Number as Sum of Distinct Primes greater than 11

Theorem
Every number greater than $45$ can be expressed as the sum of distinct primes greater than $11$.

Proof
Let $S = \set {s_n}_{n \mathop \in N}$ be the set of primes greater than $11$ ordered by size.

Then $S = \set {11, 13, 17, 19, 23, 29, 31, 37, 41, 43, \dots}$.

By Bertrand-Chebyshev Theorem:
 * $s_{n + 1} \le 2 s_n$ for all $n \in \N$.

We observe that every integer $n$ where $45 < n \le 45 + s_{11} = 92$ can be expressed as a sum of distinct elements in $\set {s_1, \dots, s_{10}} = \set {11, 13, 17, 19, 23, 29, 31, 37, 41, 43}$.

Hence the result by Richert's Theorem.

Here is a demonstration of our claim: