The function is a product of sums, so it equals 0 wherever any factor is 0. Find the minterms where each factor vanishes (each sum-term is 0 only at the single assignment that makes every literal 0).
| Factor | Makes it 0 | Minterms (ABCD) |
|---|
| A+B+Cˉ+Dˉ | A=0,B=0,C=1,D=1 | 3 |
| Aˉ+C+Dˉ | A=1,C=0,D=1 | 9,13 |
| Aˉ+B+Cˉ+Dˉ | A=1,B=0,C=1,D=1 | 11 |
| Bˉ+C | B=1,C=0 | 4,5,12,13 |
| Bˉ+Cˉ | B=1,C=1 | 6,7,14,15 |
| A+Bˉ | A=0,B=1 | 4,5,6,7 |
| Bˉ+Dˉ | B=1,D=1 | 5,7,13,15 |
Union of zeros: {3,4,5,6,7,9,11,12,13,14,15}. Hence the ones are the rest:
F=∑m(0,1,2,8,10).
Every one-minterm has B=0, so Bˉ factors out. Restricting to B=0 and looking at (A,C,D), the ones are 000,001,010,100,110 and the zeros are 011,101,111. All zeros have D=1 together with A+C, i.e. the off-set is D(A+C):
G(A,C,D)=D(A+C)=Dˉ+A+C=Dˉ+AˉCˉ.
Therefore
F=Bˉ(Dˉ+AˉCˉ),
which is option 1. Check the cover: Dˉ catches D=0 (000,010,100,110) and AˉCˉ catches 000,001 — together exactly the on-set.
Exam trick: For a POS function, don't multiply everything out. Each sum-term kills exactly one (or one block of) minterms — list those, take the complement to get the small on-set, then read the answer off a tiny K-map.