Simplify:
​$$Z =\overline{(A+\overline{B+C}+)D}$$​​

Correct option is D

1. Scope of complement: The overline is over $$(A+\overline{B+C})D$$​; parentheses determine exactly what’s being complemented.
2. Inner De Morgan: $$\overline{B+C}=\bar B\,\bar C$$​ converts an OR under a bar to an AND of complements.
3. Outer De Morgan: $$\overline{X\cdot D}=\bar X+\bar D$$​ turns the complemented product into a sum of complements.
4. Complement of a sum-with-product: $$\overline{A+\bar B\bar C}=\bar A(B+C)$$​ by chaining De Morgan: $$\overline{Y+Z}=\bar Y\,\bar Z$$​ and $$\overline{UV}=\bar U+\bar V$$​.
5. Equivalent forms: $$\bar A(B+C)+\bar D \equiv \bar A B+\bar A C+\bar D$$​; choose based on gate count/fan-in.
6. Implementation hint: Realize as one OR for B + C, one AND with $$\bar A$$​, then OR with $$\bar D$$​; inverters on A and D.

Knowledge Booster

• Why (a) $$\bar A + BC + \bar D$$​ is wrong: It asserts output 1 whenever $$\bar A=1$$ even if B = C = 0, but the correct form needs $$\bar A(B+C)$$​ (requires B or C).
• Why (b) $$\bar A B + C\bar D$$​ is wrong: It lacks the $$\bar A C$$​ term when D = 1, and its $$C\bar D$$​ part turns 1 for D = 0 regardless of A, B (too permissive/different structure than $$\bar D+\bar A(B+C$$​).
• Why (c) $$A(B+\bar C)+D$$​ is wrong: The trailing +D makes the function 1 for all D = 1, ignoring the needed $$\bar A$$​ condition; also activates for many A = 1 cases that the correct expression excludes.
• Extra check: Plug A = 0, B = 0, C = 0, D = 1: correct $$Z=\bar A(B+C)+\bar D=0$$​. Options (a)/(c) would incorrectly turn 1 due to $$\bar A$$​ or +D.
• Common pitfall: Misreading the bar’s scope. If only ($$A+\overline{B+C}$$​) were complemented (not the product), the result would be the single minterm $$\bar A\,B\,\bar C\,D$$​ —a different function.