08 - CircuitosDigitais - Simplificação Expressões Booleana

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Circuitos Lógicos Digitais 
Álgebra Booleana – Simplificação 
de Expressões 
 
Download do software LogicCircuit: http://www.logiccircuit.org/download.html 
Vídeo para apoio: https://youtu.be/4nVoLN7cBAw 
 
Simplificar as expressões: 
 
S = (A + B + C) . (A + B + C) => A+B+C = Y (substituindo por y) 
S = Y . Y 
S = Y 
S= A+B+C 
 
 
 
 
 
 
http://www.logiccircuit.org/download.html
https://youtu.be/4nVoLN7cBAw
 
S = AB + A(B+C) + B(B+C) 
S = AB + AB + AC + BB + BC 
S = AB + AC + BB + BC 
S = AB + AC + B + BC 
S = AB + AC + B 
S = B + AC 
 
 
 
 
 
 
Teorema de DeMorgan: �̅�.�̅� = 𝐴 + 𝐵̅̅ ̅̅ ̅̅ ̅̅ ̅ ou 𝐴 + 𝐵̅̅ ̅̅ ̅̅ ̅̅ ̅ = �̅�.�̅� 
S = ABC + A�̅� + A�̅� 
S = A(BC + 𝐶̅ + �̅�) 
S = A(BC + 𝐶𝐵̅̅ ̅̅ ) = A(BC + 𝐵𝐶̅̅ ̅̅ ) 
S = A(1) = A 
 
 
S = (A + �̅�).(A+C) 
S = AA + AC + �̅�A + �̅�C 
S = A1 + AC + �̅�A + �̅�C 
S = A(1 + C + �̅�) + �̅�C 
S = A(1 + �̅�) + �̅�C 
S = A(1 + �̅�) + �̅�C 
S = A(1) + �̅�C 
S = A + �̅�C 
 
S = A.B.C(A.B+�̅�(B.C+A.C)) 
S = A.B.C(A.B + 𝐶̅.B.C + 𝐶̅.A.C) 
S = A.B.C(A.B + 𝐵𝐶̅C + A𝐶̅C) 
S = A.B.C(A.B + 𝐵0 + A0) 
S = A.B.C(A.B + 0 + 0) 
S = A.B.C(A.B + 0) 
S = ABC(AB) 
S= ABC 
 
S = (A + B) . (𝑨 + 𝑩̅̅ ̅̅ ̅̅ ̅̅ ) => A+B = X 
S = X . �̅� = 0 
S = 0 
Exercícios: 
S = A.B.C + A.𝐂 + A.�̅� 
S = A(BC + 𝐶̅ + �̅�) 
S = A(BC + 𝐶𝐵̅̅ ̅̅ ) 
S = A(BC + 𝐵𝐶̅̅ ̅̅ ) 
S = A(1) 
S = A 
 
S = �̅�.𝐁.̅ 𝐂 + �̅�.B.𝐂 + A.�̅�.C 
S = �̅�. 𝐶̅. �̅� + �̅�.𝐶̅.B + A.�̅�.C 
S = �̅�.𝐶̅’.(�̅� + B) + A.�̅�.C 
S = �̅�.𝐶̅.(1) + A.�̅�.C 
S = �̅�.𝐶̅ + A.�̅�.C 
 
S = Ā.�̅� + Ā.B 
S = A’.B’ + A’.B 
S = A’(B’ + B) ➔ B = 0 ➔ 0+1 = 1 OU ➔ B = 1 ➔1+0 = 1 
S = A’.1 ➔ A’ 
 
S = (A+B+C).(Ā+�̅�+C) 
S = C + (A+B)(A’+B’) 
S = C + (A.A’ + A.B’ + B.A’ + B.B’) ➔ A.A’ = 0 ➔ A = 1 ➔ 1 . 0 = 0 
S = C + (0 + AB’ + BA’ + 0) 
S = C + AB’ + BA’ 
 
 
 
 
 
S = (A+B+C).(Ā+�̅�+C) 
 
S = C + AB’ + BA’ 
 
 
 
 
 
 
 
 
 
 
 
S = A̅. B̅. C̅ + A̅.B.C + A̅.B.C̅+ A.B̅.C̅ + A.B.C̅ 
S = A’(B’.C’ + B.C + B.C’) + A.(B’.C’ + B.C’) 
S = A’(C’.(B’ + B) + B.C) + A.( C’.(B’+B)) 
S = A’(C’.(1) + B.C) + A.C’ 
S = A’(C’ + B.C) + A. C’ ➔ C = 0 B=? ➔ C + C’.B ➔ 0 + 1. B ➔ 0 + B 
 ➔ C = 1 B = ? ➔ 1 + 0 . B ➔ 1 + 0 ➔ 1 ➔ 1 + B 
S = A’.(C’ + B) + A.C’ 
S = A’.C’ + A’.B + A.C’ 
S = C’. (A’ + A) + A’.B 
S = C’.1 + A’.B 
S = C’ + A’.B 
 
ou 
 
S = �̅�. �̅�. 𝐂 + �̅�.B.C + �̅�.B.𝐂+ A.�̅�.𝐂 + A.B.𝐂 
S = B’(A’.C’ + A.C’) + B.(A’.C + A’.C’ + A.C’) 
S = B’(C’. (A’ + A) )+ B. (A’.(C + C’) +A.C’) 
S = B’. C’ + B.(A’ + A.C’) 
S = B’.C’ + B.(A’ + C’) 
S = B’.C’ + B.A’ + B.C’ 
S = C’(B’ + B) + B.A’ 
S = C’ + B.A’ 
 
S = A'B + AB' + AB 
S = B.(A’ + A) + AB’ 
S = B + AB’ 
S = B + A 
 
 
 
S = [(A + B).C]' + [D .(C + B)]' 
S = ((A+B).C)’ + (D.(C+B)’ 
S = (A’+B’).C’ + D’.(C’+B’) 
S = (A’+B’) + C’ + D’ + (C’ + B’) 
S = A’.B’ + C’.1 + D’ + C’.B’ 
S = A’.B’ + D’ + C’(1+B’) 
S = A’.B’ + D’ + C’(1) 
S = A’.B’ + D’ + C’