Minimize the given Boolean function using only NAND gates. F(A, B, C, D) = ∑ (0, 1, 2, 9, 11, 15) + d(8, 10, 14)

To minimize the given Boolean function using only NAND gates, we will first minimize the function using standard Boolean algebra techniques, and then convert the resulting expression into one that uses only NAND gates. 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 which can be used to simplify the function but are not required to be specifically 0 or 1.

Step 1: Write the function in minterm form

First, we write the function in terms of minterms using the given indices:

Index	A	B	C	D	Minterm
0	0	0	0	0	m0 = A'B'C'D'
1	0	0	0	1	m1 = A'B'C'D
2	0	0	1	0	m2 = A'B'CD'
9	1	0	0	1	m9 = AB'C'D
11	1	0	1	1	m11 = AB'CD
15	1	1	1	1	m15 = ABCD

Don't-care conditions:

Index	A	B	C	D	Minterm
8	1	0	0	0	m8 = AB'C'D'
10	1	0	1	0	m10 = AB'CD'
14	1	1	1	0	m14 = ABCD'

Step 2: Simplify the function using Karnaugh Map (K-map)

We will use a 4-variable K-map to simplify the function. The K-map for the given function, including the don't-care conditions, is as follows:

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

From the K-map, we can group the 1s and don't-cares (d) to find the simplified expression:

• Group of 8 (including don't-care): A'B'
• Group of 4: CD
• Group of 2: AB'C'D

So the simplified expression is:

F(A, B, C, D) = A'B' + CD + AB'C'D

Step 3: Convert to NAND-only expression

To convert the expression to use only NAND gates, we use the fact that a NAND gate is equivalent to a NOT gate when both inputs are the same, and we can use De Morgan's laws to transform OR and AND operations into NAND operations.

The NAND equivalent of an AND operation is straightforward:

• AND operation: F = A • B
• NAND operation: F = (A • B)'

For an OR operation, we use De Morgan's laws:

• OR operation: F = A + B
• NAND operation: F = ((A')' • (B')')'

Now, let's convert the simplified expression into NAND-only form:

F(A, B, C, D) = A'B' + CD + AB'C'D

Using De Morgan's laws and NAND properties, we get:

F(A, B, C, D) = ((A'B')')' • (CD)' • ((AB'C'D)')'

Now, we need to express each term as a NAND operation:

• (A'B')' can be implemented as NAND(A, A) NAND(B, B)
• (CD)' can be implemented as NAND(C, D)
• (AB'C'D)' can be implemented as NAND(AB', C'D) which further breaks down to NAND(NAND(A, B), NAND(C, D))

Putting it all together, we have:

F(A, B, C, D) = NAND(NAND(NAND(A, A), NAND(B, B)), NAND(C, D), NAND(NAND(A, B), NAND(C, D)))

This is the minimized Boolean function using only NAND gates.