Inequalities and multiplication Given two inequalities the rules of multiplication are said to be (check link for the source):

1.) If $0<a<b$ and $0<c≤d$, then $0<ac<bd$
2.) If $0<a<b$ and $c≤d<0$, then $bd<ac<0$
3.) If $0<a<b$ and $c<0<d$, then $ac<0<bd$
4.) If $a<b<0$ and $c<0<d$, then $bd<0<ac$
5.) If $a<0<b$ and $c<0<d$, then no conclusion may be drawn
  about the relative positions of $ac$ and $bd$ on the number line

How do I prove the second and fourth rules "If $0<a<b$ and $c≤d<0$, then $bd<ac<0$" and "If $a<b<0$ and $c<0<d$, then $bd<0<ac$" ?
My Attempts:
1.)
$$
0<a<b \quad\&\quad 0<c<d\\0<ac<bc \quad\&\quad 0<bc<bd \implies 0<ac<bc<bd\implies 0<ac<bd
$$
3.)
$$
0<a<b \quad\&\quad c<0<d\\
0<ad<bd \quad\&\quad ac<0<ad\implies ac<0<ad<bd\implies ac<0<bd
$$
Similarly,
2.)
$$
0<a<b\quad\&\quad c<d<0\\
0>ac>bc \quad\&\quad bc<bd<0\implies ? 
$$
4.)
$$
a<b<0 \quad\&\quad c<0<d\\ad<bd<0 \quad\&\quad ac>0>ad\implies ?
$$

For 2), you can use 1):
$$d\le c < 0\quad\Rightarrow\quad 0 < -c \le -d,$$
so by 1), $0 < -ac < -bd$ and thus $bd < ac <0$.
Edit:  Note that item 2) as stated is false. I've left this here under the hypothesis that the conclusion of 2) is actually misstated.
For 4), note that $b<0$ and $d>0$, so that $bd<0$. Further, $a$, $c<0$, so that $ac>0$.
