This question requires us to find the simplified Boolean expression for a given function $F$ represented in sum-of-product form using minterms. We will use the Karnaugh map (K-map) method for minimization.

Understanding the Boolean Function and Minterms

The function is given as $F(b_3, b_2, b_1, b_0) = \sum(0, 2, 4, 8, 10, 11, 12)$. This notation means that the function $F$ is TRUE (or evaluates to 1) for the specific input combinations that correspond to the decimal values 0, 2, 4, 8, 10, 11, and 12. The variables $b_3, b_2, b_1, b_0$ represent a 4-bit binary number, where $b_3$ is the most significant bit (MSB) and $b_0$ is the least significant bit (LSB).

Let's list the binary representations for each minterm:

• Minterm 0: $0000$
• Minterm 2: $0010$
• Minterm 4: $0100$
• Minterm 8: $1000$
• Minterm 10: $1010$
• Minterm 11: $1011$
• Minterm 12: $1100$

Karnaugh Map (K-map) Minimization

The Karnaugh map is a visual tool used to simplify Boolean expressions. For a 4-variable function ($b_3, b_2, b_1, b_0$), we use a 4x4 K-map. The key is that adjacent cells in the map (horizontally, vertically, and wrapping around the edges) differ by only one variable. This adjacency property allows us to group terms that can be simplified.

Setting Up and Plotting the K-map

We construct a 4x4 grid. The rows represent combinations of $b_3b_2$ and the columns represent combinations of $b_1b_0$. The order of rows and columns must follow the Gray code sequence (00, 01, 11, 10) to maintain adjacency.

We place a '1' in each cell of the K-map that corresponds to a minterm in the function $F$. All other cells are filled with '0' (or left blank, implying '0').

Grouping the 1s

The next step is to group the adjacent '1's into the largest possible rectangular blocks. The size of these blocks must be a power of 2 (1, 2, 4, 8, etc.). Each group represents a simplified product term. We need to cover all '1's using the minimum number of groups.

Observing the K-map, we can form the following groups:

1. Group 1: A block of four '1's in the $b_1b_0=00$ column, spanning all rows ($b_3b_2=00, 01, 11, 10$). This group covers minterms {0, 4, 8, 12}. The variables $b_1$ and $b_0$ are constant at 0. The term is $\overline{b_1}\overline{b_0}$.
2. Group 2: A block of two '1's in the $b_3b_2=00$ row, covering $b_1b_0=00$ and $b_1b_0=10$. This group covers minterms {0, 2}. The variables $b_3$, $b_2$, and $b_0$ are constant at 0. The term is $\overline{b_3}\overline{b_2}\overline{b_0}$.
3. Group 3: A block of two '1's in the $b_3b_2=10$ row, covering $b_1b_0=10$ and $b_1b_0=11$. This group covers minterms {10, 11}. The variables $b_3$, $b_2$, and $b_1$ are constant at 1, 0, and 1, respectively. The term is $b_3\overline{b_2}b_1$.

These three groups collectively cover all the minterms of the function F: {0, 4, 8, 12} $\cup$ {0, 2} $\cup$ {10, 11} = {0, 2, 4, 8, 10, 11, 12}. These groups are also prime implicants, and by selecting these, we obtain the minimal sum-of-products expression.

Minimal Boolean Expression Derived

The minimized Boolean expression for $F$, derived from these groups, is the sum of the terms:

$F = \overline{b_1}\overline{b_0} + \overline{b_3}\overline{b_2}\overline{b_0} + b_3\overline{b_2}b_1$

Analysis of Provided Options

The question asks to select the correct minimized Boolean expression from the given options. Based on the standard K-map minimization process described above, the expression derived is $F = \overline{b_1}\overline{b_0} + \overline{b_3}\overline{b_2}\overline{b_0} + b_3\overline{b_2}b_1$. We will now present the option text identified as the correct answer in the problem statement.

The correct answer option text provided is:

This corresponds to the expression:

$ \overline{b_1}\,b_0 + \overline{b_2}\,\overline{b_0} + b_1 b_2 b_3 $