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I am struggling to understand this. According to truth tables, if $P$ is false, it doesn't matter whether $Q$ is true or not: Either way, $P \implies Q$ is true.

Usually when I see examples of this people make up some crazy premise for $P$ as a way of showing that $Q$ can be true or false when $P$ is something outrageous and obviously untrue, such as "If the moon is made of bacon-wrapped apple-monkey carburetors, then I am a better wakeborder than Gauss."

$P$ is clearly false, but $P \implies Q$ is true no matter what the state of $Q$ is, and I don't understand why.

Are we saying "If $P$ is false, then all bets are off and $Q$ can be anything, either true or false, and not contradict our earlier claim, and if it isn't false, it must be true"?

Otherwise why can't we say that if $P$ is false, then we can't make any claims one way or the other on whether or not it implies anything at all?

Eric Wofsey
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AJJ
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    I doubt that a really good answer suitable for a broad audience of intelligent layersons and philosophers has ever been written for this excellent question. $\qquad$ – Michael Hardy Aug 08 '16 at 19:22
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    Part of the answer is that one wants to be able to fill in the truth table with a value for $P\Rightarrow Q$ based ONLY on the truth values of $P$ and $Q$ and not on anything else about the content of $P$ and $Q$. $\qquad$ – Michael Hardy Aug 08 '16 at 19:23
  • Well most of the answers I see say something like "Well it hasn't been proven false, so we say it's true." What I want to know is whether or not this result has been arbitrarily decided by humans as a convention or if it is a consequence of some other mathematical truth / structure / something / anything. And then this answer here http://math.stackexchange.com/a/440242/70349 suggests that it is NOT mere convenience. – AJJ Aug 08 '16 at 19:23
  • $\ldots,$ and that is a lousy answer. $\qquad$ – Michael Hardy Aug 08 '16 at 19:24
  • It's just a convention, but one that has proven to be very convenient. – Eric Wofsey Aug 08 '16 at 19:24
  • This follows from the principle of explosion and its proof can be found on its Wikipedia page. – Vishal Gupta Aug 08 '16 at 19:25
  • @EricWofsey What would you say then to this answer? http://math.stackexchange.com/a/440242/70349 – AJJ Aug 08 '16 at 19:27
  • That answer is really just explaining why the convention is convenient (despite its insistence to the contrary). – Eric Wofsey Aug 08 '16 at 19:29
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    It sounds like you want "$p \implies q$" to carry some semantic content about $p$ holding because of $q$: that $q$ fails without $p,$ or that there exists no universe with $p$ but not $q$, or that $p$ is somehow 'relevant' to $q$. But it's just convenient shorthand for "$q$ or not $p$." – anomaly Aug 08 '16 at 19:30
  • @anomaly To repeat my comment on another answer: That follows directly from the definitions in the truth tables: $p \to q = (p \land q) \lor (\lnot p \land q) \lor (\lnot p \land \lnot q) = \lnot p \lor q$. It still does not address why we have both $(\lnot p \land q)$ and $(\lnot p \land \lnot q)$ in the definition. What I am asking is why we can't say that if $p$ is false, then $p \to q$ is simply undefined. – AJJ Aug 08 '16 at 19:33
  • @ArukaJ: We can say that, we just choose not to, because the standard definition has proven to be much more useful. That is, we have learned from experience that the logical statement "$q$ or not $p$" is really useful to have a shorthand for, so we've made one and named it $p\implies q$. – Eric Wofsey Aug 08 '16 at 19:37
  • Can you give an example of how it is a useful definition to have and why having it be undefined would be problematic? – AJJ Aug 08 '16 at 19:38
  • celtschk's answer (which is basically the same as the answer of Peter Smith you linked to) is an excellent example: this definition is very natural and convenient when working with quantified statements. – Eric Wofsey Aug 08 '16 at 19:48
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    Here is my answer ... http://math.stackexchange.com/a/48202/442 – GEdgar Aug 08 '16 at 20:03
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    Get over you pre-conceptions that "P \implies Q" translates as "If P than Q". That is not what it means. Instead it means "whenever P, Q". Surely it's obvious that whenever the moon is made of bacon wrapped apple monkey carburetors are precisely the same times as when both the moon is made of bacon wrapped apple monkeys carburetors and you are a better wake border than Gauss. i.e. never. – fleablood Aug 08 '16 at 20:10
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    @ArukaJ: Propositions in (first-order, etc.) logic are either true or false. There is no "undefined" option. My point is that it's really notation rather than a definition. It sounds like you're caught up in the idea of $\implies$ representing some sort of implication (which is hard to nail down); it's just shorthand for a particular, convenient disjunction. – anomaly Aug 08 '16 at 20:16
  • Are you familiar with the Principle of Explosion? (I.e., that from a contradiction one may infer anything?) – Eric Towers Aug 08 '16 at 20:48
  • @EricTowers It seems just as reasonable to conclude though that "from a contradiction one may infer nothing" – AJJ Aug 08 '16 at 21:14
  • Someone tells you, "If you don't eat your meat, you can't have any pudding". You eat your meat. Does that somehow make their statement no longer true? – David Schwartz Aug 08 '16 at 21:37
  • Informally? It doesn't make it true or false. If I don't eat my meat and you don't give me pudding, you held up your statement. If I don't eat my meat but you gave me pudding anyway, your statement was false. But if I eat my meat, we're going off-contract, here. You never told me what would happen if I did eat my meat. – AJJ Aug 08 '16 at 21:54
  • @ArukaJ Because nothing was said about what happens if you do eat your meat, there can be no "breach of contract" if you eat your meat. Only if you don't eat your meat yet are given pudding can the statement be deemed false. – Monty Harder Aug 08 '16 at 22:47

12 Answers12

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Consider the statement:

All multiples of 4 are even.

You would say that statement is true, right?

So let's formulate that in formal logic language:

$\forall x: 4|x \implies 2|x$

(Here "$a|b$" means "$a$ divides $b$", that is, $b$ is a multiple of $a$.)

Now a $\forall$ statement is true if it is true whatever you insert for the quantified variable (after all,that's "for all" means). So let's try to insert $3$:

$4|3 \implies 2|3$

But wait, $4|3$ is false! Moreover, $2|3$ is also false. So the only way for the original statement to be true is that the implication $\text{false}\implies\text{false}$ gives true.

A similar argument can be done for $\text{false}\implies\text{true}$.

celtschk
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    This is an excellent example. –  Aug 08 '16 at 20:18
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    The reason I did not accept this answer is because it does not address the confusion directly. "4|3" does not represent a multiple of 4, so now we are no longer talking about what the original statement was discussing. Now we're talking about some manipulation that is false, which may have something true come from it, may have something false come from it. What I wanted to understand was why we'd still say $P \to Q$ is true even when $\lnot P$ leads to garbage results / truth or false either way. – AJJ Aug 08 '16 at 20:20
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    @ArukaJ P=4|x and Q=2|x. Where x=3, P and Q are both false. Where x=6, P is false and Q is true. With two-value logic, we can't possibly say either of those makes P⟹Q false, so the only reasonable alternative is to say that P⟹Q is defined to be true when P is false. – Monty Harder Aug 08 '16 at 22:38
  • @MontyHarder My point is that we could also reasonably say it is undefined, as an alternative – AJJ Aug 08 '16 at 22:40
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    @ArukaJ Every time you set x to a number divisible by 4, it's also divisible by 2. How can you say that "P⟹Q" is anything but true all of the time? – Monty Harder Aug 08 '16 at 23:11
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    @ArukaJ "$\forall x: (P(x)\to Q(x) \text{ except when $P(x)\to Q(x)$ is undefined})$" is not at all a reasonable alternative. – Graham Kemp Aug 09 '16 at 02:02
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    @ArukaJ: But then you have to move the 4|x part into the quantifier (the for-all bit) somehow, in order to avoid quantifying over x which are not multiples of 4. It turns out that it's simpler to just define the material implication (the right arrow symbol) as true in this case than to extend the rules of logical quantification to allow for something like this. – Kevin Aug 09 '16 at 03:58
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    This doesn't address the original question of why not leave it undefined. It only says that if it is defined, how it should be. – DanielV Aug 09 '16 at 04:02
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This is done so that classical propositional calculus follows some natural rules. Let's try to motivate this, without getting into technical details:

The expression "$P\Rightarrow Q$" should be read "$P$ implies $Q$", or "whenever $P$ is true, $Q$ is also true".

The negation of such an expression would be a counter-example, i.e., "there is some case in which $P$ is true but $Q$ is not".

So assume $P$ is not true. The negation "$\lnot(P\Rightarrow Q)$" is not true in this case, by our interpretation above, so "$P\Rightarrow Q$" must be true.

We are basically using the rules that either an expression or its negation should be true, and that the negation of the negation of an statement is the statement itself. These are basic rules which are natural and useful, even though as a consequence we have that "$P\Rightarrow Q$" is true whenever $P$ is false.

Luiz Cordeiro
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    So we're saying that either $P \to Q$ is true or $\lnot(P \to Q)$ is true, right? Agreed there. If $P \to Q$ is false when $P$ is true and $Q$ is false (also agreed), then $\lnot(P \to Q)$ should be true. By De Morgan's laws, this means $P$ is false or $Q$ is true. Oh holy crap. – AJJ Aug 08 '16 at 19:58
  • I don't think the fact that P⇒Q is (vacuously) true if P is never true poses much of a problem in non-causal theorem spaces. I think a bigger issue occurs when the same sort of logic is used in a causal universe. Use R(A) to mean "Action A is reasonable". An invitation to assume P should mean that anything that would be reasonable if P is true, is reasonable, i.e. for all A such that P→R(A), R(A). In a causal universe, however, it would be helpful to distinguish between actions whose reasonableness would be caused by P, versus those without a forward causal link. – supercat Aug 08 '16 at 21:47
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In a more informal sense, I like to think that $P\implies Q$ means that "$Q$ is at least as true as $P$". Which means that if $P$ is something false, then anything is "more true" than $P$, and thus the statement $P\implies Q$ is true.

Alex Saad
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  • But do we presuppose that falsity is less than truth in logic? – Doug Spoonwood Aug 09 '16 at 17:58
  • @DougSpoonwood well, strictly speaking there is no defined "measure" of truth/falsity - the above answer was merely meant as a useful heuristic. But logical implication translates into set-theoretical inclusion when you take sets of elements satisfying logical properties, which means that there are at least as many elements $x$ satisfying $Q$ than $P$ if $P(x)\implies Q(x)$, so in some ways this can be made more rigorous. – Alex Saad Aug 10 '16 at 10:12
  • If there exist "at least as many" elements satisfying property Q than P does not imply that Q as bigger than P. Q and P could have equivalent size. – Doug Spoonwood Aug 10 '16 at 18:20
  • @DougSpoonwood of course - I mean "at least as many" in the non-strict sense (less than or equal to). If they have the same size then this just means that the reverse implication $Q\implies P$ also holds i.e. the properties $P(x)$ and $Q(x)$ are equivalent. – Alex Saad Aug 10 '16 at 19:12
  • You said "Which means that if P is something false, then anything is "more true" than P, and thus the statement P -> Q is true." Now, if P(x) and Q(x) can hold as equivalent, then the reverse implication can hold, which would imply that a true proposition could imply a false proposition. But, that never happens in propositional logic. Consequently, your attempt to order truth values fails. – Doug Spoonwood Aug 11 '16 at 00:10
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Intuitively, consider $P \implies Q$ to be a promise. For example: I say, "If you ever jump over the moon, then I'll pay you a thousand dollars."

Now, some time passes and you obviously haven't jumped over the moon. The question is now effectively: Have I kept my promise true, or broken it (been false)?

It seems clear that I've kept that particular promise, granted that at no time was I obligated to give you the thousand dollars. In fact, even if I had given you a thousand dollars for some other service, it wouldn't have any bearing on that particular, empty offer.

Daniel R. Collins
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    But we could also say that we haven't been able to tell if you held up the promise either way, because I didn't jump over the moon to find out – AJJ Aug 08 '16 at 21:47
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    @ArukaJ We can say definitively that Daniel has done everything he promised you he'd do (which in this case is nothing). – Monty Harder Aug 08 '16 at 22:41
  • He hasn't done anything one way or the other, so the jury's still out – AJJ Aug 08 '16 at 22:41
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    He gave you exactly the money he promised you. The jury finds in his favor, and you have to pay court costs for filing a frivolous suit against him. – Monty Harder Aug 08 '16 at 23:26
  • He didn't cheat me in that scenario so I wouldn't "file" -- he would have only cheated me if I jumped over the moon and he didn't pay me. But if I didn't jump over the moon yet, he has not yet cheated me nor has he followed through on anything. – AJJ Aug 08 '16 at 23:29
  • @ArukaJ I've turned this line of discussion into a separate answer, by using the same numbers on both sides of the equation. Suffice to say that by not jumping over the moon, you have guaranteed that Daniel has fulfilled his promise to you. Therefore his promise is truthful. – Monty Harder Aug 08 '16 at 23:54
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Consider any proof by contradiction, for example Euclid's proof of prime infinity.

If $x$ is the largest prime number ("P"), then $y=\prod_{p\,\,prime} p+ 1$ is a prime ("Q").

This is a valid implication that can be shown with perfectly valid simple arguments. However, the premise is never satisfied. But within the proof, the reader may not yet see that it is ''nonsense'' at this step. You just follow the logic step by step. The fact that the premise is never satisfied does not make the logic (used to derive the claim above) less valid. So it is reasonable to assume that the claim above is True.

Peter Franek
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    This answer reproduces a remarkably popular erroneous reading of the proof. The "Euclid number" $30031=235711*13+1$ is not prime. – Kevin Carlson Aug 08 '16 at 22:01
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    @KevinCarlson No, it doesn't. All you've proved is that if the claim is true, then $13$ cannot be the largest prime number. But that is the point: the claim is provably true, and what follows that claim is that no number can be the largest prime number. – hvd Aug 08 '16 at 22:22
  • Oh, huh, that's right. The wrong version just starts with a finite list of prime numbers, without assuming it's exhaustive. – Kevin Carlson Aug 08 '16 at 22:31
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$p\to q$ is logically equivalent to $\neg p \vee q$. Hence if $p$ is false, then $\neg p$ is true, so $\neg p \vee q$ is true. This is the situation we call vacuously true.

vadim123
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    Yes but that follows directly from the definitions in the truth tables: $p \to q = (p \land q) \lor (\lnot p \land q) \lor (\lnot p \land \lnot q) = \lnot p \lor q$. It still does not address why we have both $(\lnot p \land q)$ and $(\lnot p \land \lnot q)$ in the definition. – AJJ Aug 08 '16 at 19:29
  • @ArukaJ no, the definition you give follows from this. If $p$ is false, then $\neg p \vee x$ is independent of $x$, because "or" is satisfied by one true thing. – hobbs Aug 09 '16 at 02:12
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    @ArukaJ, the reason the definition is the way it is, is because people find this definition useful. If you don't like it, make your own definition, of your own connective. – vadim123 Aug 09 '16 at 02:37
  • How do you know that logical equivalence holds without having the truth tables for $\lor$ and ->, and $\lnot$ in the first place? – Doug Spoonwood Aug 09 '16 at 17:55
  • The definition works, because a logical equivalence exists. Were it the case that no logical equivalence existed between (p -> q) and ($\lnot$p $\lor$ q), then the definition would not qualify as sound. – Doug Spoonwood Aug 09 '16 at 17:57
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One way to look at this is by set inclusion:

"$P \implies Q$" is the same as saying that "the set of mathematical statements that have $P$ as true (call it $S_P$) is a subset of the set of statements that have $Q$ as true (call it $S_Q$)." So if $P$ is false, then $S_P = \emptyset$. It follows that whatever $S_Q$ is, $\emptyset = S_P \subset S_Q$. Hence, $P \implies Q$ is vacuously true. $Q$ by itself is not vacuously true: you need the antecendent, $P \implies$, for the "vacuous" property to make sense.

user360187
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This is what Terence Tao has to say:

This is discussed in Appendix A.2 of my book. The notion of implication used in mathematics is that of material implication, which in particular assigns a true value to any vacuous implication. One could of course use a different convention for the notion of implication, however material implication is very useful for the purpose of proving mathematical theorems, as it allows one to use implications such as “if A, then B” without having to first check whether A is true or not. Material implication also obeys a number of useful properties, such as specialisation: if for instance one knows for every x that P(x) implies Q(x), then one can specialise this to a specific value of x, say 3, and conclude that P(3) implies Q(3). Note though that by doing so a non-vacuous implication can become a vacuous implication. For instance, we know that $x \geq 5$ implies $x^2 \geq > 25$ for any real number $x$; specialising this to the real number 3, we obtain the vacuous implication that $3 \geq 5$ implies $3^2 \geq 25$.

The way I like to think of material implication is as follows: the assertion that A implies B is just saying that “B is at least as true as A”. In particular, if A is true, then B has to be true also; but if A is false, then the material implication allows B to be either true or false, and so the implication is true no matter what the truth value of B is.

Hatshepsut
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OK, here's a crack at it, based on something we had going in comments. It's an argument by analogy, which isn't really a good idea, but I think it can help you make sense of why it has to be this way:

I'm running a currency exchange and promise to give you one Cuban Convertible Peso for each US Dollar you give me.

You give me USD 10 on Monday, and I give you CUC 10 in exchange.

On Tuesday you give me nothing. I give you nothing in exchange.

On Wednesday, you give me USD 5. I give you CUC 5 in exchange.

On Thursday, you try to give me a Euro. I could just accept EUR 1 from you as a gift, offer some amount of change for it, or push it back to you and point at the sign that says I trade USD for CUC. No matter which of these I choose, I am still fulfilling my promise.

On Friday, you didn't give me anything, but I give you CUC 0.25 out of the goodness of my heart. Or maybe because you're amusing me and that's worth rewarding.

No amount of not giving me USD can change the truth of the statement "Monty is an honest currency broker who keeps his promise to give at least as much CUC for every USD you give him".

When you gave me USD 0, I gave you at least CUC 0. For the four days, you've given me USD 15 and I've given you CUC 15.25 (and maybe some more for the EUR 1). I'm absolutely keeping my promise every single day, and if you claim I'm not keeping my promise on the days you gave me USD 0, then you're lying, because I gave you at least CUC 0 on those days. You can't even argue that "the jury's out" on whether I gave you at least as many CUC each day as you gave me in USD, because 0=0. To do so would itself be untrue, and an actual jury might rule you've libeled me by doing that.

Similarly, to treat (P⟹Q) as anything but true when P is false would itself be false.

Monty Harder
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A different point of view is to think of an argument as a chain of lights. A light is lit if you know it is true.

P implies Q says that when you manage to ignite the P light then P ignites the Q light too. However, if P is not lit then it doesn't mean that you can't light Q in some other way.

Mark Joshi
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If we assume the principle of bivalence, and we want the law of identity to hold true, I answer the question by saying that the reason lies in that logical implication is not logical equivalence.

The law of identity for all propositions x, (x ⟹ x), ensures that if p is false and q is false, then (p ⟹ q) is true.

The true does not imply the false, thus if p is true and q is false, (p ⟹ q) is false.

Thus, our truth table looks like the following so far:

⟹     false  true
false  true     ?
true   false  true

Now if the question mark got filled in by false, then ⟹ would indicate logical equivalence. But, we don't logical equivalence for logical implication. We don't want to hold that {($\alpha$ ⟹ $\beta$), $\beta$} $\vdash$ $\alpha$. And we want ⟹ to work out as truth-functional. In other words (false ⟹ true) matches to some truth value. Consequently, (false ⟹ true) holds true.

Doug Spoonwood
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If the light is green, I can make a right turn. If the light is red, I might still be able to make the turn-it depends if the sign says "no turn on red" and if any cars are coming. But you don't know whether I can make the turn if all you know is that the light is red.

If you know that I can make the turn, you don't know what color the light is- it could either be green, it a red light that I'm allowed to make the turn at. But if you know that I couldn't make the turn, it's guaranteed that the light was red.

As another example, if I rob a bank, I go to jail. However, if I don't rob a bank, that doesn't guarantee that I don't get arrested- I could have committed a different crime. On the same note, if I am in jail, you don't know if I robbed a bank or not, but if I'm not in jail, I definitely haven't broken in to a bank.

Daniel M.
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