Proof. Let \(x\) be a message, let \([a, a + w)\) be its final slice, and let \(m = a + w/2\). Write \(k = {\ell\mathopen{}\left(x\right)\mathclose{}}\), and let \(t\) be the number whose bits after the point are \(c(x)\). Truncating \(m\) to \(k\) digits gives
\[t \le m < t + 2^{-k}.\]
Because \(k \ge -\log_2 w + 1\), we have \(2^{-k} \le w/2\). So \(t > m - 2^{-k} \ge m - w/2 = a\), and \(t + 2^{-k} \le m + w/2 = a + w\). This shows that the slice \([t, t + 2^{-k})\), which holds the numbers whose bits begin with \(c(x)\), lies inside the final slice of \(x\).
Suppose that \(c(x)\) were the beginning of \(c(x')\) for two different messages \(x\) and \(x'\). Every number whose digits begin with \(c(x')\) also begins with \(c(x)\), so the slice of \(c(x')\) lies inside the slice of \(c(x)\). The first lies inside the final slice of \(x'\), and the second inside the final slice of \(x\), so these two final slices would share a point. Lemma 1 rules that out, so no codeword is the beginning of another.