GATE CS 1995 Q2.5 — Boolean Algebra

We can solve the simultaneous Boolean equations step-by-step:

1. First Equation: $\overline{A} + AB = 0$

   Using the Boolean simplification identity $\overline{X} + XY = \overline{X} + Y$:

   $$\overline{A} + B = 0$$

   For the sum of two Boolean terms to be $0$, each term must be individually $0$:

   $$\overline{A} = 0 \implies A = 1$$$$B = 0$$
2. Second Equation: $AB = AC$

   Substituting $A = 1$ and $B = 0$:

   $$1 \cdot 0 = 1 \cdot C \implies 0 = C \implies C = 0$$
3. Third Equation: $AB + A\overline{C} + CD = \overline{C}D$

   Substituting $A = 1$, $B = 0$, and $C = 0$:

   $$\text{LHS} = (1 \cdot 0) + (1 \cdot \overline{0}) + (0 \cdot D) = 0 + 1 + 0 = 1$$$$\text{RHS} = \overline{0} \cdot D = 1 \cdot D = D$$

   Equating LHS and RHS gives:

   $$1 = D \implies D = 1$$

Thus, the unique solution is $A = 1, B = 0, C = 0, D = 1$.