Simplified expression/s for following Boolean function…

Simplified solution:

• Start from the given minterms: 0, 1, 2, 3, 6, 12, 13, 14, 15.

• Notice large groups: A'B' (A = 0, B = 0) covers minterms 0–3; AB (A = 1, B = 1) covers minterms 12–15.

• After these groups the only uncovered minterm is 6 (A=0, B=1, C=1, D=0). We need a product term that covers minterm 6 without introducing any minterms outside the set.

• Two valid choices for a term that include minterm 6 while not adding incorrect minterms are:

• A' C D' : (A = 0, C = 1, D = 0) covers minterms 2 and 6; both are in the function, so adding A' C D' is valid.

• B C D' : (B = 1, C = 1, D = 0) covers minterms 6 and 14; both are in the function, so adding B C D' is also valid.

• Therefore two equivalent simplified forms for F are:

• F = A'B' + AB + A' C D'

• and

• F = A'B' + AB + B C D'

• The other proposed product terms (A' C' D' or B C' D') would introduce minterm 4, which is not in the function, so those forms are invalid.

• Conclusion: the expressions containing A' C D' and containing B C D' are correct representations of F; the expressions that introduce A' C' D' or B C' D' are incorrect.