GATE CS 2025 Set 2 Q40 — Boolean Algebra

We analyze each Boolean equation individually:

• Option A: Let us express both sides in minterm form for variables $(A, B, C)$:
  • $\text{LHS} = \bar{A} B C + A \bar{B} \bar{C} + \bar{A} \bar{B} \bar{C} + A \bar{B} C + A B C = m_3 + m_4 + m_0 + m_5 + m_7 = \sum m(0, 3, 4, 5, 7)$
  • $\text{RHS} = B C + \bar{B} \bar{C} + \bar{A} \bar{B} = (m_3 + m_7) + (m_0 + m_4) + (m_0 + m_1) = \sum m(0, 1, 3, 4, 7)$
  Since $\text{LHS} \neq \text{RHS}$ (for example, at $A=0, B=0, C=1$, $\text{LHS} = 0$ while $\text{RHS} = 1$), Option A is incorrect.
• Option B: By the standard Consensus Theorem ($X Y + \bar{X} Z + Y Z = X Y + \bar{X} Z$ with $X = A, Y = B, Z = C$): $$A B + \bar{A} C + B C = A B + \bar{A} C$$ Thus, Option B is correct.
• Option C: Expanding the left-hand side: $$(A+C)(\bar{A}+B) = A\bar{A} + AB + \bar{A}C + BC = AB + \bar{A}C + BC$$ Applying the Consensus Theorem to the result yields $AB + \bar{A}C$. Thus, Option C is correct.
• Option D: Let $F$ be the expression inside the complement: $$F = (A+\bar{B}+\bar{D})(C+D)(\bar{A}+C+D)(A+B+\bar{D})$$ Grouping the terms:
  • $(A+\bar{B}+\bar{D})(A+B+\bar{D}) = (A+\bar{D}) + B\bar{B} = A+\bar{D}$
  • $(C+D)(\bar{A}+C+D) = C+D$ (by absorption law)
  Thus, $F = (A+\bar{D})(C+D)$. Taking the complement of $F$ using De Morgan's laws: $$\bar{F} = \overline{(A+\bar{D})(C+D)} = \overline{(A+\bar{D})} + \overline{(C+D)} = \bar{A}D + \bar{C}\bar{D}$$ Thus, Option D is correct.