1. **State the problem:** Minimise the Boolean expression $$\overline{A} + \overline{B} + \overline{C} + \overline{A} \cdot \overline{C} \cdot D + A \cdot \overline{B} \cdot C \cdot D + A \cdot \overline{B} \cdot C + \overline{A} \cdot \overline{C} \cdot A \cdot \overline{C} \cdot (A + B + C + D)$$ using Boolean algebra rules and De Morgan's theorem. 2. **Simplify terms step-by-step:** - Note that $$\overline{A} \cdot \overline{C} \cdot A = 0$$ because $$A \cdot \overline{A} = 0$$. - Therefore, the term $$\overline{A} \cdot \overline{C} \cdot A \cdot \overline{C} \cdot (A + B + C + D) = 0$$. 3. **Rewrite the expression without the zero term:** $$\overline{A} + \overline{B} + \overline{C} + \overline{A} \cdot \overline{C} \cdot D + A \cdot \overline{B} \cdot C \cdot D + A \cdot \overline{B} \cdot C$$ 4. **Use absorption and consensus rules:** - Since $$\overline{A} + \overline{C}$$ is already present, the term $$\overline{A} \cdot \overline{C} \cdot D$$ is absorbed by $$\overline{A} + \overline{C}$$. - So the expression reduces to: $$\overline{A} + \overline{B} + \overline{C} + A \cdot \overline{B} \cdot C \cdot D + A \cdot \overline{B} \cdot C$$ 5. **Factor terms with $$A \cdot \overline{B} \cdot C$$:** $$A \cdot \overline{B} \cdot C \cdot D + A \cdot \overline{B} \cdot C = A \cdot \overline{B} \cdot C (D + 1) = A \cdot \overline{B} \cdot C$$ 6. **Final simplified expression:** $$\overline{A} + \overline{B} + \overline{C} + A \cdot \overline{B} \cdot C$$ 7. **Interpretation:** The expression is minimal because the first three terms cover all cases where any of $$A, B, C$$ is zero, and the last term covers the case when $$A=1, B=0, C=1$$. **Answer:** $$\boxed{\overline{A} + \overline{B} + \overline{C} + A \cdot \overline{B} \cdot C}$$