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This question already has an answer here:

I don't know very much about formal logic, and I'm trying to understand the concept of vacuously true statements. Consider the truth table below: $$\begin{array} {|c|} \hline P & Q & P\implies Q & Q\implies P & P\iff Q \\ \hline T & T & T & T & T \\ \hline T & F & F & \color{blue} T & F\\ \hline F & T & \color{blue}T & F & F\\ \hline F & F & \color{blue} T & \color{blue}T & T \end{array}$$.

The blue letters are definitions. To see why these definitions are the correct choices (as opposed to $F$s), suppose we changed the lower left $\color{blue} T$ to an $F$ (this would then force the lower right $\color{blue}T$ to an $F$), so $P\iff Q$ would be $F$ for $P$ and $Q$ both $F$, which isn't want we want. So I see why this makes sense to choose the lower entries as $\color{blue}T$.

However, it isn't clear to me why $P\implies Q$ true for $P$ false and $Q$ true is the sensible choice (same for $Q\implies P$ true for $Q$ false and $P$ true). For if the $\color{blue}T$s in the middle rows were switched to $F$s, then $P\iff Q$ would still be $F$. So I don't see what the problem would be. Can someone please explain?

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marked as duplicate by GEdgar, SchrodingersCat, user91500, Alex M., Tom-Tom Nov 30 '15 at 14:18

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  • $\begingroup$ @MauroALLEGRANZA Those are all great answers. I would only add: $p\implies q$ is true iff $Q$ is at least as true as $P$, that is, iff $\operatorname{truthValue}(P) \le \operatorname{truthValue}(Q)$, whereas $p\iff q$ is true iff $\operatorname{truthValue}(P) = \operatorname{truthValue}(Q)$. Obviously, then, $false\implies true$ is (should be) true, as $0\le 1$. $\endgroup$ – BrianO Nov 29 '15 at 12:45
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Here is why we say $P \implies Q$ is true if $P$ is false and $Q$ is true:

Let $P$ be the statement "it's raining outside" and $Q$ be the statement "the car is wet".

In order for $P \implies Q$ to be true, what needs to happen is: every time it rains outside, it better follow that the car is wet. That's all you need to check.

So the only time $P \implies Q$ is false is when we get that it's raining outside and the car isn't wet.

When it's not raining, we don't have that it's raining outside and the car is not wet. Since we don't have this, that means $P \implies Q$ isn't false, since the only time it is false is when we get that it's raining outside and the car isn't wet. If it's not raining outside, we don't get this, so the statement is not false, so it must be true.

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    $\begingroup$ Another example I always like to give: suppose before a football match I say “If Brazil beats Germany, I’ll eat my hat.” As long as Brazil doesn’t win, I don’t have to do anything. My promise is vacuously kept. $\endgroup$ – Peter LeFanu Lumsdaine Nov 29 '15 at 15:03
  • $\begingroup$ Cute example, but in the mathematical sense of implication, time is not a consideration. Neither is causality. See my answer. $\endgroup$ – Dan Christensen Nov 29 '15 at 15:21
  • $\begingroup$ I didn't mean to suggest that your answer was wrong. Only that the passage of time is, e.g. the suggestion of a historical pattern, is not an essential part of an implication in mathematics as it may be in scientific applications. In mathematics, an implication may apply even to a single instant in time. $\endgroup$ – Dan Christensen Nov 29 '15 at 15:40
  • $\begingroup$ @DanChristensen Good point. Regardless of which setting/context, or if time is a factor or not, $P \implies Q$ is true if in every scenario where $P$ is true, $Q$ will be true. $\endgroup$ – layman Nov 29 '15 at 16:12
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    $\begingroup$ Yes, and in every scenario where $P$ is false, it wouldn't matter if $Q$ was true or false, $P\implies Q$ would be true. $\endgroup$ – Dan Christensen Nov 29 '15 at 17:38
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$P \Rightarrow Q$ is logically equivalent to $\sim P \vee Q$ (where $\sim$ is '"not" and $\vee$ is "or"), and $\sim P \vee Q$ is true whenever $P$ is false. Another thing to think about is providing counterexamples. $P\Rightarrow Q$ can only be proved false if you give an example where $P$ is true and $Q$ is not true. Any example where $P$ is false is not a counterexample, so $P\Rightarrow Q$ is vacuously true in that case.

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  • $\begingroup$ Thank you, I've edited my answer. $\endgroup$ – kccu Nov 29 '15 at 15:59
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Consider the implication "If it is raining, then it is cloudy."

$$Raining\implies Cloudy$$

In the mathematical sense of implication, this is not a claim of a causal relationship between rain and cloudiness: e.g. that cloudiness somehow causes rain, or that rain somehow causes cloudiness. This is also not a claim of any kind of historical pattern over time. An implication can be applied to a single instant in time, e.g. to logical relationships that may be evident even in a single still photograph. In the mathematical sense of implication, it is only being claimed that, at a given instant in time, it is not both raining and not cloudy.

$$Raining \implies Cloudy \equiv \neg[Raining \land \neg Cloudy]$$

There is no notion of cause and effect in mathematics. There is no passage of time. These are the realms of science. In mathematics, time is just another variable of no particular significance.

$$\begin{array}{c|c|c|c} \space&Raining&Cloudy&Raining\implies Cloudy\\\hline 1&T&T&T\\ 2&T&F&F\\ 3& F&T&T\\ 4&F&F&T \end{array}$$

From this truth table, we see that, at any given instant in time...

  • If $Raining\implies Cloudy$ is true and $Raining$ is true (line 1), then $Cloudy$ will be true.
  • If $Raining \implies Cloudy$ is true and $Cloudy$ is false (line 4), then $Raining$ is false.
  • If $Raining \implies Cloudy$ is false (line 2) then $Raining$ is true and $Cloudy$ is false
  • If $Raining$ is false (lines 3, 4), then $Raining \implies Cloudy$ will be true
  • If $Cloudy$ is true (lines 1, 3) , then $Raining \implies Cloudy$ will be true

Follow-up: The truth table for material implication does not depend solely on the above definition. See my answer at How does one know if $A \implies B$ (an implication) is true without knowing if $B$ (the consequent is true) is true?

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  • $\begingroup$ I'm having trouble based on your comment on my answer, and your own answer posted, understanding where I was wrong in my answer or in my understanding. In my mind, the statement $P \implies Q$ means if you look out in the "world", and see statement $P$ happening, then statement $Q$ must also happen (if there were time, it could happen just after $P$ happened). What's wrong with this way of interpreting an implication? That $P \implies Q$ being true means whenever $P$ happens, $Q$ must happen. $\endgroup$ – layman Nov 29 '15 at 15:27
  • $\begingroup$ Sorry, I didn't mean to suggest that your answer was wrong. Only that the passage of time is, e.g. the suggestion of a historical pattern, is not an essential part of an implication in mathematics as it may be in science. $\endgroup$ – Dan Christensen Nov 29 '15 at 15:36
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One of the "basic" pattern of inference is modus ponens (its latin name gives us a "flavour" about how old it is).

Modus ponens formalize the inference step of the form : "if $A$ then $C$, $A$; hence $C$.

It is useful to define a conenctive "supporting" this basic move : the conditional.

Suppose we claim "if $A$, then $C$. This means to rule out the case where the antecedent $A$ holds and yet the consequent $C$ doesn't; i.e. we have "defined" the second row in your table above.

Thus, having set the first, second and fourth row, the third must follow, in order at least to avoid to "conflate" conditional with bi-conditional.

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Maybe it'll make a little more sense if you see the table knowing the following logical equivalence: $$(P \implies Q) \equiv (\neg P \lor Q)$$ $$(Q \implies P) \equiv (\neg Q \lor P)$$ i.e. "$P$ implies $Q$" is logically equivalent to "either not $P$, or $Q$"

Now that we know that this is the case, let's redraw the truth table using the right hand side of the equivalence given above, and hopefully it will seem a bit more intuitive:

$$\begin{array} {|c|c|} \hline P & Q & \neg P & \neg Q & P\implies Q & Q\implies P & P\iff Q \\ \hline P & Q & \neg P & \neg Q &\neg P \lor Q & \neg Q \lor P & (\neg P \lor Q) \land(\neg Q \lor P) \\ \hline \color{red} T & \color{red} T & F & F & \color{red} T & \color{red} T & \color{red} T \\ \hline \color{red} T & F & F & \color{blue} T & F & \color{blue} T & F\\ \hline F & \color{red} T & \color{blue} T & F & \color{blue}T & F & F\\ \hline F & F & \color{blue} T & \color{blue} T & \color{blue} T & \color{blue}T & \color{blue} T \\ \hline\end{array}$$.

I used red to indicate things that might make the statement "normally" true, and blue to indicate things that might make the statement vacuously true.

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The reason "If $P$ then $Q$" is considered to be true in the case when $P$ is false is simple:

It's a true statement because it's not false.

Let me unpack that. What would it mean for "If $P$ then $Q$" to be false? It would mean that $P$ is true, but $Q$ isn't. That's really the only way an implication can fail: If the hypothesis holds, but the conclusion doesn't.

In the situation where $P$ is false, the implication can't fail to deliver, because it's never invoked in the first place.

Consider this example: "If $n$ is an even prime number greater than $10$, then the last digit of $n$ is $6$." To falsify that statement, you would need to exhibit an even prime number greater than $10$ whose last digit isn't $6$. But that can't happen -- not because all such numbers end in $6$, but because there are no even prime numbers greater than $10$, so there's no way the claim can fail.

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The blue letters aren't definitions. The entire column is the definition.

The reason some rows of the definition seem questionable is that they are--relative to the most common uses of if/then in English. The most common uses of if/then can't be formalized in terms of propositional logic; no truth table can capture the meanings of these uses.

What some of the other answers give are reasons that the standard material conditional truth table seems reasonable if we must represent if/then using only propositional logic. (But be careful; sometimes fallacious arguments are given for the standard truth table, even in logic textbooks.)

Another motivation for the material conditional is that it's exactly what you need in many cases in predicate logic, e.g. to represent the logic of "All ravens are black", i.e. "Everything is such that if it is a raven then it's black."

More advanced logic is needed to capture the meanings of commonplace uses of if/then in English--e.g. some uses of if/then can be captured in modal logic, maybe, as Robert Stalker argued. Work in this area is ongoing. The Stanford Encyclopedia of Philosophy has articles relevant to this topic:

http://plato.stanford.edu/entries/conditionals/

http://plato.stanford.edu/entries/logic-conditionals/

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