1. **State the problem:** Simplify the Boolean expression $$F_2 = (AB + A'C)(B + C' + D)(A + B' + D') + (A + C)(B'D + C'A')$$. 2. **Recall Boolean algebra rules:** - Complementarity: $$AA' = 0$$ - Identity: $$A + 0 = A$$, $$A1 = A$$ - Distributive law: $$A(B + C) = AB + AC$$ - Absorption: $$A + AB = A$$ - De Morgan's laws: $$(AB)' = A' + B'$$ 3. **Simplify the first part:** $$P = (AB + A'C)(B + C' + D)(A + B' + D')$$ First, expand $(AB + A'C)(B + C' + D)$: $$= AB(B + C' + D) + A'C(B + C' + D)$$ $$= ABB + ABC' + ABD + A'CB + A'CC' + A'CD$$ Since $$BB = B$$ and $$CC' = 0$$: $$= AB + ABC' + ABD + A'CB + 0 + A'CD$$ Note $$AB + ABC' = AB(1 + C') = AB$$ (since $$1 + C' = 1$$): $$= AB + ABD + A'CB + A'CD$$ Factor $$AB$$: $$= AB(1 + D) + A'CB + A'CD = AB + A'CB + A'CD$$ Now multiply by $(A + B' + D')$: $$P = (AB + A'CB + A'CD)(A + B' + D')$$ Distribute: $$= AB(A + B' + D') + A'CB(A + B' + D') + A'CD(A + B' + D')$$ Calculate each term: - $$AB(A + B' + D') = ABA + ABB' + ABD' = AB + ABB' + ABD'$$ Note $$ABB' = A B B' = 0$$, so: $$= AB + 0 + ABD' = AB + ABD'$$ - $$A'CB(A + B' + D') = A'CB A + A'CB B' + A'CB D' = 0 + A'CB B' + A'CB D'$$ Since $$A' A = 0$$. - $$A'CB B' = A' C B B' = 0$$ - $$A'CB D'$$ remains. - $$A'CD(A + B' + D') = A'CD A + A'CD B' + A'CD D' = 0 + A'CD B' + 0$$ Since $$A' A = 0$$ and $$D D' = 0$$. So: $$P = AB + ABD' + A'CB D' + A'CD B'$$ 4. **Simplify the second part:** $$Q = (A + C)(B'D + C'A')$$ Distribute: $$= A B' D + A C' A' + C B' D + C C' A'$$ Note $$A C' A' = 0$$ (since $$A A' = 0$$) and $$C C' A' = 0$$: $$Q = A B' D + C B' D$$ Factor $$B' D$$: $$Q = B' D (A + C)$$ 5. **Combine:** $$F_2 = P + Q = AB + ABD' + A'CB D' + A'CD B' + B' D (A + C)$$ 6. **Final simplified expression:** $$F_2 = AB + ABD' + A' C B D' + A' C D B' + B' D (A + C)$$ This is a simplified form showing the main terms clearly.