This is a very common example in math classes, though in my experience usually presented as a proof that "all horses are the same color".
No cat has eight tails, and every cat has one more tall than no cat. Ergo...
> 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.
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.