1. **Problem statement:** Simplify the Boolean expression $\overline{A}B + \overline{A}\overline{B}\overline{C} + A\overline{B}\overline{C} + ABC$ and construct its circuit diagram and truth table. 2. **Expression:** $F = \overline{A}B + \overline{A}\overline{B}\overline{C} + A\overline{B}\overline{C} + ABC$ 3. **Group and simplify terms:** - Combine $\overline{A}\overline{B}\overline{C}$ and $A\overline{B}\overline{C}$ using the distributive law: $$\overline{A}\overline{B}\overline{C} + A\overline{B}\overline{C} = \overline{B}\overline{C}(\overline{A} + A) = \overline{B}\overline{C}$$ - So expression becomes: $$F = \overline{A}B + \overline{B}\overline{C} + ABC$$ 4. **Further simplify if possible:** - Consider the term $ABC$; no obvious overlap with $\overline{A}B$ or $\overline{B}\overline{C}$. - This is the simplified form. 5. **Circuit diagram components:** - Inputs: A, B, C - Gates for terms: - $\overline{A}B$: NOT A, AND with B - $\overline{B}\overline{C}$: NOT B, NOT C, AND both - $ABC$: AND A, B, C - OR gate combines these three outputs. 6. **Truth table:** List all $2^3=8$ combinations for A, B, C and compute $F$. | A | B | C | $\overline{A}$ | $\overline{B}$ | $\overline{C}$ | $\overline{A}B$ | $\overline{B}\overline{C}$ | $ABC$ | $F$ | |---|---|---|--------------|--------------|--------------|------------|----------------|-------|-------| | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 7. **Final answer:** $$F = \overline{A}B + \overline{B}\overline{C} + ABC$$ Circuit contains NOT gates for $A$, $B$, and $C$, AND gates for each product term, and an OR gate to combine outputs.