Which of the following Boolean algebraic equation(s) is/are CORRECT?

Let us verify each option using Boolean algebra rules:

• Option (A):

  $$\text{LHS} = \bar{A} B C + A \bar{B} \bar{C} + \bar{A} \bar{B} \bar{C} + A \bar{B} C + A B C$$

  Combine terms:

  $$(\bar{A} B C + A B C) + (A \bar{B} \bar{C} + \bar{A} \bar{B} \bar{C}) + A \bar{B} C = B C + \bar{B} \bar{C} + A \bar{B} C$$

  Using absorption ($BC + A\bar{B}C = BC + AC$):

  $$= BC + \bar{B}\bar{C} + AC$$

  $$\text{RHS} = BC + \bar{B}\bar{C} + \bar{A}\bar{B}$$

  For $A=0, B=0, C=1$, $\text{LHS} = 0$, but $\text{RHS} = \bar{A}\bar{B} = 1$. Thus, $\text{LHS} \neq \text{RHS}$. (Incorrect)

• Option (B):

  $$\text{LHS} = A B + \bar{A} C + B C$$

  By the Consensus Theorem, $BC$ is redundant because $A$ and $\bar{A}$ are present in $AB$ and $\bar{A}C$.

  $$\text{LHS} = A B + \bar{A} C = \text{RHS} \quad \text{(Correct)}$$

• Option (C):

  $$\text{LHS} = (A + C)(\bar{A} + B) = A \bar{A} + A B + \bar{A} C + B C = 0 + A B + \bar{A} C + B C$$

  By the Consensus Theorem, $AB + \bar{A}C + BC = AB + \bar{A}C = \text{RHS}$. (Correct)

• Option (D): Let $F = (A + \bar{B} + \bar{D})(C + D)(\bar{A} + C + D)(A + B + \bar{D})$. Combine terms:

  • $(A + \bar{B} + \bar{D})(A + B + \bar{D}) = (A + \bar{D}) + \bar{B}B = A + \bar{D}$

  • $(C + D)(\bar{A} + C + D) = C + D$ (by absorption)

  Therefore:

  $$F = (A + \bar{D})(C + D)$$

  Applying De Morgan's Laws to find the complement:

  $$\overline{F} = \overline{(A + \bar{D})(C + D)} = \overline{A + \bar{D}} + \overline{C + D} = \bar{A} D + \bar{C} \bar{D} = \text{RHS} \quad \text{(Correct)}$$

Therefore, options B, C, and D are correct.