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Minimización de funciones
Booleanas
• Métodos algebraicos
F1 (B,C)= 1+B’+C
F1 (B,C)= 1
1.-Identidades
2.- Factorización
AB’ + AB = A(B’+B)= A
3.- Propiedad Distributiva
X+YZ = (X+Y) (X+Z)
X (Y+Z) = XY +XZ
4.-Teorema del consenso
AB+A’C+BC = AB+A’C
5.-Teorema de Dmorgan
(AB)’=A’+ B’ (A+B)’=A’ B’
A+B =(A’ B’)’ AB =(A’+B’)’
AND OR
A A=A A + A=A
A 0 =0 A + 0 = A
A 1 =A A + 1 =1
A A’ =0 A+A’ =1
F2 (D,C)= DC’(0)
F2 (D,C)= 0
1.-Identidades
2.- Factorización
AB’ + AB = A(B’+B)= A
3.- Propiedad Distributiva
X+YZ = (X+Y) (X+Z)
X (Y+Z) = XY +XZ
4.-Teorema del consenso
AB+A’C+BC = AB+A’C
5.-Teorema de Dmorgan
(AB)’=A’+ B’ (A+B)’=A’ B’
A+B =(A’ B’)’ AB =(A’+B’)’
AND OR
A A=A A + A=A
A 0 =0 A + 0 = A
A 1 =A A + 1 =1
A A’ =0 A+A’ =1
F3 (A, B) = A’+B+A
F3 (A, B) = 1
1.-Identidades
2.- Factorización
AB’ + AB = A(B’+B)= A
3.- Propiedad Distributiva
X+YZ = (X+Y) (X+Z)
X (Y+Z) = XY +XZ
4.-Teorema del consenso
AB+A’C+BC = AB+A’C
5.-Teorema de Dmorgan
(AB)’=A’+ B’ (A+B)’=A’ B’
A+B =(A’ B’)’ AB =(A’+B’)’
AND OR
A A=A A + A=A
A 0 =0 A + 0 = A
A 1 =A A + 1 =1
A A’ =0 A+A’ =1
F4 (A,,B,C) = A+A’BC
F4 (A,,B,C)=(A+A’)(A+BC)
F4 (A,,B,C)=A+BC
1.-Identidades
2.- Factorización
AB’ + AB = A(B’+B)= A
3.- Propiedad Distributiva
X+YZ = (X+Y) (X+Z)
X (Y+Z) = XY +XZ
4.-Teorema del consenso
AB+A’C+BC = AB+A’C
5.-Teorema de Dmorgan
(AB)’=A’+ B’ (A+B)’=A’ B’
A+B =(A’ B’)’ AB =(A’+B’)’
AND OR
A A=A A + A=A
A 0 =0 A + 0 = A
A 1 =A A + 1 =1
A A’ =0 A+A’ =1
F4 (A,,B,C) = A+A’BC
F4 (A,,B,C)=(A+A’)(A+BC)
F4 (A,,B,C)=A+BC
1.-Identidades
2.- Factorización
AB’ + AB = A(B’+B)= A
3.- Propiedad Distributiva
X+YZ = (X+Y) (X+Z)
X (Y+Z) = XY +XZ
4.-Teorema del consenso
AB+A’C+BC = AB+A’C
5.-Teorema de Dmorgan
(AB)’=A’+ B’ (A+B)’=A’ B’
A+B =(A’ B’)’ AB =(A’+B’)’
AND OR
A A=A A + A=A
A 0 =0 A + 0 = A
A 1 =A A + 1 =1
A A’ =0 A+A’ =1
1.-Identidades
2.- Factorización
AB’ + AB = A(B’+B)= A
3.- Propiedad Distributiva
X+YZ = (X+Y) (X+Z)
X (Y+Z) = XY +XZ
4.-Teorema del consenso
AB+A’C+BC = AB+A’C
5.-Teorema de Dmorgan
(AB)’=A’+ B’ (A+B)’=A’ B’
A+B =(A’ B’)’ AB =(A’+B’)’
AND OR
A A=A A + A=A
A 0 =0 A + 0 = A
A 1 =A A + 1 =1
A A’ =0 A+A’ =1

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simplificar funciones

  • 2. F1 (B,C)= 1+B’+C F1 (B,C)= 1 1.-Identidades 2.- Factorización AB’ + AB = A(B’+B)= A 3.- Propiedad Distributiva X+YZ = (X+Y) (X+Z) X (Y+Z) = XY +XZ 4.-Teorema del consenso AB+A’C+BC = AB+A’C 5.-Teorema de Dmorgan (AB)’=A’+ B’ (A+B)’=A’ B’ A+B =(A’ B’)’ AB =(A’+B’)’ AND OR A A=A A + A=A A 0 =0 A + 0 = A A 1 =A A + 1 =1 A A’ =0 A+A’ =1
  • 3. F2 (D,C)= DC’(0) F2 (D,C)= 0 1.-Identidades 2.- Factorización AB’ + AB = A(B’+B)= A 3.- Propiedad Distributiva X+YZ = (X+Y) (X+Z) X (Y+Z) = XY +XZ 4.-Teorema del consenso AB+A’C+BC = AB+A’C 5.-Teorema de Dmorgan (AB)’=A’+ B’ (A+B)’=A’ B’ A+B =(A’ B’)’ AB =(A’+B’)’ AND OR A A=A A + A=A A 0 =0 A + 0 = A A 1 =A A + 1 =1 A A’ =0 A+A’ =1
  • 4. F3 (A, B) = A’+B+A F3 (A, B) = 1 1.-Identidades 2.- Factorización AB’ + AB = A(B’+B)= A 3.- Propiedad Distributiva X+YZ = (X+Y) (X+Z) X (Y+Z) = XY +XZ 4.-Teorema del consenso AB+A’C+BC = AB+A’C 5.-Teorema de Dmorgan (AB)’=A’+ B’ (A+B)’=A’ B’ A+B =(A’ B’)’ AB =(A’+B’)’ AND OR A A=A A + A=A A 0 =0 A + 0 = A A 1 =A A + 1 =1 A A’ =0 A+A’ =1
  • 5. F4 (A,,B,C) = A+A’BC F4 (A,,B,C)=(A+A’)(A+BC) F4 (A,,B,C)=A+BC 1.-Identidades 2.- Factorización AB’ + AB = A(B’+B)= A 3.- Propiedad Distributiva X+YZ = (X+Y) (X+Z) X (Y+Z) = XY +XZ 4.-Teorema del consenso AB+A’C+BC = AB+A’C 5.-Teorema de Dmorgan (AB)’=A’+ B’ (A+B)’=A’ B’ A+B =(A’ B’)’ AB =(A’+B’)’ AND OR A A=A A + A=A A 0 =0 A + 0 = A A 1 =A A + 1 =1 A A’ =0 A+A’ =1
  • 6. F4 (A,,B,C) = A+A’BC F4 (A,,B,C)=(A+A’)(A+BC) F4 (A,,B,C)=A+BC 1.-Identidades 2.- Factorización AB’ + AB = A(B’+B)= A 3.- Propiedad Distributiva X+YZ = (X+Y) (X+Z) X (Y+Z) = XY +XZ 4.-Teorema del consenso AB+A’C+BC = AB+A’C 5.-Teorema de Dmorgan (AB)’=A’+ B’ (A+B)’=A’ B’ A+B =(A’ B’)’ AB =(A’+B’)’ AND OR A A=A A + A=A A 0 =0 A + 0 = A A 1 =A A + 1 =1 A A’ =0 A+A’ =1
  • 7. 1.-Identidades 2.- Factorización AB’ + AB = A(B’+B)= A 3.- Propiedad Distributiva X+YZ = (X+Y) (X+Z) X (Y+Z) = XY +XZ 4.-Teorema del consenso AB+A’C+BC = AB+A’C 5.-Teorema de Dmorgan (AB)’=A’+ B’ (A+B)’=A’ B’ A+B =(A’ B’)’ AB =(A’+B’)’ AND OR A A=A A + A=A A 0 =0 A + 0 = A A 1 =A A + 1 =1 A A’ =0 A+A’ =1