Collatz Sequence Proof 3rd Way

Abstract of I discovered a fact in Collatz Sequence: S (n) = {a,b,c,…,w} ⇔ LS(n) = LS(a) = LS(b) = LS(c) = …. = LS (w) … (Fact1) n, a, b, c, w, r ∈ N_+, LS(n) = {4,2,1} when n ∈ {1,2,3,4,5, …} Let LS(r) = {4,2,1} when n ∈ Z = {1, 2, 3, 4, 5, …, r}, when r is even or odd. Part a) If (r+1) ∈ (N_even), and I proved LS(r+1) = {4, 2, 1} ⇒ LS (n) = {4,2,1} for all even #s Part b) If (r+1) ∈ (N_odd), and 2(r + 1) ∈ (N_even), Then I proved LS(r+1) =LS(2(r + 1) ) = {4, 2, 1} by part a. Therefore LS(N) = {4, 2, 1} for all n ∈ N_+