Boolean Algebra

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
$$A \land 1 = A$$

$$A \lor 1 = 1$$

$$A \land 0 = 0$$

$$A \lor 0 = A$$

$$A \land A = A$$

$$A \lor A = A$$

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

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

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

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

De Morgan Gesetze
de-wp:De_Morgansche_Gesetze

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

6
$$X = (( A \land B ) \lor C) \land (( A \lor B ) \lor D )) <=> (A \land B) \lor (C \land D)$$

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

7
$$X = ( C \lor D \lor F ) \land ( C \lor D \lor G ) <=> C \lor D \lor (F \land G)$$

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. for A in false true; do for B in false true do   for C in false true; do     x=0 (($A && $B) || ($A && $C)) || x=1 y=0 ($A && ($B || $C)) || y=1 if [ $x -eq $y ] then echo "ok" else echo "FALSCH - Belegung: A: $A, B: $B, C: $C" fi    done done done
 * 1) !/bin/bash

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