Note that Step 9 can be read as introducing an entity that is known to exist, or introducing a variable to be universally quantified over. It isn't until Step 13 that we see that Step 9 should be read as the former.
TFA will inform you of this though, it's intended to be a game of reasoning. Posting the answer is spoiling the fun for people who read the comments first
Why isn't there a mechanism for hiding spoilers in your comments? I came to the comments to check whether my guess was right (it wasn't), and I'm very glad for the spoiler.
You don't need to come to the comments to check your guess, the website lets you check your guess yourself by clicking on the step.
> See if you can figure out in which step the fallacy lies. When you think you've figured it out, click on that step and the computer will tell you whether you are correct or not, and will give an additional explanation of why that step is or isn't valid.
There is a problem in step 9, but the problem is not related to introducing a third person. The existence of a third person makes step 9 correct.
The problem is that we establish the base cases S(0) and S(1), and then we establish the inductive step S(n) -> S(n+1) subject to the restriction that n > 1. But this proof never demonstrates that S(1) implies S(2). The proof is completely correct that S(2) implies S(3), S(3) implies S(4), and so on.
When step 9 says "Let R be someone else in G other than P or Q", the existence of a person in G other than P or Q has not been demonstrated. G is an arbitrary group of k+1 people, and the highest k for which our statement is known to be true is 1. Thus, the minimum size of G is 2, and R may not exist.
Please take a look at the principle of induction that is explained on that page.
In short: You fix k and assume that S(k) holds. If you can show that this always implies that S(k+1) holds, all you need is a base case (e.g. S(1) being true) to "recursively" prove that S(n) holds for all n >= 1 (or whatever your base case was).
Here the fault is that the induction step (proving S(k+1) from S(k)) requires k >= 2 but only specifically S(1) was proven.
That's the way induction works.
Assume the statement is true for n
Deduce that it then must be true for n+1
Prove it for n=1
Now it's proven for every n
No, not quite, the conclusion is that given a group of k people where all people have the same age, then a group of k+1 people also must have the same age. Since we know that in a group of 1 all k people have the same age, then if the proof held then a group of 2 people would have the same as well, then 3, etc. This is a common tactic in induction, you make an assumption, prove that if true it implies a general result, and then give an explicit case where the assumption holds, proving the general result.
You're right if we were speaking normal English, but "assuming" doesn't mean the same thing in math jargon. In math, you would say, "Assuming A, then B", to mean, "If A is true, then B is true."
It's way more confusing than simply "If A, then B", and I'm not sure why the wording hasn't fallen out of favor, but that's what it is.
This is easier to understand in constructive type theory. You can't pattern match on `k` to recur, but you can pattern match on `succ(k)` to recur on `k`.
Even when doing classic logic, I start with my constructive intuition and the sprinkle in the continuation-passing-style spooky magic when needed to get back the classical craziness.
