Given the following truth table where A , B are inputs and Y the output :
`{:(A,B,Y),(0,0,1),(1,0,1),(0,1,1),(1,1,0):}`
The output Y is :

Options

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

Answer

To solve the problem, we need to analyze the given truth table and determine which logical gate it represents based on the inputs A and B and the output Y.

### Step-by-Step Solution:

1. **Understand the Truth Table**:
The truth table provided is:
```
A | B | Y
---------
0 | 0 | 1
1 | 0 | 1
0 | 1 | 1
1 | 1 | 0
```

2. **Identify the Output Pattern**:
- When both inputs A and B are 0 (0,0), the output Y is 1.
- When A is 1 and B is 0 (1,0), the output Y is 1.
- When A is 0 and B is 1 (0,1), the output Y is 1.
- When both A and B are 1 (1,1), the output Y is 0.

3. **Compare with Logical Gates**:
- **AND Gate**: Outputs 1 only when both inputs are 1. (Does not match)
- **OR Gate**: Outputs 1 when at least one input is 1. (Does not match)
- **NAND Gate**: Outputs 0 only when both inputs are 1. (Matches)
- **NOR Gate**: Outputs 1 only when both inputs are 0. (Does not match)
- **XOR Gate**: Outputs 1 when inputs are different. (Does not match)
- **XNOR Gate**: Outputs 1 when inputs are the same. (Does not match)

4. **Conclusion**:
The output Y matches the behavior of a NAND gate, which can be expressed as:
\[ Y = \overline{A \cdot B} \]
or equivalently,
\[ Y = A.B' \]
where \( B' \) is the negation of B.

### Final Answer:
The output Y is represented by a NAND gate.

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