Minimize the given Boolean function using K map and implement the simplified function using only NAND gates. F(A, B, C, D) = ∑ (0, 1, 2, 9, 11, 15) + d(8, 10, 14) Method F(A, B, C, D) = ∑ (0, 2, 3, 6, 7, 8, 9, 10)

To minimize the given Boolean function using a Karnaugh map (K-map) and then implement the simplified function using only NAND gates, we'll follow these steps:

Step 1: Construct the K-map

A K-map for a 4-variable function has 16 cells, corresponding to the 16 possible combinations of the variables A, B, C, and D. We will fill in the K-map with 1s for the minterms of the function and Xs for the don't-care conditions.

The given function is: [ F(A, B, C, D) = ∑ (0, 1, 2, 9, 11, 15) + d(8, 10, 14) ]

Here, ∑ represents the sum of minterms, and d represents the don't-care conditions.

The K-map can be represented as follows:

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

Step 2: Group the 1s and Xs

We need to group the 1s and Xs in the K-map in the largest possible power-of-two rectangles (1, 2, 4, 8, etc.) to simplify the function. Each group should contain only 1s and Xs and should be as large as possible to simplify the function to the greatest extent.

Here are the groups:

1. Group 1: Cells (0, 1, 9, 11) - This is a group of four that can be grouped vertically.
2. Group 2: Cells (2, 10, 14) - This is a group of three, including don't-care conditions, which can be grouped horizontally.
3. Group 3: Cell (15) - This is a single cell that cannot be grouped with others.

Step 3: Write the simplified Boolean function

Each group corresponds to a product term in the simplified Boolean function, where we only include variables that are the same for all 1s in the group.

1. Group 1 corresponds to ( \overline{C}D ) since C is 0 and D is 1 for all minterms in the group.
2. Group 2 corresponds to ( A\overline{D} ) since A is 1 and D is 0 for all minterms in the group.
3. Group 3 corresponds to ( ABCD ) since all variables are 1 for the minterm in the group.

The simplified Boolean function is: [ F(A, B, C, D) = \overline{C}D + A\overline{D} + ABCD ]

Step 4: Implement the simplified function using NAND gates

To implement the function using only NAND gates, we need to apply De Morgan's Theorems and the properties of NAND gates. A NAND gate is equivalent to an AND gate followed by a NOT gate. Therefore, we can convert AND and OR operations into NAND operations.

The simplified function can be rewritten using NAND operations as follows:

[ F(A, B, C, D) = \overline{\overline{\overline{C}D} \cdot \overline{A\overline{D}} \cdot \overline{ABCD}} ]

This is because the NAND operation is the complement of the AND operation, and by double negation (a NOT of a NOT), we can revert back to the original value.

Step 5: Draw the NAND gate implementation

Now we can draw the circuit using only NAND gates:

1. For ( \overline{C}D ), we use a NAND gate with inputs C and D, and then another NAND gate to invert the output (since NAND is AND followed by NOT).
2. For ( A\overline{D} ), we use a NAND gate with inputs A and D, and then another NAND gate to invert the output.
3. For ( ABCD ), we use a series of NAND gates to first create the AND function and then invert it.

The final circuit would have three NAND gates for the three terms, and then one more NAND gate to combine them, with the output of this final gate being the function F.

Please note that the actual drawing of the circuit is not possible in this text-based format, but the above description should give you a clear idea of how to implement the simplified function using NAND gates.