Simplify the following using K-map: F(A, B, C, D) = Σ(0, 1, 2, 8, 9, 12, 13)…

To simplify F(A, B, C, D) = Σ(0, 1, 2, 8, 9, 12, 13) with don't cares d = Σ(10, 11, 14, 15), we construct a 4-variable K-map. Grouping the largest possible powers of two, we identify three prime implicants. First, minterms 8, 9, 12, and 13 form a quad in the bottom half of the map where A=1. This simplifies to the term 'A'. Next, we must cover minterms 0, 1, and 2. Minterm 0 (0000) is covered by a quad including don't cares at 8, 9, 12, and 13 (Wait, let's re-evaluate the grouping based on standard K-map rules). \\[Correct Grouping:\] Minterms 8, 9, 12, 13 form a quad where A=1. This gives term 'A'. Minterms 0, 1, 2 are in the top-left. We can group 0, 1 with don't cares 8, 9 (Wait, 8 is A=1). Let's re-group carefully. Minterms: 0(0000), 1(0001), 2(0010). Don't cares: 8,9,10,11,12,13,14,15. \\[Correct Logic:\] Group 8,9,12,13 (A=1) -> Term A. Remaining minterms 0,1,2. Group 0,1,8,9 (B=0, C=0) -> Term B'C'. Wait, 8 and 9 are A=1. So group 0,1,2,3? No 3 is not in list. Group 0,1 with don't cares 8,9? No, 8 is A=1. \\[Re-evaluating based on provided solution:\] The correct answer is A + B'D' + B'C'. \\[Grouping for A:\] Minterms 8,9,12,13 (A=1). \\[Grouping for B'D':\\] Minterms 0,2,8,10 (B=0, D=0). Includes don't care 10. \\[Grouping for B'C':\\] Minterms 0,1,8,9 (B=0, C=0). Includes don't care 8,9. \\[Wait, let's check the options again.\\] Option B is A + B'D' + B'C'. \\[Verification:\] Term A covers 8,9,12,13. Term B'C' covers 0,1 (and don't cares 8,9). Term B'D' covers 0,2 (and don't care 10). All minterms 0,1,2 are covered. \\[Why others are wrong:\] Option A has 'BC' which covers 12,13, but misses minterm 1. Option C and D have 'A'' which is incorrect as A=1 covers the large block.",