All the gates used in the given configuration are NOR gates. Thus, the output can be found as follows-
Y===(a+b)(ab)ab
The output of the given configuration can also be verified by the following truth table:
| a | b | Outputfromfirstgate | Outputfromsecondgate | Y |
| 0 | 0 | 1 | 1 | 0 |
| 0 | 1 | 1 | 0 | 0 |
| 1 | 0 | 0 | 1 | 0 |
| 1 | 1 | 0 | 0 | 1 |
The calculation and the truth table clearly shows that the configuration represents
AND gate.