×
Create a new article
We currently have 3,189 articles on s23. Type your article name above or create one of the articles listed here!

3,189Articles
in:

# 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) :)

Cookies help us deliver our services. By using our services, you agree to our use of cookies.
Cookies help us deliver our services. By using our services, you agree to our use of cookies.