Comment by kr99x

I've been down that road, and it's unfortunately a dead end. You can generate an infinite number of reducing forms, each of which itself covers an infinite number of integers, like 4k + 5 → 3k + 4. Each one covers a fraction of the integers 1/(2^x) where x is the number of division steps in its reducing sequence (and the right hand side is always 3^y where y is the number of multiplying steps). You can't just make 1/2 + 1/4 + 1/8 and so on though (the easy path to full coverage) because sometimes the power of 3 overwhelms the power of 2. There is no 8k → 9k form, because that's not a reduction for all k, so you instead have to go with 16k → 9k. This leaves a "gap" in the coverage, 1/2 + 1/4 + 1/16th. Fortunately, when this happens, you start to be able to make multiple classes for the same x and y pair and "catch up" some, though slower. As an amateur I wrote a whole bunch about this only to eventually discover it doesn't matter - even if you reach 1/1th of the integers by generating these classes out to infinity, it doesn't work. An infinite set of density 1 implies a complementary set of density 0, but a set of density 0 doesn't have to be empty! There can still be finitely many non-reducing numbers which are not in any class, allowing for alternate cycles - you would only eliminate infinite growth as a disproof option.

Mind you, it's almost certain Collatz is true (generating these classes out to 3^20 nets you just over 99% coverage, and by 3^255 you get 99.9999999%) but this approach doesn't work to PROVE it.