To determine the minimum number of NAND gates required to implement the given Boolean expression, we must first simplify the expression using Boolean algebra laws.

Boolean Expression Simplification

The given Boolean expression is:

\( F = A + A\bar{B} + AB\bar{C} \)

Step-by-Step Expression Simplification

We will simplify the expression using fundamental Boolean algebra identities. The key is to look for common terms or patterns that allow us to reduce the complexity of the expression.

1. Simplify the first two terms: Consider the first part of the expression: \( A + A\bar{B} \). This form matches the Absorption Law, which states that \( X + X\bar{Y} = X \). In our case, let \( X = A \) and \( Y = B \). Applying this law, \( A + A\bar{B} \) simplifies to \( A \). So, the expression now becomes: \( F = A + AB\bar{C} \)
2. Simplify the remaining terms: Now we have the expression \( A + AB\bar{C} \). This form matches another variant of the Absorption Law, which states that \( X + XY = X \). Here, let \( X = A \) and \( Y = B\bar{C} \). Applying this law, \( A + A(B\bar{C}) \) simplifies to \( A \).

Therefore, the fully simplified Boolean expression is:

\( F = A \)

Minimum NAND Gates for Implementation

The simplified expression \( F = A \) means that the output of the digital circuit is simply the input variable \( A \). If the input variable \( A \) is already available as a signal, then no logic gates are necessary to generate this output.

When a Boolean expression simplifies to one of its input variables, it implies that the output can be directly taken from that input line. There is no need for any logical operation or gate to produce this result. Since no logic operation is required, no NAND gates (or any other type of gate) are needed.

Conclusion on Minimum NAND Gates

The original Boolean expression \( A + A\bar{B} + AB\bar{C} \) simplifies completely to \( A \). This simplification indicates that the desired output is simply the input signal \( A \) itself. Consequently, no logic gates, including NAND gates, are required to implement this function. The input signal \( A \) directly serves as the output.

Thus, the minimum number of NAND gates required for this implementation is 0.