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    Boolean Algebra: Difference between revisions

    Content added Content deleted
    imported>mutante
    mNo edit summary
    imported>Took
    Line 83: Line 83:
    betrachte A=0, B=1, C=1, D=0
    betrachte A=0, B=1, C=1, D=0
    dann gilt: der linke ausdruck ist wahr, der rechte aber nicht.
    dann gilt: der linke ausdruck ist wahr, der rechte aber nicht.

    besser
    <math>X = (( A \land B ) \lor C) \land (( A \lor B ) \lor D ))
    <=> (A \land B) \lor ( C \land (A \lor B \lor D) )</math>


    === 7 ===
    === 7 ===

    Revision as of 23:52, 14 November 2006

    Axiome

    Kommutativ

    <math>A \land B = B \land A</math>

    <math>A \lor B = B \lor A</math>

    Assoziativ

    <math>(A \land B) \land C = A \land B \land C</math>

    <math>(A \lor B) \lor C = A \lor B \lor C</math>

    Distributiv

    <math>(A \land B) \lor (A \land C) = A \land (B \lor C)</math>

    <math>(A \lor B) \land (A \lor C) = A \lor (B \land C)</math>

    Vereinfachungsregeln

    <math>A \land 1 = A</math>

    <math>A \lor 1 = 1</math>


    <math>A \land 0 = 0</math>

    <math>A \lor 0 = A</math>


    <math>A \land A = A</math>

    <math>A \lor A = A</math>


    <math>\bar{A} \land A = 0</math>

    <math>\bar{A} \lor A = 1</math>


    <math>A \land (A \lor B) = A</math>

    <math>A \lor (A \land B = A</math>

    De Morgan Gesetze

    de-wp:De_Morgansche_Gesetze

    Examples

    1

    <math>Q = ( A \land B ) \lor ( A \land C )

       <=> A \land (B \lor C)</math>
    

    2

    <math>Q = ( C \lor B ) \land ( A \lor C )

       <=> C \lor (B \land A)</math>
    

    3

    <math>Y = ( A \land B) \lor ( C \land D) \lor ( D \land A ) \lor ( E \land C) <=>

     (A \land (B \lor D)) \lor ( C \land (D \lor E))</math>
    

    4

    <math>Z = ( A \land B ) \lor ( B \land A )

       <=> A \land B</math>
    

    5

    <math>Y = (\bar{C} \lor D \lor F) \land (\bar{C} \lor E \lor G)

       <=> \bar{C} \land ((D \land F) \land (G \lor E))</math>
    

    6

    <math>X = (( A \land B ) \lor C) \land (( A \lor B ) \lor D ))

     <=> (A \land B) \lor (C \land D)</math>
    
    ne das is nicht richtig so!
    betrachte A=0, B=1, C=1, D=0
    dann gilt: der linke ausdruck ist wahr, der rechte aber nicht.
    

    besser <math>X = (( A \land B ) \lor C) \land (( A \lor B ) \lor D ))

     <=> (A \land B) \lor ( C \land (A \lor B \lor D) )</math>
    

    7

    <math>X = ( C \lor D \lor F ) \land ( C \lor D \lor G )

     <=> C \lor D \lor (F \land G)</math>
    

    8

    <math>U = ( A \lor B ) \land ( A \land C )

       <=> ( A ) \land (A \land C) <=> A \land A \land C <=> A \land C</math>
    

    9

    <math>Q = (B \land C) \lor (B \land \bar{C})

       <=> B \land (C \lor \bar{C}) = B \land 1 = B</math>
    

    10

    <math>Y = ( G \lor \bar{F}) \land (G \lor F)

       <=> G \lor (F \land \bar{F}) = G \lor 0 = G</math>
    


    Are those right? I dont think so. Please check. --done :) --took 23:33, 14 November 2006 (CET) thank you, help is appreciated :) mutante 00:07, 15 November 2006 (CET)

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