The document discusses techniques for minimizing Boolean functions including:
1. Algebraic manipulation using factorization, duplicating terms, consensus theorem, and distributive property.
2. Karnaugh maps for factorization.
3. Examples are provided demonstrating factorization and simplification of Boolean functions using the discussed techniques.
11. m S P E CS CP CE
0 0 0 0 0 0 0
1 0 0 1 0 0 1
2 0 1 0 0 1 0
3 0 1 1 0 1 0
4 1 0 0 1 0 0
5 1 0 1 1 0 0
6 1 1 0 1 0 0
7 1 1 1 1 0 0
CS (S, P, E ) = S P’ E’ + S P’ E
12. m S P E CS CP CE
0 0 0 0 0 0 0
1 0 0 1 0 0 1
2 0 1 0 0 1 0
3 0 1 1 0 1 0
4 1 0 0 1 0 0
5 1 0 1 1 0 0
6 1 1 0 1 0 0
7 1 1 1 1 0 0
CS (S, P, E ) = S P’ E’ + S P’ E + S P E’
13. m S P E CS CP CE
0 0 0 0 0 0 0
1 0 0 1 0 0 1
2 0 1 0 0 1 0
3 0 1 1 0 1 0
4 1 0 0 1 0 0
5 1 0 1 1 0 0
6 1 1 0 1 0 0
7 1 1 1 1 0 0
CS (S, P, E ) = S P’ E’ + S P’ E + S P E’ + S P E
14. CS (S, P, E ) = S P’ E’ + S P’ E + S P E’ + S P E
CS (S, P, E ) = S P’ (E’+E)+ S P (E’+E)
CS (S, P, E ) = S P’ + S P
CS (S, P, E ) = S (P’+P)
CS (S, P, E ) = S
15. m S P E CS CP CE
0 0 0 0 0 0 0
1 0 0 1 0 0 1
2 0 1 0 0 1 0
3 0 1 1 0 1 0
4 1 0 0 1 0 0
5 1 0 1 1 0 0
6 1 1 0 1 0 0
7 1 1 1 1 0 0
CS (S, P, E ) = S P’ E’ + S P’ E + S P E’ + S P E
CS (S,) = S
16. m S P E CS CP CE
0 0 0 0 0 0 0
1 0 0 1 0 0 1
2 0 1 0 0 1 0
3 0 1 1 0 1 0
4 1 0 0 1 0 0
5 1 0 1 1 0 0
6 1 1 0 1 0 0
7 1 1 1 1 0 0
CS (S, P, E ) = S P’ E’ + S P’ E + S P E’ + S P E = S
CS (S,) = S
17. m S P E CS CP CE
0 0 0 0 0 0 0
1 0 0 1 0 0 1
2 0 1 0 0 1 0
3 0 1 1 0 1 0
4 1 0 0 1 0 0
5 1 0 1 1 0 0
6 1 1 0 1 0 0
7 1 1 1 1 0 0
CP (S, P, E ) = S’ P E’ + S’ P E
CP (S, P, E ) = S’ P (E’+E)
CP (S, P, E ) = S’ P
CP (S, P ) = S’ P
19. m A B C X
0 0 0 0 1
1 0 0 1 1
2 0 1 0 1
3 0 1 1 0
4 1 0 0 0
5 1 0 1 0
6 1 1 0 0
7 1 1 1 0
FX (A,B,C) = A’ B’ C’ + A’ B’ C + A’ B C’
FX (A,B,C) = A’ B’ C’ + A’ B’ C
+ A’ B C’ + A’ B’ C’
FX (A,B,C) = A’ B’ + A’ C’
FX (A,B,C) = A’ (B’ + C’)
28. F= A’ B + A B’ + A B + A’ C’
F= B + A + A’ C’
F= B + (A+ A’)(A+ C’)
F= B + A+ C’ F= A+B+C’
29. Actividad
Usando como recursos
• Factorización
• Duplicando un termino ya existente
• Teorema del consenso
• Propiedad distributiva
• Identidades
• Teorema de Dmorgan
Resuelva las siguientes funciones
30. 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
1+ B’+ C
2
DC’(0)
3
A’+B+A
4
A+ A’ BC
5
A’BC+A’BC’
31. 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
32. 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
33. 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
34. 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