Equivalence of two definitions of mutual information
You can define the mutual information as $I(X;Y) = H(X) - H(X|Y)$. This definition is symmetric in $X,Y$ since
$$
H(X)-H(X|Y) = H(X) - \sum_y \Pr[Y=y] H(X|Y=y) = \\
\sum_x \Pr[X=x] \log \frac{1}{\Pr[X=x]} - 
\sum_y \Pr[Y=y] \sum_x \Pr[X=x|Y=y] \log \frac{1}{\Pr[X=x|Y=y]} = \\
\sum_{x,y} \Pr[X=x,Y=y] \log \frac{1}{\Pr[X=x]} -
\sum_{x,y} \Pr[X=x,Y=y] \log \frac{\Pr[Y=y]}{\Pr[X=x,Y=y]} = \\
\sum_{x,y} \Pr[X=x,Y=y] \log \frac{\Pr[X=x,Y=y]}{\Pr[X=x]\Pr[Y=y]},
$$
and this expression is symmetric in $X,Y$.

This calculation also shows that if $X,Y$ are independent then $I(X;Y) = 0$, since then $\Pr[X=x,Y=y] = \Pr[X=x] \Pr[Y=y]$.

Moreover, if we write
$$
I(X;Y) = \sum_{x,y} \Pr[X=x] \Pr[Y=y] \frac{\Pr[X=x,Y=y]}{\Pr[X=x] \Pr[Y=y]} \log \frac{\Pr[X=x,Y=y]}{\Pr[X=x] \Pr[Y=y]},
$$
then the convexity of $z\log z$ implies that $I(X;Y) \geq z\log z$, where
$$
z = \sum_{x,y} \Pr[X=x] \Pr[Y=y] \frac{\Pr[X=x,Y=y]}{\Pr[X=x] \Pr[Y=y]} = \sum_{x,y} \Pr[X=x,Y=y] = 1.
$$
Thus $I(X;Y) \geq 0$.