Nonsense from combining two iffs ($\iff$)
Solution 1:
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
$$ \frac{p(E \cap F)}{p(F)}=p(E) \qquad \mathrm{iff} \qquad p(E | F)=p(E)$$
Solution 2:
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 $c$.
Solution 3:
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 $2 = 1+1$.
Solution 4:
[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 $z=p(E\mid F).$