Question

Consider the following 4-variable Boolean function

\[ F(A, B, C, D) = \Sigma m(0, 1, 2, 3, 8, 9, 10, 11) \]
Consider \(A\) as MSB, \(D\) as LSB. Which one of the following options represents the minimal sum of products form for the above function?

Note: + is OR operation, . is AND operation, ' is NOT operation

• A. \(A' + B' + C' + D'\)
• B. \(B'\)
• C. \(A'.B' + A.B\)
• D. \(A'\)

Hint:

Write out all eight minterms in binary and look for a single variable that is 0 in every one of them.

Answer 1

The Correct Option is B

Step 1: Write out each minterm in terms of A, B, C, D.
A minterm number equals \(8A + 4B + 2C + D\) when A is the MSB and D is the LSB. Listing the given minterms in binary (A B C D):
0 = 0000, 1 = 0001, 2 = 0010, 3 = 0011, 8 = 1000, 9 = 1001, 10 = 1010, 11 = 1011.

Step 2: Look for a pattern common to every listed minterm.
Reading the B column (second bit) of every one of these eight binary codes: 0000 has B=0, 0001 has B=0, 0010 has B=0, 0011 has B=0, 1000 has B=0, 1001 has B=0, 1010 has B=0, 1011 has B=0. Every single minterm in the list has \(B=0\).

Step 3: Check that this is the complete set of minterms with \(B=0\).
For a 4-variable function, fixing \(B=0\) leaves A, C, D free, so there are \(2^3 = 8\) minterms with \(B=0\): these are exactly the minterms formed by A taking value 0 or 1 and C, D taking all four combinations, which gives minterm numbers \(0,1,2,3\) (A=0) and \(8,9,10,11\) (A=1). This is precisely the eight minterms listed in the problem, with none missing and none extra.

Step 4: Conclude the minimal expression.
Since \(F=1\) exactly when \(B=0\), and \(F=0\) for every minterm with \(B=1\) (since none of 4,5,6,7,12,13,14,15 appear in the sum), the function is simply the complement of B:
\[ F(A,B,C,D) = B' \] This uses only 1 literal, which is the smallest possible size for a non-trivial 4-variable function, so it cannot be simplified further; it is already the minimal SOP form.

Step 5: Check why the other options are wrong.
Option (A), \(A'+B'+C'+D'\), is true for far more minterms than just B=0; for instance it evaluates to 1 at minterm 15 (A=B=C=D=1) as long as at least one variable is 0 in general, but actually check minterm 4 (A=0,B=1,C=0,D=0): A'=1, so the expression is 1, yet minterm 4 is not in the given list, so this expression is 1 where F should be 0, making option (A) incorrect (it is also structurally a sum term with 4 literals, not even a product term, so it does not fit an SOP form for this problem).
Option (C), \(A'B' + AB\), equals 1 whenever A and B agree (both 0 or both 1). Check minterm 4 (A=0,B=1): A'B'=0, AB=0, so this expression gives 0, correctly excluding minterm 4; but check minterm 12 (A=1,B=1,C=0,D=0): A'B'=0, AB=1, so the expression gives 1, while minterm 12 is not in the given list where F should be 1, so option (C) is wrong (it wrongly includes minterms 12 to 15 for A=B=1). Also it uses 2 product terms and 4 literals, more complex than option (B) even where it happens to match.
Option (D), \(A'\), equals 1 whenever A=0, which covers minterms 0 to 7, but the list only wants 0 to 3 to be 1 among the A=0 group, so it wrongly makes minterm 4 (A=0,B=1,C=0,D=0) equal to 1 when the true function value there is 0 (since 4 is not in the given minterm list). So option (D) is wrong.

Final Answer:
The function depends only on B, and equals 1 exactly when B=0. \[ \boxed{F(A,B,C,D) = B' \ \text{(Option B)}} \]