Results 1 to 5 of 5

Math Help - Boolean Algebra

  1. #1
    Junior Member
    Joined
    Feb 2008
    Posts
    37

    Boolean Algebra

    Simplify this Boolean Expression

    A & ~((~A) & (~B)).

    According to the truth table I did it should be equivalent to ~B but I can't get the boolean algebra to work out to that.

    A & ~((~A) & (~B))
    = A & (~~A or ~~B) Demorgan's law
    = A & (A or B) Double negation
    = (A & A) or (A & B) Distributive Property


    And this is where I get stuck. If A&A = just A then you get into an infinate loop of distrubutive properties and never reach your goal.
    Last edited by algebrapro18; February 8th 2008 at 05:47 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,617
    Thanks
    1581
    Awards
    1
    I get that the expression is equivalent A alone.
    Do a truth table on the final expression: A \vee \left( {A \wedge B} \right).

    You might think of it as sets: A \cup \left( {A \cap B} \right) \equiv A.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,716
    Thanks
    634
    Hello, algebrapro18!

    Simplify this Boolean expression: . A \;\cap \;\sim\![(\sim\!A) \cap (\sim\!B)]

    According to the truth table I did,
    it should be equivalent to \sim B . . . . no, it's equivalent to {\color{blue}A}


    A \;\cap \;\sim\![(\sim\!A) \cap (\sim\!B)]

    . . = \;A \;\cap \;(\sim\sim\!A \;\cup \;\sim\sim\!B) \qquad\text{DeMorgan's law}

    . . = \;A \;\cap\;(A \cap B)\qquad\text{Double negation}

    . . = \;(A\cap A) \cup (A \cap B)\qquad\text{Distributive Property} . . . . All this is correct!

    And this is where I get stuck.
    If A \cap A \:=\:A, then you get into an infinite loop
    of distrubutive properties and never reach your goal. . . . . This is true

    There should be (or we must establish) a theorem

    . . which states: . A \cap (A \cup B) \;=\; A \cup (A \cap B) \;=\;A

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Feb 2008
    Posts
    37
    Yes it is equal to just A alone and we can't uses sets we have to use Boolean Algebra, so there is no graphing or ven diagrams it has to be stuff like Demorgans and Distribution and stuff. My prof. was VERY clear about that.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,617
    Thanks
    1581
    Awards
    1
    Here is a proof of the absorption law. [You may have different symbols to use.]
    \left( {x \wedge y} \right) \vee x = \left( {x \wedge y} \right) \vee \left( {x \wedge 1} \right)\,\,\mbox{ idenity}.
    \left( {x \wedge y} \right) \vee x = x \wedge \left( {y \vee 1} \right) \,\,\mbox{ distributive law}.
    \left( {x \wedge y} \right) \vee x = x \wedge 1 \,\,\mbox{ idenity law}.
    \left( {x \wedge y} \right) \vee x = x \,\,\mbox{ idenity law}.

    That proves that Aor(A&B)=(A&B)orA=A.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] boolean algebra help
    Posted in the Statistics Forum
    Replies: 1
    Last Post: October 10th 2011, 02:06 PM
  2. Boolean Algebra
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: September 28th 2010, 11:18 PM
  3. Boolean Algebra Help
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 3rd 2010, 04:16 AM
  4. Boolean algebra Help
    Posted in the Discrete Math Forum
    Replies: 11
    Last Post: January 11th 2010, 01:15 PM
  5. Boolean algebra help
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: April 20th 2009, 04:13 AM

Search Tags


/mathhelpforum @mathhelpforum