Boolean simplification — minimize F = ABD + CD + ACD + ABC + ABCD Using Boolean algebra (no K-map required), find the most simplified SOP form from the options provided.

Introduction / Context:
Algebraic simplification reduces hardware by removing redundant terms. For the given four-variable function, we apply absorption and factoring to reach a minimal or near-minimal sum-of-products (SOP) that matches one of the choices.

Given Data / Assumptions:

• F = ABD + CD + ACD + ABC + ABCD.
• Variables A, B, C, D are Boolean.
• Operators: + is OR, concatenation is AND.

Concept / Approach:
Use absorption: X + XY = X, and redundancy removal where a more specific product term is covered by a more general one. Combine terms where a factorization can eliminate literals without altering coverage of minterms.

Step-by-Step Solution:
Start: F = ABD + CD + ACD + ABC + ABCD.Observe ABCD is contained in ABD (and also in ACD and ABC). Apply absorption: remove ABCD.Now F = ABD + CD + ACD + ABC.Notice CD + ACD = CD(1 + A) = CD. Absorb ACD into CD.Thus F simplifies to F = ABD + ABC + CD.

Verification / Alternative check:
Factor the remaining two AB terms: F = AB(C + D) + CD. Re-expanding yields ABC + ABD + CD, identical to the simplified expression. A truth table or logic simulator would confirm equivalence with the original function across all 16 input combinations.

Why Other Options Are Wrong:

• F = CD + AD: AD covers minterms where B = 0 that are not in the original, so it overstates the function.
• F = BC + AB: Omits required D-dependent coverage and includes minterms not present.
• F = AC + AD: Again introduces minterms not guaranteed by the original and misses specific AB-dependent ones.

Common Pitfalls:

• Dropping CD entirely after absorbing ACD; CD must remain.
• Over-aggressive factoring leading to terms like AD or AB alone, which add unwanted minterms.

Final Answer:
F = ABD + ABC + CD