Prove that $\neg(p \wedge q) \Leftrightarrow \neg p \vee \neg q$ using truth table

Question

Prove that $\neg(p \wedge q) \Leftrightarrow \neg p \vee \neg q$ using truth table

Here is my table but I get a contradiction..

See, the yellow mark things must be same but they aren't :(

What did I do wrong?

Answer 1

You got the definition of $\land$ wrong. In fact, $p \land q = 1$ if and only if both $p = 1$ and $q = 1$. So, $0 \land 0 = 0$.

Answer 2

The rows in the column below the proposition $(p \land q)$ should be $1\;$ if and only if both $\;p=1 \land q = 1.\;$ Otherwise, $0$. So only the bottom row in the column below $(p\land q)$ should be $1$.

With that knowledge, all the rows, except the bottom row of the column $\lnot (p\land q)$ should be $1$, with only the bottom row of that column equal to $0$. The rest is plain to see.