例 ((X NAND Y) NAND X) NAND Y
- 出題(応用情報技術者試験ドットコム)
- 前提
- $X\cap Y \neq\phi$
- 否定論理積(NAND)
| X | Y | X NAND Y |
|---|---|---|
| 0 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
More than 1 year has passed since last update.
Register as a new user and use Qiita more conveniently
- You get articles that match your needs
- You can efficiently read back useful information
- You can use dark theme