Each of two evidences increases prior probability but both decrease it. May this only happen if two evidences are dependent? I noticed this while working on another problem. My intuition is that the statement is true, but I am not sure. 
Let A is an event. Evidence 1 and 2 are $E_1$ & $E_2$ correspondingly.
$$P(A|E_1) = \frac{P(E_1|A)}{P(E_1)} P(A)$$
$$P(A|E_2) = \frac{P(E_2|A)}{P(E_2)} P(A)$$
Since either $E_1$ & $E_2$ increases the chance of A. $\frac{P(E_1|A)}{P(E_1)}>1$ & $\frac{P(E_2|A)}{P(E_2)}>1$.
$$P(A|E_1,E_2) = \frac{P(E_2|A,E_1)}{P(E_2|E_1)}P(A|E1) = \frac{P(E_2|A,E_1)}{P(E_2|E_1)}\frac{P(E_1|A)}{P(E_1)} P(A)$$
If $E_1$ & $E_2$ are independent:
$$ \frac{P(E_2|A,E_1)}{P(E_2|E_1)} = \frac{P(E_2|A)}{P(E_2)} >1 \Rightarrow P(A|E_1, E_2) > P(A)$$
This makes that both $E_1$ and $E_2$ increase the chance of A.
Thus, to preclude the chance of A, it must be that $E_1$ and $E_2$ are dependent, $\frac{P(E_2|A,E_1)}{P(E_2|E_1)}<1$ and it overpowers $\frac{P(E_1|A)}{P(E_1)}$.
I am not confident with the contradiction I make. Did I miss anything?
Thanks,
 A: The following scenario seems to contradict your conjecture.
Roll two standard dice, one red, one blue. Let $A$ be the event “The dice total 6 or 7.” Let $E_1$ be “The red die comes up 4.” Let $E_2$ be “The blue die comes up 4.”
Then $E_1$ and $E_2$ are independent, but I think $P(A)=\frac{11}{36}$, $P(A|E_1)=P(A|E_2)=\frac{1}{3}>P(A)$, and $P(A|E_1,E_2)=0<P(A)$.
A: I'm concerned about this step: You claim that if $E_1$ and $E_2$ are independent, then
$$
\frac{P(E_2 \mid A, E_1)}{P(E_2 \mid E_1)} = \frac{P(E_2 \mid A)}{P(E_2)}
$$
presumably by equating numerator with numerator, and denominator with denominator.  I see the denominators being equal, but not the numerators.  For instance, suppose $E_1, E_2, A$ all have probability $1/2$, and all are pairwise independent—but, either exactly one of them occurs, or all three of them jointly occur, all with probability $1/4$.  Then $P(E_2 \mid A, E_1) = 1$, but $P(E_2 \mid A) = P(E_2) = 1/2$.
Here's a counterexample to the original problem:

Please excuse the revolting color scheme; I'm using an old tool and I'm still working out how to add more colors.  Broadly speaking, $A$ is represented by the bottom half of the square, $E_1$ is a bottom-heavy trapezoid toward the left, and $E_2$ is a bottom-heavy trapezoid toward the right.  They intersect as shown above.
Graphically, the basic idea is that $E_1$ and $E_2$ are individually bottom-heavy, so they increase the probability of $A$.  But their intersection, the joint event $E_1, E_2$, is top-heavy, so it decreases the probability of $A$.  It remains only to check that $E_1$ and $E_2$ are indeed independent.
The probability table is as follows:
$$
\begin{array}{c|c}
& P(\cdot) \\ \hline
\emptyset & \frac{75}{256} \\
E_1 & \frac{15}{256} \\
E_2 & \frac{15}{256} \\
A & \frac{25}{256} \\
E_1, E_2 & \frac{23}{256} \\
E_1, A & \frac{45}{256} \\
E_2, A & \frac{45}{256} \\
E_1, E_2, A & \frac{13}{256}
\end{array}
$$
Note first that
$$
P(A) = \frac{25+45+45+13}{256} = \frac{1}{2}
$$
$$
P(E_1) = P(E_2) = \frac{15+23+45+13}{256} = \frac{3}{8}
$$
Now,
$$
P(E_1, E_2) = \frac{23+13}{256} = \frac{9}{64} = P(E_1)P(E_2)
$$
so $E_1$ and $E_2$ are independent.  Next,
$$
P(A \mid E_1) = \frac{P(A, E_1)}{P(E_1)} = \frac{45+13}{15+23+45+13}
              = \frac{29}{48} > \frac{1}{2} = P(A)
$$
and similarly for $P(A \mid E_2)$.  But, on the other hand,
$$
P(A \mid E_1, E_2) = \frac{P(A, E_1, E_2)}{P(E_1, E_2)} = \frac{13}{23+13}
                   = \frac{13}{36} < \frac{1}{2} = P(A)
$$
So $E_1$ and $E_2$ individually increase the probability of $A$, but jointly decrease it.
