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I want to prove that some function $f(x)$ assigns the same quantity $y$ to any pair of identical objects in a finite set of $|N|$ objects. Given the structure of the problem, I'm using induction. So far, I'm proving that my claim is true under the $2$ objects case; and I then do the inductive step. However, for the case $|N|=1$, the statement is vacuously true. Therefore, my original statement is vacuously true for the case $|N|=1$.

Then, my question is the following one: can I use the case of $|N|=1$, which holds vacuously, as the first step of my inductive proof?

I hope my question makes (now) sense.

Thank you all very much in advanced for your time.

EoDmnFOr3q
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    This is not clear ... you say that the Property X is a property of individual objects (which makes sense in the context of trying to do an inductive proof) .. but then you say: "For all pairs of objects meeting condition A, Property X must hold" ... are you saying that if there is a pair that satisfies condition A, then both objects of that pair have Porperty X? – Bram28 Apr 20 '18 at 15:51
  • Thank you for your comment. Now that you point it out, I see that the (original) question was poorly written. Thank you for pointing this out. I'm sorry for the confusion. Please, see the edited question. I hope it is more clear now. – EoDmnFOr3q Apr 20 '18 at 15:55
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    Sorry, I am still confused. You write "Such a property reads as follows: For all pairs of identical objects, property X must hold" ... that almost sounds like some meta-property ... which I don't think you meant. Are you just trying to say that "We are given that: For all pairs of identical objects, property X must hold"? And by that you mean that whenever two object are identical they both have the property? But later you write: "any Property relating two objects ..." Eh, are we talking about a property or a relation? Your question is still very confusing! – Bram28 Apr 20 '18 at 16:10
  • OK, now I see that the question is still poorly written. I'm sorry my mistake prevented you from understanding the question since the beginning. Please, see the edited question. – EoDmnFOr3q Apr 20 '18 at 16:15

2 Answers2

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The short answer is yes -- vacuous truth is truth. It's no different from any other truth.

However, you have to be careful to make sure your inductive step isn't implicitly relying on anything which doesn't apply in the non-vacuous case.

For example, suppose I wanted to prove, for a finite set of $N$ colored objects, that every pair of objects is blue. I prove this as follows: for $N=1$, this is vacuously true. Now suppose among every set of $N$ objects, every pair is blue. Among a set of $N+1$ objects, each pair of objects belongs to a subset of $N$ objects, so they must be blue; therefore every pair of objects in the set of $N+1$ is blue. QED.

Clearly this argument can't be sound. The problem is the claim "Among a set of $N+1$ objects, each pair of objects belongs to a subset of $N$ objects." This isn't true when $N=1$ and $N+1=2$. The (only) pair of objects in the set of $2$ doesn't belong to a subset of size $1$. The claim will hold only for $N\geq2$.

This isn't a problem with using vacuous truth per se, but it's an example of the type of thing that could go wrong when you try to jump from the vacuous case to the non-vacuous case. If you make sure to avoid this sort of fallacy, inductive arguments with a vacuous base case are perfectly okay.

BallBoy
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  • Thank you for your answer. Could you please elaborate on the idea that I should "make sure [my] inductive step isn't implicitly relying on the truth's being non-vacuous"? I'm not sure I get what you mean. – EoDmnFOr3q Apr 20 '18 at 16:02
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    @Héctor Here's a particularly egregious example. Suppose I wanted to prove, for a finite set of $N$ colored objects, that every pair of objects is blue. I prove this as follows: for $N=1$, this is vacuously true. Now suppose among every set of $N$ objects, every pair is blue. Among a set of $N+1$ objects, each pair of objects belongs to a subset of $N$ objects, so they must be blue; therefore every pair of objects in the set of $N+1$ is blue. QED. – BallBoy Apr 20 '18 at 16:07
  • Thank you A LOT for this BEAUTIFUL EXAMPLE. I would like to encourage you to add it your main answer, albeit I will accept it whether you do it or not. Now, what I would REALLY appreciate is that you pointed out where in the "proof" of your example the mistake is. Because the inductive reasoning must be wrong somewhere, right? – EoDmnFOr3q Apr 20 '18 at 16:09
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    @Héctor Lots of examples of that argument can be found on this site. See e.g. https://math.stackexchange.com/questions/2222480/show-that-all-horses-are-of-the-same-color – Bram28 Apr 20 '18 at 16:12
  • Thank you a lot for this link; I hadn't seen it before. – EoDmnFOr3q Apr 20 '18 at 16:14
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    @Héctor I've added the example to my answer and explained the fallacy. – BallBoy Apr 20 '18 at 16:17
  • Thank you very much for this! :D – EoDmnFOr3q Apr 20 '18 at 16:19
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It sounds like you are saying that whenever two objects are identical, they both have the property.

Well, that makes the problem trivial: every object is identical to itself, and therefore has the property: no induction needed!

Bram28
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