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    s23
    3,187Articles
    in:

    Axiome[edit]

    Kommutativ[edit]

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

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

    Assoziativ[edit]

    <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[edit]

    <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[edit]

    <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[edit]

    de-wp:De_Morgansche_Gesetze

    Examples[edit]

    1[edit]

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

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

    2[edit]

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

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

    3[edit]

    <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[edit]

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

       <=> A \land B</math>
    

    5[edit]

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

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

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

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

    6[edit]

    <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[edit]

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

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

    CDEFG -> ABCDE

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

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

    8[edit]

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

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

    9[edit]

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

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

    10[edit]

    <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)

    Automatic Proof[edit]

    with this little script u can check all of them by yourself. Just enter both conditions in line 9 and 11 and run it.

    #!/bin/bash
    for A in false true;
     do
      for B in false true
       do
        for C in false true;
         do
          for D in false true;
          do
           for E in false true;
           do
          x=0
          # 1
          # (($A && $B) || ($A && $C)) || x=1
          # 2
          # ($C || $B) && ($A || $C) || x=1
          # 3
          # (($A && $B) || ($C && $D) || ($D && $A) || ($E && $C)) || x=1
          # 4
          # (($A && $B) || ($B && $A)) || x=1
          # 5
          # ((! $A || $B || $D ) && (! $A || $C || $E)) || x=1
          # 6
          # ((($A && $B) || $C) && (($A || $B) || $D)) || x=1
          # 7
          # (($A || $B || $D) && ($A || $B || $E)) || x=1
          # 8
          # (($A || $B) && ($A && $C)) || x=1
          # 9
          # (($B && $C) || ($B && ! $C)) || x=1
          # 10
          (($A || ! $B) && ($A || $B)) || x=1
          y=0
          # 1
          # ($A && ($B || $C)) || y=1
          # 2
          # ($C || ($B && $A)) || y=1
          # 3
          # (($A && ($B || $D)) || ($C && ( $D || $E))) || y=1
          # 4
          # ($A && $B) || y=1
          # 5
          # (! $A || (($B || $D) && ($E || $C))) || y=1
          # 6
          # (($A && $B) || ($C && ($A || $B || $D))) || y=1
          # 7
          # ($A || $B || ($D && $E)) || y=1
          # 8
          # ($A && $C) || y=1
          # 9
          # $B || y=1
          # 10
          $A || y=1
          if [ $x -eq $y ]
           then
            echo "ok"
           else
            echo "FALSCH - Belegung: A: $A, B: $B, C: $C, D: $D E: $E"
           fi
         done
       done
      done
     done
    done
    
    
    

    wow cool, Hilfe zur Selbsthilfe ist natürlich viel besser :) mutante

    Ok, all checked with script above :) mutante 12:00, 16 November 2006 (CET) :)

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