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This question already has an answer here:

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.

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marked as duplicate by Cameron Buie, user147263, Joel Reyes Noche, Mike Pierce, Daniel Jun 30 '15 at 4:18

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I assume you mean $P$ false and $Q$ false means that "$P\Rightarrow Q$" is true, right? $\endgroup$ – Peter Jun 29 '15 at 18:47
  • $\begingroup$ The last time I answered this question: math.stackexchange.com/questions/1317408/… Let me know if this helps, or what else might help you understand. $\endgroup$ – Ian Jun 29 '15 at 19:10
  • $\begingroup$ Just remember that, in mathematics, if not in everyday usage, $P\implies Q$ just means that $\neg[P\land \neg Q]$. $\endgroup$ – Dan Christensen Jun 30 '15 at 2:20
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$P \implies Q$ 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.

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  • $\begingroup$ or as the legal profession might say, it's true unless you can prove it false beyond all reasonable doubt. $\endgroup$ – JamesM Jun 29 '15 at 19:04
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    $\begingroup$ One way of viewing the first point: $(P \Rightarrow Q) \wedge \neg P \wedge Q$ and $(P \Rightarrow Q) \wedge \neg P \wedge \neg Q$ should have the same truth value, since $(P \Rightarrow Q) \wedge \neg P$ tells us nothing about the truth of $Q$. If we claim that they're both false, then $P \Rightarrow Q$ becomes $P \wedge Q$, which I don't think makes any sense. $\endgroup$ – Ian Jun 29 '15 at 19:15
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    $\begingroup$ @Ian: That is confusing. Assume $(P\implies Q)\land \lnot P$ is true, then $(P\implies Q)\land \lnot P \land Q$ is $Q$ and $(P\implies Q)\land \lnot P \land \lnot Q$ is $\lnot Q$. If your claim is right, then $(P\implies Q)\land \lnot P$ is always false. That is obviously wrong, for $P$ is false and $Q$ arbitrary. $\endgroup$ – user251257 Jun 29 '15 at 22:14
  • $\begingroup$ I gave the proper values for $P$ and $Q$ such that $(P\implies Q)\land\lnot P\land Q$ has a different truth value than $(P\implies Q)\land\lnot P\land \lnot Q$. His intuition might be right, but his logic is flawed. $\endgroup$ – user251257 Jun 29 '15 at 22:34
  • $\begingroup$ I said that badly, sorry. I should have said that the truth values of $P \Rightarrow Q$ in the $\neg P \wedge Q$ row and in the $\neg P \wedge \neg Q$ row should be the same, because $(P \Rightarrow Q) \wedge \neg P$ tells us nothing about the truth of $Q$. Then the rest of what I said makes sense. $\endgroup$ – Ian Jun 29 '15 at 23:44
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To make it short (in classical logic): $P\implies Q$ is the same as $\lnot P \lor Q$ and your textbook is right.

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    $\begingroup$ I think that this just substitutes one question for another: why is $P \implies Q$ the same as $\lnot P \lor Q$? That is to say, someone who doesn't understand why the truth table is as it is is probably looking for motivation, not just a bare definition, to explain it. $\endgroup$ – LSpice Jun 29 '15 at 22:30
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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.

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