To simplify the Boolean expression using a Karnaugh map (K-map), we'll proceed step by step.

Step 1: Create the Karnaugh Map Layout

We are given a Boolean function in four variables $A$, $B$, $C$, and $D$. The K-map for four variables will be a 4x4 grid, where the rows and columns represent different combinations of two variables each.

For convenience, let's assign the variables $A$ and $B$ to the rows, and $C$ and $D$ to the columns. The labeling of the K-map cells will follow Gray code, where only one bit changes between adjacent cells.

Here's how the grid is labeled:

AB\CD	00	01	11	10
00	0	1	3	2
01	4	5	7	6
11	12	13	15	14
10	8	9	11	10

Step 2: Fill in the Karnaugh Map

We are provided with two sets of information:

• $f^{-1}(1) = \{2, 3, 5, 7, 11, 13, 14, 15\}$, which represents the minterms where the function is 1.
• $f^{-1}(0) = \{0, 1, 4, 6, 8, 9, 10, 12\}$, which represents the minterms where the function is 0.

Now, we'll place 1s in the cells corresponding to $f^{-1}(1)$ and 0s in the cells corresponding to $f^{-1}(0)$.

AB\CD	00	01	11	10
00	0	0	1	1
01	0	1	1	0
11	0	1	1	1
10	0	0	1	0

Step 3: Group the 1s in the Karnaugh Map

Now we group the adjacent 1s in the K-map. The goal is to create the largest possible rectangular groups of size $2^n$ (1, 2, 4, 8, etc.), ensuring that each group contains only 1s.

1. Group 1: A block of four 1s in cells corresponding to minterms {2, 3, 6, 7}. This group can be simplified to: $B'C$ because $A = 0$ (for rows 00 and 01) and $C = 1$.

2. Group 2: A block of four 1s in cells corresponding to minterms {11, 13, 14, 15}. This group simplifies to: $A D$ because $B = 1$ (for rows 10 and 11) and $D = 1$.

Step 4: Write the Final Simplified Expression

The simplified Boolean expression is the sum of the terms found in the grouping step: $f(A, B, C, D) = B'C + A D$

Final Simplified Boolean Expression:

$f(A, B, C, D) = B'C + A D$

Would you like more details about the grouping process or a breakdown of how we arrived at each term?

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Here are five follow-up questions to explore:

1. How does the Karnaugh map change if you switch one of the variables in rows with those in columns?
2. What happens if the function had “don’t care” conditions in addition to 0s and 1s?
3. How can we apply Quine–McCluskey method instead of a K-map for this function?
4. How would the process differ if we had five variables instead of four?
5. Can you simplify the same function without using a Karnaugh map?

Tip: When grouping in a Karnaugh map, always aim to create the largest possible groups, as this will minimize the Boolean expression.