# Boolean Algebra

## Axiome

### Kommutativ

$A \land B = B \land A$

$A \lor B = B \lor A$

### Assoziativ

$(A \land B) \land C = A \land B \land C$

$(A \lor B) \lor C = A \lor B \lor C$

### Distributiv

$(A \land B) \lor (A \land C) = A \land (B \lor C)$

$(A \lor B) \land (A \lor C) = A \lor (B \land C)$

## Vereinfachungsregeln

$A \land 1 = A$

$A \lor 1 = 1$

$A \land 0 = 0$

$A \lor 0 = A$

$A \land A = A$

$A \lor A = A$

$\bar{A} \land A = 0$

$\bar{A} \lor A = 1$

$A \land (A \lor B) = A$

$A \lor (A \land B = A$

## Examples

### 1

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


### 2

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


### 3

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


### 4

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


### 5

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

CDEFG -> ABCDE


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


### 6

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


### 7

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


CDEFG -> ABCDE

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


### 8

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


### 9

$Q = (B \land C) \lor (B \land \bar{C})  <=> B \land (C \lor \bar{C}) = B \land 1 = B$


### 10

$Y = ( G \lor \bar{F}) \land (G \lor F)  <=> G \lor (F \land \bar{F}) = G \lor 0 = G$


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) :)