The problem asks for the simplified Boolean expression of the given complex expression:

$(a \land c' \land d') \lor (a \land b' \land d) \lor (a \land c \land d')$

We can simplify this expression using Boolean algebra rules, such as the Distributive, Absorption, and Complement Laws.

Step-by-step simplification:

1. Factor out the common term:
   All three terms share a common factor of $a$. So we factor it out: $a \land [(c' \land d') \lor (b' \land d) \lor (c \land d')]$

2. Simplify the inner expression:
   Now, simplify the expression inside the brackets using Boolean rules: $(c' \land d') \lor (b' \land d) \lor (c \land d')$ We notice that $d'$ appears in both $(c' \land d')$ and $(c \land d')$, so we can factor $d'$ from them: $d' \land (c' \lor c) \lor (b' \land d)$

3. Apply the Complement Law:
   $c' \lor c$ equals 1 (as per the complement law), so the expression simplifies to: $d' \lor (b' \land d)$

4. Final simplified expression:
   Substituting this back into the original expression gives: $a \land (d' \lor (b' \land d))$

This is the simplified Boolean expression.

Checking the options:

The correct answer corresponds to option (iv) $a \lor (b \land d')$.

Would you like more details on any of the simplification steps or Boolean algebra rules used?

5 Relative Questions:

1. What is the Absorption Law in Boolean algebra, and how does it apply to simplifying expressions?
2. How does the Distributive Law help in simplifying Boolean expressions?
3. Can you explain why $c' \lor c$ simplifies to 1 in Boolean logic?
4. What is the Complement Law, and how is it useful in Boolean algebra?
5. How would this simplification change if the variables represented different logic gates (AND, OR, NOT)?

Tip:

In Boolean algebra, always look for common factors in terms to simplify expressions efficiently.