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This question already has an answer here:

How can I convince students that "P implies Q" is true when P is false, independent of what truth value Q takes?

Is there any real life or a convincing argument for this?

I have given the analogy of if being elected then you will abolish the death penalty which seems ok but rather crass. Are there any other examples?

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marked as duplicate by Paracosmiste, Xander Henderson, kcrisman, Chris Cunningham, JoeTaxpayer Apr 25 at 0:46

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    $\begingroup$ The problem with "If I am elected, then I will abolish the death penalty" is, if you are not elected, then how can you abolish the death penalty? $\endgroup$ – Joel Reyes Noche Apr 17 at 10:48
  • $\begingroup$ How is implication defined for the students? Both truth table and the common $A \Rightarrow B \iff \neg A \vee B$ would make things very clear. $\endgroup$ – Jasper Apr 17 at 11:01
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    $\begingroup$ So you begin by giving the truth table? $\endgroup$ – Jasper Apr 17 at 11:25
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    $\begingroup$ I'm sure this had been asked before on math.se. You can use $x>8 \Rightarrow x>0$. $\endgroup$ – Paracosmiste Apr 17 at 20:51
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    $\begingroup$ The mathematical concepts bear only a superficial resemblance to the real world. Yes, math is inspired by real-world problems but abstracts from them and establishes concepts which try to do without natural language as much as possible, for good reason. So "P implies Q" is a math operation with two boolean arguments resulting in a bool. Just present the truth table. "Any resemblance to existing real-world situations is accidental and unintentional." $\endgroup$ – Peter A. Schneider Apr 18 at 13:36
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I make the statement "If it is raining, then I have an umbrella." Did I lie?

If it is raining and I do not have an umbrella, then I lied.

If it is raining and I do have an umbrella, then I didn't lie.

If it is not raining, then it doesn't matter whether or not I have an umbrella; I still did not lie.

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    $\begingroup$ I use exactly this example when I teach this concept. Moreover, you can use it to explain the logical negation: $\neg (P\implies Q) \iff (P \wedge \neg Q)$, because the only situation in which you can call me a liar is if you see me walking around in the rain without an umbrella. $\endgroup$ – Brendan W. Sullivan Apr 17 at 18:19
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    $\begingroup$ If it is not raining, did you tell the truth? $\endgroup$ – immibis Apr 18 at 3:24
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    $\begingroup$ @immibis, yes.. $\endgroup$ – Joel Reyes Noche Apr 18 at 5:32
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    $\begingroup$ This answer has the implication that all statements that are not false (not a lie) are true. Probably fine for introducing the concept... but @immibis has a good point. $\endgroup$ – Rick Apr 18 at 13:12
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    $\begingroup$ @Rick And exactly there is the problem for the casual reader. It's called vacuous truth. A condition which is always wrong yields true for any consequence. $\endgroup$ – Peter A. Schneider Apr 18 at 13:32
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You could say it means "Whenever P is true, Q is true". So "If it rains, I will bring an umbrella" means "Every time it rains, I bring an umbrella". It's not possible to disprove this statement by looking at what happens when it doesn't rain.

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If I do the work, I will get paid.

  • If I do the work and get paid, all is good.
  • If I do the work and not get paid, it's bad!
  • If I don't do the work and don't get paid, I cannot complain - it's all good.
  • If I don't do the work but still get paid - now this is good!

I know it's not exactly how one should interpret implication, but it worked for me. :)

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You can introduce implication and equivalence side by side to make the difference clear.

Implication: If A, then also B. (But if not A, this statement does not tell us anything. See the umbrella example in Joel's answer.)

Equivalence: If and only if A, then B. This includes that not A implies not B.

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    $\begingroup$ I've never encountered "If and only if A, then B" in writing or in speech. Is it common in your context (country/text book/field)? Otherwise, I'd consider "A if and only if B" or equivalently (by symmetry) "B if and only if A" far better because yours sounds much more asymmetric to me -- which $\Leftrightarrow$ is exactly not. $\endgroup$ – ComFreek Apr 17 at 13:53
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    $\begingroup$ Hmm. You're right, but I've tried to keep the similarity to the implication. I also think that "B if and only if A" is way more common. $\endgroup$ – Jasper Apr 17 at 14:36
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It's obvious if you view a proposition as a subset of all values in the proposition domain for which the proposition is true.

For example, a proposition "this umbrella is red" selects only red umbrellas form the set of all umbrellas that exist in the world.

A proposition "I own only red umbrellas" can be viewed as implication - P is "I own this umbrella", Q is "this umbrella is red", and it can be interpreted as "the set of umbrellas that I own is a subset of red umbrellas", that is, implication corresponds to "being a subset" relation in the set interpretation (similarly, conjunction corresponds to set intersection and disjunction corresponds to set union).

False proposition corresponds to an empty set - there are no values in the proposition domain for which it's true. From set theory axioms we know that empty set must be a subset of any set. This immediately gives you

"P implies Q" is true when P is false

It's essentially a formality, but a convenient one - for example if you don't own any umbrellas at all you can safely say "I own only umbrellas hand-made by unicorns from rainbows" and it's assumed to be true because it can't be disproved.

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  • $\begingroup$ This explains the truth values for implications assuming students already accept that implication (of propositions) should correspond to inclusion (of sets). Is it easy to convince them to accept that? Will it cause a problem when they see $\subseteq$ defined (in set theory) using $\implies$? $\endgroup$ – Andreas Blass Apr 18 at 10:49
  • $\begingroup$ I don't think it will cause any problems, you just have to explain that ⊆ is exactly the same thing as ⟹ when viewed from a certain angle. $\endgroup$ – artem Apr 18 at 23:43
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My approach in the past has always been to work via negation. However, I've recently read one of Tim Gowers' 'discussions' that showed me a new side to the problem.

Basically he points out that in everyday usage (and in proofs too) we are looking for causal relationships. We want to say that 'if A then B' is true when we can understand A as leading to B. But when dealing with more formal logic, we are actually only concerned with evidence. We say 'if A then B' is true when 'in every case where A is true, B is also true', regardless of whether this is something you would expect or just plain chance.

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