Nonsense from combining two iffs ($\iff$) E and F are independent iff $\frac{p(E \cap F)}{p(F)}=p(E)$. 
Also, E and F are independent iff $p(E | F)=p(E)$
Why do I get nonsense if I combine the two? I'd get: E and F are independent iff $p(E | F) = \frac{p(E \cap F)}{p(F)}$
Why doesn't the iff carry over when I combine the two equations?
 A: [This answer assumes the OP's argument is different from that assumed by the other answers.]
There's a logical fallacy in your argument, which seems to be of the following form:


*

*$A\iff(x=y)$

*$A\iff(z=y)$, where we define $z::=x$

*Hence, $A\iff [(x=y)\land (z=y)]$  


So far, so good, but the next step, eliminating $y$ in the conjunction, is incorrect:


*Therefore, $A\iff(x=z)$.  


That step would be correct if 
$$[(x=y)\land (z=y)]\ \ \iff\ \  (x=z),
$$
but this fails simply because
$$[(x=y)\land (z=y)]\require{cancel}\ \ \cancel\impliedby\ \  (x=z).
$$ 
NB: In your case, $A$=("$E,F$ are independent"), $x=\frac{p(E \cap F)}{p(F)},$ $y=p(E),$ and $z=p(E\mid F).$
A: To address the more general question, here’s an example:
Suppose Kim Jong Un tells you, “In a democracy, what the leader promised is what the people voted for, and what the people voted for is what the government does.  Just bear with me and take that as a definition.  Therefore, I rule a democracy, because what the leader promised is what the government does.”
A: You combined them wrong. From $A \Leftrightarrow B$ and $A \Leftrightarrow C$, you can infer $B \Leftrightarrow C$.
But you've misidentified what $A,B,C$ are. They aren't the sides of the equation, but rather


*

*$A$ is the statement "$E$ and $F$ are independent"

*$B$ is the formula $\frac{p(E \cap F)}{p(F)}=p(E)$

*$C$ is the formula $p(E | F)=p(E)$


so the correct combination is
$$ \frac{p(E \cap F)}{p(F)}=p(E) \qquad \mathrm{iff} \qquad p(E | F)=p(E)$$
A: Let $a=\frac{p(E \cap F)}{p(F)}$, $b=p(E | F)$, and $c=p(E)$.  You know that the equations $a=c$ and $b=c$ are equivalent to each other (since they are both equivalent to "$E$ and $F$ are independent").  That is, whenever $a=c$ is true, $b=c$ is also true, and conversely.
There is no reason for $a=b$ to be equivalent to these two statements, though.  If $a=c$ is true, then $b=c$ is also true, and so $a=b$ is true as well.  So we have an implication in one direction.  But the other direction doesn't work: if we know $a=b$, that doesn't tell us anything about whether $a=c$ or $b=c$ are true.  It's entirely possible that $a$ and $b$ are equal to each other, but not to $c$.
A: Here is a simple mathematical example to clearly elucidate the logical error:

An integer $n$ is an even prime iff $n = 2$.
An integer $n$ is an even prime iff $n = 1+1$.
(WRONG!) An integer $n$ is an even prime iff $2 = 1+1$.

