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>
De Morgan Gesetze
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} \land ((D \land F) \land (G \lor E))</math>
6
<math>X = (( A \land B ) \lor C) \land (( A \lor B ) \lor D ))
<=> (A \land B) \lor (C \land D)</math>
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.
7
<math>X = ( C \lor D \lor F ) \land ( C \lor D \lor G )
<=> C \lor D \lor (F \land G)</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)