The original statement is, "If I am in Paris, then I am in France". The inverse statement is, "If I am not in Paris, then I am not in France".
If $\lnot A = \lnot B =$ True, then $\lnot A \implies \lnot B$ = True.
If $\lnot A = \lnot B =$ False, then $\lnot A \implies \lnot B$ = True.
Here is where my misunderstanding occurs:
If $\lnot A =$ True and $\lnot B =$ False, then $\lnot A \implies \lnot B$ = True.
After all, the statement, "If I am not in Paris, then I am in France", is a logical statement: One can be somewhere else in France.
If $\lnot A =$ False and $\lnot B =$ True, then $\lnot A \implies \lnot B$ = False.
After all, the statement, "If I am in Paris, then I am in not France", is an illogical statement: If one is in Paris, then they must necessarily also be in France.
I would greatly appreciate it if someone could please take the time to clarify my misunderstanding.
Thank you.
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$\begingroup$I would like to show you this in truth tables, hopefully it resolves some of the confusion. Of course, you should read the other answers as well, Patrick Stevens answer I like personally.
The connective "$\Rightarrow$" is defined according to this truth table:
$$ \begin{array}{c|l|c|} P & \text{Q} & \text{P $\Rightarrow$ Q} \\ \hline T & T & T \\ T & F & F \\ F & T & T \\ F & F & T \end{array} $$ This should be in your book. As you can see, $P \Rightarrow Q$ is false only when $P$ is true and $Q$ is false, otherwise $P \Rightarrow Q$ is true.
For the contrapositive, we can draw the following table:
$$ \begin{array}{c|l|c|c} P & Q & \neg Q & \neg P & \neg Q \Rightarrow \neg P \\ \hline T & T & F & F & T \\ T & F & T & F & F\\ F & T & F & T & T\\ F & F & T & T & T \end{array} $$
As you can see, for any valuation of $P, Q$ (for any assignment of $T,F$ to $P,Q$), whenever $ P \Rightarrow Q$ is true, so is $\neg Q \Rightarrow \neg P $.
So as you can see, if $\neg Q$ is false, and $\neg P$ is true, then $\neg Q \Rightarrow \neg P$ is true, by the third line in the above table. This is exactly in line with what is says in your book.
For the converse (or inverse) , we have:
$$ \begin{array}{c|l|c|c} P & \text{Q} & \text{P $\Rightarrow$ Q} & Q \Rightarrow P \\ \hline T & T & T & T \\ T & F & F & T \\ F & T & T & F \\ F & F & T & T \end{array} $$
As you can see, they don't match for all valuations, in particular, if $P$ is true, and $Q$ is false, then one implication is true and the other isn't.
Finally: if $A$ is the statement "you are in Paris" and $B$ is the statement "you are in France" and you know $A \Rightarrow B$ is true, then if you are not in France, then you are not in Paris ($\neg B \Rightarrow \neg A$).
And of course, if you are not in Paris ($\neg A$ is true) then you could be on Mars for all we know. You are not necessarily in France.
EDIT: I would like to encourage you to look at more mathematical examples, similar to other peoples' examples in their answers, natural language is full of vague statements that are not always as clear cut as mathematical ones.
$\endgroup$ 13 $\begingroup$The negation of $(A\Rightarrow B)$ is not $(\lnot A\Rightarrow\lnot B)$, you have also to reverse the implication.
In fact $(A\Rightarrow B)$ is equivalent to $(\lnot B\Rightarrow \lnot A)$ : "If I'm not in France then I'm not in Paris".
This comes from $(A\Rightarrow B)$ is equivalent to $(\lnot A\lor B)$ and you can clearly see that $(A\Rightarrow B)=(\lnot A\lor B)=(\lnot \lnot B\lor \lnot A)=(\lnot B\Rightarrow \lnot A)$.
Now if you negate the implication you get $\lnot(A\Rightarrow B)=\lnot(\lnot A\lor B)=(A\land \lnot B)$ : "I am in Paris and I am not in France" which is obviously false since you negated a true proposition to start with. But this $(A\land \lnot B)$ does not have a signification with some $\Rightarrow$, it's just something different.
So all these are true :
- If I am in Paris then I am in France (proposition)
- If I am not in France then I am not in Paris (contraposition)
- If I am not In Paris then either I am not in France either I am in France but not in Paris (negation)
As you can see, the negation doesn't tell much, just that well... you are not in Paris.
$\endgroup$ 6 $\begingroup$Let $P$ be the statement "$A$ implies $B$".
The inverse (or converse) of $P$ is the statement "NOT $A$ implies NOT $B$".
The contrapositive of $P$ is the statement "NOT $B$ implies NOT $A$".
The following is always true: $P$ holds if and only if the contrapositive of $P$ holds. That is, $P$ is equivalent to its contrapositive.
The following need not be true: if $P$ holds, then the inverse of $P$ holds. (Let $A$ be "$x=2$", and $B$ be "$x$ is a natural number", so $P$ is the true statement "if $x=2$, then $x$ is a natural number", and the inverse of $P$ is the false statement "if $x$ is a natural number, then $x=2$".) That is, $P$ is not equivalent to its inverse (indeed, $P$ does not even imply its inverse).
The following need not be true: if the inverse of $P$ holds, then $P$ holds. (Just swap $A$ and $B$ around in the previous case.) This is another reason why $P$ is not equivalent to its inverse: $P$ is not even implied by its inverse.
To address specifically "If $\neg A$ is True, and $\neg B$ is False, then $\neg A \Rightarrow \neg B$ is True": you're saying that something which is True implies something which is False. To see why this can't be right, could you similarly say that $1+1=2$ implies $1=2$?
Notice also that everything is implied by a false statement (this is "vacuous truth"), so if $\neg A$ is false, then $\neg A \Rightarrow X$ for any statement $X$.
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