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The Boolean expression provided is:

\[
Y = f(A, B, C, D) = \Sigma m(1, 3, 7, 8, 10, 12, 13, 15)
\]

This is a sum of minterms representation, where the given minterms correspond to specific binary combinations of the variables \( A \), \( B \), \( C \), and \( D \).

### Step 1: Generate the Karnaugh Map (K-Map)
We can simplify this Boolean function by placing these minterms into a 4-variable K-map and applying grouping to simplify the expression.

The binary representations of the minterms are:

- m(1)  = 0001
- m(3)  = 0011
- m(7)  = 0111
- m(8)  = 1000
- m(10) = 1010
- m(12) = 1100
- m(13) = 1101
- m(15) = 1111

We will now populate the K-map.

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

### Step 2: Simplification Using the K-Map
From the K-map, we can group adjacent 1's:

- Group 1 (4 cells): m(12), m(13), m(15), m(7) → \( C \)
- Group 2 (2 cells): m(8), m(12) → \( A \overline{B} \)
- Group 3 (2 cells): m(3), m(7)

Simplify the Boolean algebra expression Y = f(A, B, C, D) = Σ m(1, 3, 7, 8, 10, 12, 13, 15) and draw the logic circuit for the reduced expression.

This guide walks through simplifying a Boolean algebra expression using a Karnaugh Map and creating the corresponding logic circuit. The problem includes minterms (1, 3, 7, 8, 10, 12, 13, 15) and focuses on key topics like Boolean algebra and circuit design.

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