truth table for (~pv¬q) <-> (p^q)
MathBot Answer:
$\left(\neg p \vee \neg q\right) \leftrightarrow \left(p \wedge q\right)$ is a contradiction.
Legend
| $p$ | $q$ | $\neg p$ | $\neg q$ | $\neg p \vee \neg q$ | $p \wedge q$ | $\left(\neg p \vee \neg q\right) \leftrightarrow \left(p \wedge q\right)$ |
|---|---|---|---|---|---|---|
| 1 | 1 | 0 | 0 | 0 | 1 | 0 |
| 1 | 0 | 0 | 1 | 1 | 0 | 0 |
| 0 | 1 | 1 | 0 | 1 | 0 | 0 |
| 0 | 0 | 1 | 1 | 1 | 0 | 0 |
T/F