The given truth table relates Y to A and B.
`{:("A","B","Y"),(0,0,1),(0,1,0),(1,0,0),(1,1,0):}`
Then, Y is given by

Options

A. `A+B`
B. `AB`
C. `bar(AB)`
D. `bar(A+B)`

Answer

To solve the problem, we need to analyze the given truth table and determine the logical expression that represents the output \( Y \) based on the inputs \( A \) and \( B \).

### Given Truth Table:
\[ \begin{array}{|c|c|c|} \hline A & B & Y \\ \hline 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 0 \\ \hline \end{array} \]

### Step 1: Identify the output conditions
From the truth table, we can see the output \( Y \) is 1 only when both \( A \) and \( B \) are 0. In all other cases, the output is 0.

### Step 2: Write the logical expression
To express the output \( Y \) in terms of \( A \) and \( B \):
- The only case where \( Y = 1 \) is when \( A = 0 \) and \( B = 0 \).
- This can be represented using the logical expression \( Y = \overline{A + B} \) (the NOR operation), which means \( Y \) is true when both \( A \) and \( B \) are false.

### Step 3: Verify the expression
Now, let's verify the expression \( Y = \overline{A + B} \):
- If \( A = 0 \) and \( B = 0 \): \( Y = \overline{0 + 0} = \overline{0} = 1 \) (matches the truth table)
- If \( A = 0 \) and \( B = 1 \): \( Y = \overline{0 + 1} = \overline{1} = 0 \) (matches the truth table)
- If \( A = 1 \) and \( B = 0 \): \( Y = \overline{1 + 0} = \overline{1} = 0 \) (matches the truth table)
- If \( A = 1 \) and \( B = 1 \): \( Y = \overline{1 + 1} = \overline{1} = 0 \) (matches the truth table)

### Conclusion
The logical expression that represents \( Y \) based on the inputs \( A \) and \( B \) is:
\[ Y = \overline{A + B} \]