1. **State the problem:** We are given the Boolean function $F = AC + AB$ and asked to deduce its Karnaugh map (K-map) and find the simplified expression. 2. **Identify variables:** The function has three variables: $A$, $B$, and $C$. 3. **Write the truth table:** List all combinations of $A$, $B$, and $C$ and evaluate $F = AC + AB$. | A | B | C | AC | AB | F=AC+AB | |---|---|---|----|----|---------| | 0 | 0 | 0 | 0 | 0 | 0 | | 0 | 0 | 1 | 0 | 0 | 0 | | 0 | 1 | 0 | 0 | 0 | 0 | | 0 | 1 | 1 | 0 | 0 | 0 | | 1 | 0 | 0 | 0 | 0 | 0 | | 1 | 0 | 1 | 1 | 0 | 1 | | 1 | 1 | 0 | 0 | 1 | 1 | | 1 | 1 | 1 | 1 | 1 | 1 | 4. **Construct the 3-variable K-map:** Use $A$ as row variable and $B,C$ as column variables in Gray code order. | A \ BC | 00 | 01 | 11 | 10 | |--------|----|----|----|----| | 0 | 0 | 0 | 0 | 0 | | 1 | 0 | 1 | 1 | 1 | 5. **Group the 1s:** The three 1s in the bottom row can be grouped together as a single group of size 3 (actually size 3 is not standard, so group the three 1s as one group of 2 and one group of 2 with overlap). Groups: - Group 1: Cells (1,01) and (1,11) correspond to $A=1$, $B=1$ (since $C$ changes), so term $AB$. - Group 2: Cells (1,01) and (1,10) correspond to $A=1$, $C=1$ (since $B$ changes), so term $AC$. 6. **Final simplified expression:** $F = AB + AC$ (which is the original expression, so it is already simplified). **Answer:** The K-map confirms the function $F = AB + AC$ is already simplified.