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Resolva os circuitos booleanos: S = (ABC)(\overlin...

Criado em: 1 de junho de 2025

Respondido usando O3 por Chat01

Pergunta

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Resolva os circuitos booleanos:
S = (ABC)(\overline{A} + \overline{B} + \overline{C})
S = (A + B + C)(\overline{A} + \overline{B} + C)
S = \overline{A}BC + A\overline{B}C + \overline{A}B\overline{C} + ABC + AB\overline{C}
S = \overline{A}B + A\overline{B} + AB
S = (A + \overline{B} + AB).(A + \overline{B}).(\overline{A}B)
S = (A + \overline{B} + AB).(AB + \overline{A}C + BC)
V = (W + X + Y).(\overline{W} + \overline{X} + Y).(\overline{Y} + Z).(W + Z)
y = A\overline{B}D + AB\overline{D}
z = (\overline{A} + B)(A + B)
x = ACD + \overline{A}\overline{B}CD

Python

python
A,B,C,D,W,X,Y,Z=sp.symbols('A B C D W X Y Z')
expr1 = A&B&C & (~A | ~B | ~C)
expr1_simplified=sp.simplify_logic(expr1,form='dnf')
expr1_simplified, sp.simplify_logic(expr1,'cnf')


(False, False)

Python

python
expr2 = (A|B|C) & (~A|~B|C)
sp.simplify_logic(expr2,'dnf'), sp.simplify_logic(expr2,'cnf')


(C | (A & ~B) | (B & ~A), (A | B | C) & (C | ~A | ~B))

Python

python
expr3 = (~A&B&C) | (A&~B&C) | (~A&B&~C) | (A&B&C) | (A&B&~C)
sp.simplify_logic(expr3,'dnf'), sp.simplify_logic(expr3,'cnf')


(B | (A & C), (A | B) & (B | C))

Python

python
expr4 = (~A&B) | (A&~B) | (A&B)
sp.simplify_logic(expr4,'dnf')


A | B

Python

python
expr5 = (A | ~B | (A&B)) & (A | ~B) & (~A & B)
sp.simplify_logic(expr5,'dnf'), sp.simplify_logic(expr5,'cnf')


(False, False)

Python

python
expr6 = (A | ~B | (A&B)) & ((A&B) | (~A & C) | (B & C))
sp.simplify_logic(expr6,'dnf'), sp.simplify_logic(expr6,'cnf')


((A & B) | (C & ~A & ~B), (A | C) & (A | ~B) & (B | ~A))

Python

python
W,X,Y,Z=sp.symbols('W X Y Z')
expr7 = (W|X|Y) & (~W|~X|Y) & (~Y|Z) & (W|Z)
sp.simplify_logic(expr7,'dnf'), sp.simplify_logic(expr7,'cnf')


((Y & Z) | (X & Z & ~W) | (W & ~X & ~Y),
 (W | Z) & (Z | ~Y) & (W | X | Y) & (Y | ~W | ~X))

Python

python
yexpr = A&~B&D | A&B&~D
sp.simplify_logic(yexpr,'dnf'), sp.simplify_logic(yexpr,'cnf')


((A & B & ~D) | (A & D & ~B), A & (B | D) & (~B | ~D))

Python

python
zexpr = (~A | B) & (A | B)
sp.simplify_logic(zexpr,'dnf')


B

Python

python
xexpr = A&C&D | (~A & ~B & C & D)
sp.simplify_logic(xexpr,'dnf')


(A & C & D) | (C & D & ~B)

Pensamento

Simplifying Boolean Expressions

For Expression 10, x = ACD + ¬A¬BCD, I factor out CD to get: CD(A + ¬A¬B). Simplifying inside the parentheses with a well-known rule (X + ¬XY = X + Y), I get A + ¬B. So, the expression simplifies to CD(A + ¬B). The minimal form is as CD(A + ¬B). Now, the next step is to craft my response in Portuguese, explaining the simplification process in a clear, step-by-step manner.

Simplifying Boolean Expressions

The user seems to want confirmation of their circuit simulations. They shared images of logic gates corresponding to different Boolean expressions, and I will simplify these expressions and compare them to the user's expected simulations. I'll also mention truth table results along the way. For example, some expressions simplify to 0 or 1, and I’m providing the step-by-step algebraic manipulation.

I’ll organize it as a list:

  1. (ABC)(¬A + ¬B + ¬C) = 0
  2. (A + B + C)(¬A + ¬B + C) = C + (A ⊕ B)

Thought for 3m 4s

Resposta

Soluções e formas mínimas

#Expressão originalPassos-chave de simplificação¹Forma mínima / comentário
1S=(ABC)(A‾+B‾+C‾)S=(ABC)(\overline A+\overline B+\overline C)• O produto só pode ser 1 quando A=B=C=1A=B=C=1.<br>• Nesse ponto, A‾+B‾+C‾=0\overline A+\overline B+\overline C=0.<br>• Logo o produto é sempre 0.S=0S = 0 (nível baixo constante)
2S=(A+B+C)(A‾+B‾+C)S=(A+B+C)(\overline A+\overline B+C)Expansão parcial → eliminação de termos redundantesS=C+AB‾+A‾B=C+(A⊕B)S = C + A\overline B + \overline A B = C + (A\oplus B)
3S=A‾BC+AB‾C+A‾BC‾+ABC+ABC‾S=\overline ABC+A\overline BC+\overline AB\overline C+ABC+AB\overline CAgrupar por BB e por ACAC → absorçãoS=B+AC=(A+B)(B+C)S = B + AC = (A+B)(B+C)
4S=A‾B+AB‾+ABS=\overline AB+A\overline B+ABAB+AB‾=AAB+A\overline B=A → A+A‾B=A+BA+\overline AB=A+BS=A+BS = A + B
5S=(A+B‾+AB)(A+B‾)(A‾B)S=(A+\overline B+AB)(A+\overline B)(\overline AB)Termo final força A=0, B=1A=0,\,B=1; isso zera o 1.º fatorS=0S = 0
6S=(A+B‾+AB)(AB+A‾C+BC)S=(A+\overline B+AB)(AB+\overline AC+BC)Usar X+XY=XX+XY=X e consensoS=AB+A‾ B‾ CS = AB + \overline A\,\overline B\,C
7V=(W+X+Y)(W‾+X‾+Y)(Y‾+Z)(W+Z)V=(W+X+Y)(\overline W+\overline X+Y)(\overline Y+Z)(W+Z)Mapas de Karnaugh ou consensoV=YZ+XZW‾+WX‾Y‾V = YZ + XZ\overline W + W\overline X\overline Y
8y=AB‾D+ABD‾y=A\overline BD+AB\overline DColher AA → A(B‾D+BD‾)=A(B⊕D)A(\overline BD+B\overline D)=A(B\oplus D)y=A(B⊕D)y = A(B\oplus D)
9z=(A‾+B)(A+B)z=(\overline A+B)(A+B)(B+A‾)(B+A)=B+(A‾A)=B(B+\overline A)(B+A)=B+(\overline A A)=Bz=Bz = B
10x=ACD+A‾ B‾ CDx=ACD+\overline A\,\overline B\,CDFatorar CDCD → CD(A+B‾)CD(A+\overline B)x=CD(A+B‾)x = CD(A+\overline B)

¹ Leis usadas com mais frequência

  • Absorção X+XY=XX + X Y = X  X(X+Y)=XX(X+Y) = X
  • Complementação X+X‾=1,  XX‾=0X+\overline X = 1,\; X\overline X = 0
  • Distributiva X(Y+Z)=XY+XZX(Y+Z) = XY+XZ
  • Consenso XY+X‾Z+YZ=XY+X‾ZXY + \overline X Z + YZ = XY + \overline X Z
  • XOR B⊕D=BD‾+B‾DB\oplus D = B\overline D + \overline B D

Com estas reduções você pode redesenhar cada circuito usando muito menos portas lógicas (ou, nos casos 1 e 5, nenhuma!), facilitando implementação e teste.

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