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Math Help - Proof on discrete maths, help :)

  1. #1
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    Proof on discrete maths, help :)

    I 'm trying to proove that :

    (a AND -b) OR (b AND -c) OR (c AND -a) = (a OR b OR c) AND (-a OR -b OR -c)

    i ve put it on truth table and its working, but i have to do it actually without truth table, so i tryed boolean algebra.

    I think that i am making a fatal mistake but i dont know where it is. So i started from this with boolean algebra --> (a OR b OR c)AND(-a OR -b OR -c) trying to reach 1st part, but actually this is what i achived :

    aa' + ab' + ac' + ba' +bb' +bc' + ca' + cb' + cc' =
    ab' + ac' + ba' +bc' + ca' +cb' =
    (a XOR c) + (b XOR c) + (a XOR b)

    + = logical OR
    multiplies = logical AND


    Thanks in advance for your help
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  2. #2
    MHF Contributor

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    Quote Originally Posted by primeimplicant View Post
    I 'm trying to proove that :

    (a AND -b) OR (b AND -c) OR (c AND -a) = (a OR b OR c) AND (-a OR -b OR -c)

    i ve put it on truth table and its working, but i have to do it actually without truth table, so i tryed boolean algebra.

    I think that i am making a fatal mistake but i dont know where it is. So i started from this with boolean algebra --> (a OR b OR c)AND(-a OR -b OR -c) trying to reach 1st part, but actually this is what i achived :

    aa' + ab' + ac' + ba' +bb' +bc' + ca' + cb' + cc' =
    ab' + ac' + ba' +bc' + ca' +cb' =
    (a XOR c) + (b XOR c) + (a XOR b)

    + = logical OR
    multiplies = logical AND
    Thanks in advance for your help
    You do understand, don't you, that "aa' " means "a is true and a is false"? In "boolean" aa'= bb'= cc'= 0.

    The left side of (a AND -b) OR (b AND -c) OR (c AND -a) = (a OR b OR c) AND (-a OR -b OR -c) reduces to ab'+ bc'+ a'c and the right side to (a+ b+ c)(a'+ b'+ c').

    Now, apply booean algebra to those.
    Last edited by mr fantastic; October 30th 2009 at 12:39 PM.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    You do understand, don't you, that "aa' " means "a is true and a is false"? In "boolean" aa'= bb'= cc'= 0.
    Ehm, apparently i understand that, thats why from this point :
    aa' + ab' + ac' + ba' +bb' +bc' + ca' + cb' + cc'

    i m going here :

    ab' + ac' + ba' +bc' + ca' +cb' (i erase apparently aa', bb', cc' )

    BUT in boolean algebra for example YZ' + ZY' = Y XOR Z ,

    and the above statement --> ab' + ac' + ba' +bc' + ca' +cb'

    equals (a XOR c) + (b XOR c) + (a XOR b) ,

    I cant exactly understand where should i apply boolean, left side of equation or right side?

    I am supposed to start from one side and result the other side.

    I choose to start from right side but i didnt result left side .

    What am i doing sooo wrong ? I think that i apply really bad boolean on right side, but where exactly?

    Thanks again for your reply
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  4. #4
    Member oldguynewstudent's Avatar
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    Quote Originally Posted by primeimplicant View Post
    I 'm trying to proove that :

    (a AND -b) OR (b AND -c) OR (c AND -a) = (a OR b OR c) AND (-a OR -b OR -c)

    i ve put it on truth table and its working, but i have to do it actually without truth table, so i tryed boolean algebra.

    I think that i am making a fatal mistake but i dont know where it is. So i started from this with boolean algebra --> (a OR b OR c)AND(-a OR -b OR -c) trying to reach 1st part, but actually this is what i achived :

    aa' + ab' + ac' + ba' +bb' +bc' + ca' + cb' + cc' =
    ab' + ac' + ba' +bc' + ca' +cb' =
    (a XOR c) + (b XOR c) + (a XOR b)

    + = logical OR
    multiplies = logical AND


    Thanks in advance for your help
    Try using logical equivalences:

    (a \wedge -b) \vee (b \wedge -c) \vee (c \wedge -a) =

    (a \vee b \vee c) \wedge (a \vee -c \vee -a) \wedge (-b \vee b \vee c) \wedge (-b \vee -c \vee -a) Distributive property

    = (a \vee b \vee c) \wedge T \wedge T \wedge (-b \vee -c \vee -a) Negation and Domination Laws

    = (a \vee b \vee c) \wedge (-a \vee -b \vee -c) Domination and Commutative Laws
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