Bi conditional Statements:Did the author make a mistake in the passage? The book I am reading called How to prove it mentions the following, which I think it might be wrong but not sure.

One of the reasons it's so easy to confuse a conditional statement
  with its converse is that in everday speech we sometimes use a
  conditional statement when we mean to convey is actually a
  bi-conditional statement... Suppose a child is told by his parents,
  "if you don't eat your dinner, you won't get any dessert." The child
  certainly expects that if he does eat his dinner, he will get dessert,
  although that's not literally what his parents said. In other words,
  the child interprets the statement as meaning "Eating your dinner is a
  necessary and sufficient condition for getting dessert."

Do you think the example he gave was a good example? I do not think so because it does not sound like we assume a biconditional statement in everyday language. It's more like we mean (P-->Q)V(~P-->~Q)(Edit: it should be (P-->Q)/\(~P-->~Q) since (P-->Q)V(~P-->~Q) is a tautology) in everyday speech, instead the biconditional (P-->Q)/\ (Q-->P) which the author says.Maybe I am getting this wrong or do not understand it well or he gave a bad example. Please help make this clear for me, thanks.
 A: The example does illustrate the author’s point; it’s very similar to examples that I’ve used in my classes to make the same point. Let $P$ be you do not eat your dinner, and let $Q$ be you will not get any dessert. The original statement then appears to be $P\to Q$. However, as the author points out, the child understands the statement to mean not only $P\to Q$, but also $\neg P\to\neg Q$: if you do eat your dinner, you will get dessert. In other words, the child understands the original English statement to mean $(P\to Q)\land(\neg P\to\neg Q)$ (and that’s undoubtedly what the parent meant).
Now $\neg P\to\neg Q$ is logically equivalent to $Q\to P$, as you can check by examining its truth table. (It’s called the contrapositive of $Q\to P$.) Thus, $(P\to Q)\land(\neg P\to\neg Q)$ is logically equivalent to $(P\to Q)\land(Q\to P)$, which in turn is logically equivalent to the biconditional $P\leftrightarrow Q$. 
In other words, the author is correct: in that context we understand the statement if you don’t eat your dinner, you won’t get any dessert as 

if you don’t your dinner, you won’t get any dessert; and if you do eat your dinner, you will get some dessert,

which is logically equivalent to the biconditional

if you don’t your dinner, you won’t get any dessert; and if dont’ get any dessert, you didn’t eat your dinner.

A: As it turns out, the following statements are equivalent (in classical logic).


*

*$(P \rightarrow Q) \wedge (Q \rightarrow P)$

*$(P \rightarrow Q) \wedge (\neg P \rightarrow \neg Q)$


To see this, note that, by the law of contraposition (applied twice), we have
$$(P \rightarrow Q) \implies (\neg Q \rightarrow \neg P) \implies (\neg\neg P \rightarrow \neg\neg Q).$$
However, by double negation elimination, the expression on the right is logically equivalent to the one on the left. Thus
$$(P \rightarrow Q) \iff (\neg Q \rightarrow \neg P),$$
from which the equivalence of interest follows.
