Boolean Algebra

Axiome

Kommutativ

A ∧ B = B ∧ A

A ∨ B = B ∨ A

Assoziativ

(A ∧ B) ∧ C = A ∧ B ∧ C

(A ∨ B) ∨ C = A ∨ B ∨ C

Distributiv

(A ∧ B) ∨ (A ∧ C) = A ∧ (B ∨ C)

(A ∨ B) ∧ (A ∨ C) = A ∨ (B ∧ C)

Vereinfachungsregeln

A ∧ 1 = A

A ∨ 1 = 1

A ∧ 0 = 0

A ∨ 0 = A

A ∧ A = A

A ∨ A = A

¬A ∧ A = 0

¬A ∨ A = 1

A ∧ (A ∨ B) = A

A ∨ (A ∧ B = A

De Morgan Gesetze

1

 Q = ( A ∧ B ) ∨ (A ∧ C)
   = A ∧ (B ∨ C)

2

 Q = ( C ∨ B) ∧ (A ∨ C)
   = C ∨ (B ∧ A)

3

 Y = ( A ∧ B) ∨ (C ∧ D) ∨ ( D ∧ A ) ∨ ( E ∧ C) =
 (A ∧ (B ∨ D)) ∨ (C ∧ (D ∨ E))

4

 Z = ( A ∧ B ) ∨ ( B ∧ A) 
   = A ∧ B

5

 Y = (¬C ∨ D ∨ F) ∧ (¬C ∨ E ∨ G) 
   = ¬C ∧ ((D ∧ F) ∧ (G ∨ E))

6

 X = (( A ∧ B ) ∨ C) ∧ (( A ∨ B ) ∨ D ))
 = (A ∧ B) ∨ (C ∧ D)

7

 X = ( C ∨ D ∨ F ) ∧ (C ∨ D ∨ G )
 = (C ∨ D)  ∨ (F ∧ G)

8

 U = ( A ∨ B) ∧ ( A ∧ C )
   = ?

9

 Q = (B ∧ C) ∨ (B ∧ ¬C) 
   = B ∧ (C ∨ ¬C) = B ∧ 0 = 0

10

 Y = ( G ∨ ¬F) ∧ (G ∨ F) 
   = G ∨ (F ∧ ¬F) = G ∨ 0 = G

Are those right? I dont think so. Please check.