sign is DeMorgans Theorm.
    I had forgotten the name of the rule, i just remembered the method. It's 2 years since i've solved these kind of questions ;)


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    #7
    Webmonkey
    Registered Users Posts: 9,579 ✭✭✭Webmonkey


    :)


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    #8
    Misticles
    Registered Users Posts: 13,746 ✭✭✭✭Misticles


    i got part 1 to equal part 2


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    #9
    Mutant_Fruit
    Closed Accounts Posts: 4,943 ✭✭✭Mutant_Fruit


    Try putting in
    A = 1
    B = 1
    C = 0

    The left side evaluates to true, the right side evaluates to false (if i'm doing this right).


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    #10
    Webmonkey
    Registered Users Posts: 9,579 ✭✭✭Webmonkey


    Yep exact same for me


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    #11
    techguy
    Registered Users Posts: 2,234 ✭✭✭techguy


    Thanks guys, the notes I got 1st time round were wrong..
    it should be:
    (' = not)
    (A'+B'+AC)' = (B'+A'+C)' (not de morgans but another rule..)


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    #12
    Mutant_Fruit
    Closed Accounts Posts: 4,943 ✭✭✭Mutant_Fruit


    (A'+B'+AC)' = (B'+A'+C)'

    Break it down to just this:
    A' + AC = A' + C

    Look at the left side, here there's both an A and A'. If A is true, the output relies solely on C being true or false. If A is false, the output is false no matter what C is.

    So, you can break this down to just A' + AC = A' + C as was shown above. That make sense?

    Simlarly, what would A'B + AB' break down into? *hopes he got the example right*


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