Bayes Theorem with multiple conditions I don't understand how this equation was derived.
$P(I|M_{1}\cap M_{2}) \leq \frac{P(I)}{P(I')}\cdot \frac{P(M_{1}|I)P(M_{2}|I)}{P(M_{1}|I')P(M_{2}|I')}$
This equation was from the paper "Trial by Probability" where the case of OJ Simpson was given as an example problem. The defendant is under trial for double murder and two evidences are introduced against him.
$M_{1}$ is the event the defendant's blood matches a drop of blood found in a crime scene. $M_{2}$ is the event a victim's blood matches blood on a sock belonging to the defendant. Assuming guilt, the occurrence of one evidence increases the probability of the other.
$I$ is the event a defendant is innocent while $I'$ is when he is guilty.
We are trying to get the CEILING of the probability that the defendant is innocent given the two evidences.
Values for some variables were given but what I'm interested in is how the equation was derived. I tried but got nowhere.
Yes, I've already checked the 'Questions that may already have your answer'.
 A: By Bayes' Theorem: \begin{align*}P(I\mid M_1\cap M_2)&=\frac{P(I)P(M_1\cap M_2\mid I)}{P(M_1\cap M_2)}\\&=\frac{P(I)P(M_1\cap M_2\mid I)}{P(I)P(M_1\cap M_2\mid I)+P(I')P(M_1\cap M_2\mid I')}.\end{align*} Now the paper you provided argues that 

If $I$ is true,
  then $M_1$ and $M_2$ are independent. But assuming guilt, the occurrence of one would increase the probability of the other.

So $$P(M_1\cap M_2\mid I)=P(M_1\mid I)P(M_2\mid I),\label{eq1}\tag{1}$$ and $$ P(M_1\cap M_2\mid I')=P(M_1\mid M_2\cap I')P(M_2\mid I')\geq P(M_1\mid I')P(M_2\mid I').\label{eq2}\tag{2}$$ Hence, \begin{align*}P(I\mid M_1\cap M_2)&=\frac{P(I)P(M_1\mid I)P(M_2\mid I)}{P(I)P(M_1\cap M_2\mid I)+P(I')P(M_1\cap M_2\mid I')}&& \text{(Substitute with \eqref{eq1})}\\&\leq\frac{P(I)P(M_1\mid I)P(M_2\mid I)}{P(I')P(M_1\cap M_2\mid I')}&& \text{(Lesser Denominator)}\\&\leq\frac{P(I)}{P(I')}\cdot\frac{P(M_1\mid I)P(M_2\mid I)}{P(M_1\mid I')P(M_2\mid I')}.&& \text{(Substitute with \eqref{eq2})}
\end{align*}
To derive \eqref{eq2}, note \begin{align*}\frac{P(M_1\cap M_2\mid I')}{P(M_2\mid I')}&=\frac{P(M_1\cap M_2\cap I')/P(I')}{P(M_2\cap I')/P(I')}\\&=\frac{P(M_1\cap M_2\cap I')}{P(M_2\cap I')}\\&=P(M_1\mid M_2\cap I')\end{align*} and since the occurrence of $M_2$ would increase the probability of $M_1$: $$P(M_1\mid M_2\cap I')\geq P(M_1\mid I')$$
