Calculo Diferencial

a) Demostrar si A ∩ (B ’BC) entonces A ∩ (B A)’ C
Recordar:
* A B A ∩ B’ = ∅
* A B = (A ∩ B’) ∪ (A’ ∩ B) = (A’ ∪ B’) ∩ (A ∪ B)
1° A ∩ (B’ C)
A ∩ [(B ∩ A’) ∪ (B’ ∩ A)]’ [B’ ∪ C)’] ∩ [B ∪ C)]}
A ∩ [(B ∩ A’)’ ∩ (B’ ∩ A)’] ∩ [B’ ∪ C)’] ∩ [B ∪ C)]}’ = ∅
Propiedad de Morgan: (A ∩ B)’= A’ ∪ B’
{A ∩ (B’ ∪ A) ∩ (B ∪ A’)} ∩ [B ∩ C)] ∪ [B’ ∩ C)’]} = ∅
Propiedad de absorción: A ∩ (A ∪ B) = A
A ∪ (A’ ∩B) = A ∪ B
A ∩ B ∩ [B ∩ C)] ∪ [B’ ∩ C)’]} = ∅
Propiedad distributiva: A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
A ∩ {B ∩ [B ∩ C)] ∪ [B ∩ [B’ ∩ C)’]]} = ∅Propiedad de absorción
Ley del complemento: A ∩ A’ = ∅
A ∩ B ∩ C) ∪ ∅ = ∅
B ∩ {A ∩ [(A ∩ C’) ∪ (A’ ∩ C)]} = ∅
Propiedad distributiva
B ∩ {[A ∩ (A ∩ C’)] ∪ [A ∩ (A’ ∩ C)]} =∅
B ∩ A ∩ C’ = ∅
Cambio de variable
B ∩ A = X
→ X ∩ C’ = ∅
2° A ∩ (B A)’ C
A ∩ [(B ∩ A’) ∪ (B’ ∩ A)]’ C
A ∩ [(B ∩ A’)’ ∩ (B’ ∩ A)’] C
Propiedad de Morgan ydistributiva
[A ∩ (B’ ∪ A)] ∩ [A ∩ (B ∪ A’)] C
Propiedad de Absorción
A ∩ A ∩ B C
A ∩ B ∩ C’ = ∅
Cambio de variable
A ∩ B = X
→ X ∩ C’ = ∅
b) Usando elementos demostrarque {[A’ – (B’ – C)]’ ∩ (C’ – B)’} A ∩ (B ∪ C)}
{[A’ – (B’ – C)]’ ∩ (C’ – B)’} A∩ (B ∪ C)}
Recuerda que: A – B = A ∩ B’
{[A’ – (B’ ∩ C’)]’∩ (C’ ∩ B’)’} A∩ (B ∪C)}
Propiedad de Morgan: (A ∩ B)’ = A’ ∪ B’
{[A’ ∩ (B’ ∩ C’)’]’ ∩ (C ∪ B)} A∩ (B ∪ C)}
{[A’ ∩ (B ∪ C)]’ ∩ (C ∪ B)} A∩ (B ∪ C)}
Propiedad de absorción
{[A ∪ (B∪C)’] ∩(B ∪ C)} A∩ (B ∪ C)}
A ∩ (B ∪ C) A ∩ (B ∪ C)
Cambio de variable
A ∩ (B ∪ C) = M → M ∩ M’ =...