|
|
|
|
|
|
by thaumasiotes
1657 days ago
|
|
|
You've gotten confused somewhere. > However if A ⋂ B is not empty, it follows from knowing f(A) is constant. That f(A) is constant is supposedly a conclusion, not a premise. It is not a valid conclusion to draw from the stated premises. > f(A ⋂ B) = f(A) is saying that every result of f(A ⋂ B) is equal to every result of f(A) As I just responded above, interpreting the claim this way doesn't get it to make any more sense. When f(A ⋂ B) = f(A), there is no basis from which to conclude that f is constant. You have no information about whether f is or isn't constant. This is not the claim made in the false proof, nor does it have anything to do with the claim made in the false proof. It is a creation of poetically's own mind, and it makes no sense. |
|
|
You seem to have forgotten what we are talking about. A ⋃ B is an arbitrary set of horses of size n+1, A and B are subsets of A ⋃ B, each with a single different element removed and thus of size n. We already know that f() is constant over A and over B because they are of size n and we have assumed that all sets of horses of size n have the same color (which I already pointed out in my last comment.) We also know that f() is constant over A ⋂ B becasue it is a subset of sets we already know f() is constant over. The only thing that needs to be proven is that f() is constant over the set A ⋃ B. That proof is impossible if A ⋂ B is empty. We can only assert A ⋂ B is not empty if we assume that A ⋃ B must have a size of at least 3. From assuming n+1>=3 we are able to correctly conclude that n>1 implies the inductive step is valid.