No proof exists.

Suppose for the sake of contradiction that the Collatz conjecture is true, which would imply every positive natural number eventually reaches \(1\). Since \(n {\gt} 4 \ge 1\), there exists some \(j \ge 0\) such that \(C^j(n) = 1\). Because \(n\) is part of a cycle of length \(k \ge 1\), we have \(C^{j \cdot k}(n) = n\). Since \(j \cdot k \ge j\), we can write \(j \cdot k = (j \cdot k - j) + j\), allowing us to decompose the iteration: \(n = C^{j \cdot k}(n) = C^{j \cdot k - j}(C^j(n)) = C^{j \cdot k - j}(1)\). However, it is a known property of the Collatz map that iterating starting from \(1\) only yields values in the set \(\{ 1, 2, 4\} \). Thus, \(C^{j \cdot k - j}(1) \le 4\), which implies \(n \le 4\). This contradicts our initial assumption that \(n {\gt} 4\), so the conjecture must be false.

Suppose for the sake of contradiction that the Collatz conjecture is true, meaning every positive integer reaches \(1\). For our starting value \(n \ge 1\), there must exist some index \(j \ge 0\) such that \(C^j(n) = 1\). Let \(M\) be the maximum value reached in the first \(j\) steps of the orbit, that is, \(M = \max _{0 \le i \le j} C^i(n)\). Because the orbit of \(n\) is unbounded, there exists some index \(k\) such that \(C^k(n) {\gt} M + 4\). If we had \(k \le j\), then \(C^k(n) \le M {\lt} M + 4\), which is a contradiction. Therefore, we must have \(k {\gt} j\). We can then decompose the iteration as \(C^k(n) = C^{k-j}(C^j(n)) = C^{k-j}(1)\). Since iterating the Collatz map from \(1\) only produces values in \(\{ 1, 2, 4\} \), we have \(C^{k-j}(1) \le 4\). This implies \(C^k(n) \le 4\), which contradicts \(C^k(n) {\gt} M + 4 \ge 4\). Thus, the assumption that the conjecture is true must be false.

To prove that the Collatz conjecture is true under these assumptions, we must show that for an arbitrary target number \(m\) (distinct from the hypothesis parameter \(n\)), either \(m = 0\) or there exists some index \(j\) such that \(C^j(m) = 1\). If \(m = 0\), this condition trivially holds. Assume \(m \ge 1\). By the boundedness assumption, there exists some bound \(B\) such that \(C^k(m) \le B\) for all \(k \ge 0\). Because the infinite sequence \((C^k(m))_{k=0}^\infty \) only takes values in the finite set \(\{ 1, 2, \ldots , B\} \), the Pigeonhole Principle implies that there must be a collision. That is, there exist indices \(i {\lt} j\) such that \(C^i(m) = C^j(m)\). Let \(c = C^i(m)\). Then \(C^{j-i}(c) = C^j(m) = C^i(m) = c\), meaning \(c\) is part of a cycle of length \(j-i \ge 1\). By our no-cycle assumption, no number strictly greater than \(4\) can be part of a cycle, so we must have \(c \le 4\). Since \(m \ge 1\), all of its iterates are strictly positive, giving \(c \in \{ 1, 2, 3, 4\} \). We can verify that every number in this set eventually reaches \(1\): \(C^0(1) = 1\), \(C^1(2) = 1\), \(C^7(3) = 1\) (since \(3 \mapsto 10 \mapsto 5 \mapsto 16 \mapsto 8 \mapsto 4 \mapsto 2 \mapsto 1\)), and \(C^2(4) = 1\). In all cases, there exists some \(j'\) such that \(C^{j'}(c) = 1\). Therefore, starting from \(m\), after \(i\) steps we reach \(c\), and after \(j'\) more steps we reach \(1\), yielding \(C^{i+j'}(m) = 1\). Thus, the Collatz conjecture is true.