Implies in a truth table, unclear. In my textbook, we have the following truth table:


*

*$P$ true and $Q$ true means that "$P \implies Q$" is true.

*$P$ true and $Q$ false means that "$P \implies Q$" is false.

*$P$ false and $Q$ true means that "$P \implies Q$" is true.

*$P$ false and $Q$ false means that "$P \implies Q$" is true.


I'm confused by what this means? I don't see why $P$ true and $Q$ false is the only combination that gets us that false and the rest of them get us true. Could anyone help clarify what's going on? Why is what is in the textbook correct? Thanks.
 A: $P \implies Q$ should be read as saying that whenever $P$ is true, $Q$ is true. If $P$ is false, then $P \implies Q$ says nothing about the truth value of $Q$.
For example, if I say "If it is raining, then I will bring an umbrella.", (where $P$ is "it is raining" and $Q$ is "I will bring an umbrella", I have said nothing about what happens if it is not raining. I will have spoken truly if it is not raining and I do not bring an umbrella, and I will have spoken truly if it is not raining and I do bring an umbrella.
The only case in which I have not spoken truly is if it is raining and I fail to bring an umbrella -- the case where $P$ is true and $Q$ is false.
A: To make it short (in classical logic): $P\implies Q$ is the same as $\lnot P \lor Q$ and your textbook is right.
A: Although it may not help much, and although it's sneakily substituting constructive for classical logic, I enjoy thinking about this as a computer-science theorist might:  namely, interpret the variables $P$ and $Q$ not as statements but as sets, and $P \implies Q$ as the set of functions from $P$ to $Q$.  Then we have richer information than mere truth (the set is non-empty) and falsity (the set is empty), but can return to the mere truth perspective when appropriate.
Now $\lnot P$ implies that $P$ is uninhabited, i.e., as a set, there's nothing in it.  (Here's where I'm being a little sneaky; you asked about $P$ being false, and I'm talking about its negation $\lnot P$ being true.  These are equivalent from a classical point of view, but a constructive point of view requires the finer distinction that I am making.)  Then, you might think, there is no function from $P$ to $Q$, so that $P \implies Q$ should be false; but actually that's not quite right.  A function, remember, just has to tell you what to do with the elements of its domain, and if the domain doesn't have any elements then there's nothing to do; so there is, trivially, a (in fact exactly one) function from the empty set to any other set.  (This is true even if the other set is empty; we never even look in it.)  That is, $P \implies Q$ has got some elements (one of them), and so, from the Boolean point of view, is true.
Contrast this with the situation where $P$ and $\lnot Q$ hold, i.e., there's something in the set $P$ but nothing in the set $Q$.  In that case, there's no way to specify a rule for assigning elements of $Q$ to elements of $P$, because there are no elements of $Q$ to assign; so $P \implies Q$ is false in this case, as your textbook says.
