You can't prove that S(n) is true for all values of n by just going through each possible n and proving it, since n can be anything so you'll be there a while.
The way this works is, you attempt to show that _if_ S(n) is true, _then_ S(n+1) is also true. Then you show that S(1) is true. Since S(n) being true implies that S(n+1) is also true, you can show that S(2) is true since 2 = 1+1 (you've proven the case where n=1). Then since S(2) is true, so is S(3), and S(4), S(5), ... and so on.
The way the links proof works is:
You show that S(1) is true. The set of 1 person {A} must have everyone in that set being the same age (there's only one person in it after all).
The next step is to show that if S(n) is true, then S(n+1) is also true. To do this we can take an example set of people {P,Q,R}, in this case we are saying that this is the group of n+1 people (so n=2). We can take {Q,R} (the set of everyone except P) and know that they are the same age (since we have assumed S(n) to be true, as long as we can show this relation works, we only need to prove one value of n to prove all of them, and we've already done this with S(1)). We can also do this for the set {P,R} using the same reasoning.
Since Q=R (Q and R are the same age) and P=R, we can prove that P=Q since P=R=Q. This shows that as long as S(n) is true, so is S(n+1).
Where this proof breaks down is this proof only works for a set of people larger than 2. So we haven't actually proven that S(1) being true implies S(2) is true, since this relation does not hold, we can't rely on S(n) being true.
You can't prove that S(n) is true for all values of n by just going through each possible n and proving it, since n can be anything so you'll be there a while.
The way this works is, you attempt to show that _if_ S(n) is true, _then_ S(n+1) is also true. Then you show that S(1) is true. Since S(n) being true implies that S(n+1) is also true, you can show that S(2) is true since 2 = 1+1 (you've proven the case where n=1). Then since S(2) is true, so is S(3), and S(4), S(5), ... and so on.
The way the links proof works is:
You show that S(1) is true. The set of 1 person {A} must have everyone in that set being the same age (there's only one person in it after all).
The next step is to show that if S(n) is true, then S(n+1) is also true. To do this we can take an example set of people {P,Q,R}, in this case we are saying that this is the group of n+1 people (so n=2). We can take {Q,R} (the set of everyone except P) and know that they are the same age (since we have assumed S(n) to be true, as long as we can show this relation works, we only need to prove one value of n to prove all of them, and we've already done this with S(1)). We can also do this for the set {P,R} using the same reasoning.
Since Q=R (Q and R are the same age) and P=R, we can prove that P=Q since P=R=Q. This shows that as long as S(n) is true, so is S(n+1).
Where this proof breaks down is this proof only works for a set of people larger than 2. So we haven't actually proven that S(1) being true implies S(2) is true, since this relation does not hold, we can't rely on S(n) being true.