1. **State the problem:** Simplify the Boolean expression $$A + ABC + AB\bar{E} + (A + B + \bar{C})\bar{B}C$$. 2. **Recall Boolean algebra rules:** - $X + XY = X$ (Absorption law) - $X + X\bar{Y} = X + Y$ (Consensus theorem) - $\bar{\bar{X}} = X$ (Double negation) - $X + X = X$ (Idempotent law) - $XY + X\bar{Y} = X$ (Distributive law) 3. **Simplify the first part:** $$A + ABC + AB\bar{E}$$ Using absorption, $A + ABC = A$ because $A$ covers $ABC$. So, $$A + AB\bar{E} = A$$ Again, $A$ covers $AB\bar{E}$. 4. **Simplify the second part:** $$(A + B + \bar{C})\bar{B}C$$ Distribute: $$A\bar{B}C + B\bar{B}C + \bar{C}\bar{B}C$$ Note $B\bar{B} = 0$, so second term is 0. Also, $\bar{C}C = 0$, so third term is 0. So, $$A\bar{B}C$$ 5. **Combine both parts:** $$A + A\bar{B}C$$ Using absorption, $A + A\bar{B}C = A$. **Final simplified expression:** $$A$$