# 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?

• Why would you expect that to work? – Eric Wofsey Mar 4 '18 at 6:28
• @EricWofsey I consider iff to indicate equality, so I was intuitively expecting the "iff" to carry over and I'm not grasping why it isn't – JobHunter69 Mar 4 '18 at 6:29
• You have combined illogically. What we do have is : $E, F$ are independent iff $p(E)=p(E|F)=p(E\cap F)/p(F).$.... For $any$ $E$ we have $p(E|F)=p(E\cap F)/p(F)$ so $E,F$ are independent iff $p(E)$ is equal to either of them. – DanielWainfleet Mar 5 '18 at 9:42

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 $c$.

• In fact, $a$ and $b$ in this case are equal to each other by definition, regardless of whether $E$ and $F$ are independent. – Bolton Bailey Mar 5 '18 at 3:48
• What you did is equivalent to saying, "E and F are independent if E and F are independent." – richard1941 Mar 12 '18 at 15:37

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

$$\frac{p(E \cap F)}{p(F)}=p(E) \qquad \mathrm{iff} \qquad p(E | F)=p(E)$$

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 $$2 = 1+1$$.

• I don't feel like this example is much different than my question, so it doesn't elucidate anything for me. – JobHunter69 Mar 5 '18 at 19:44

[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:

1. $A\iff(x=y)$
2. $A\iff(z=y)$, where we define $z::=x$
3. Hence, $A\iff [(x=y)\land (z=y)]$

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

1. 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 $z=p(E\mid F).$

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.”