Boolean Algebra

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>

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} \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

<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>

CDEFG -> ABCDE

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

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

Automatic Proof

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) :)