
Prove ...is a wellorder
Given P={1,2,3,...} with usual < ordering
Let f(n)= the number of distinct prime divisors of n
(ie: f(1)=0, f(2)=1, f(3)=1, f(4)=1, f(5)=1, f(6)=2,...)
Now, define $\displaystyle \triangleleft$ on P where:
a $\displaystyle \triangleleft$ b iff f(a) < f(b) OR [f(a) = f(b) and a < b]
So, I need to prove that P with $\displaystyle \triangleleft$ is a wellordering; prove transitive, irreflexive, comparable, and that every nonempty subset has a least element. The last part here is my problem, everything else is fine but I need help proving the least element part.
I know that if you break up P into sets so that all numbers with 1 unique prime divisor goes into a subset, then all the numbers with 2 unique prime divisors go into a disjoint subset, and so on, where each subset is ordered as usual by <. Then, take the union of these subsets and it can be shown that it still holds the usual ordering of <, but we are now dealing with $\displaystyle \triangleleft$ ordering so it is a wellordering (since < is a well ordering). As a restatement of my issue, I need help writing this out and actually making it solid. I've been working on it for a while...
Cheers and thank you!

You can prove $\displaystyle \triangleleft$ is a well ordering using that $\displaystyle <$ is a well ordering.
Let $\displaystyle X$ be a non empty subset of $\displaystyle P.$ Then $\displaystyle f(X)$ is a non empty subset of $\displaystyle P,$ which has a least element $\displaystyle b$ for $\displaystyle <.$
Consider $\displaystyle X\cap f^{1}(b)\subseteq P,$ which is also a non empty subset (why?) of $\displaystyle P.$ Once again, it has a least element for $\displaystyle <,$ let say $\displaystyle a.$
Prove that $\displaystyle a$ is the least element for $\displaystyle \triangleleft$ in $\displaystyle X.$