Bertrand–Chebyshev Step Forward

 Consider, the last prime, $p$ for which

$p+q=2n$

for $n\gt 2$ where $q \le p$ is also prime.  

In addition, $S_n: p+q=2n,\, p\ne q,\,n\gt 2$ is all true up to $n$, that is to say $S_n$ is true for all even numbers below and including $2n$.

This starting condition is easily satisfied by enumerating the first $S_n$.  Then for the interval $2n$ to $4n$ in which a prime exist by Bertrand–Chebyshev Theorem, this prime is the next prime, $p^{next}$ in succession.  Since $S_{n}$ exist for all $n$.  By replacing one of the prime pair with $p^{last}$ or $p^{next}$ the interval $(0,2n]$ will translate to $(2n,4n]$ and so, $S_{n}$ exist for all $n$ up to $4n$.

Then we move to the next interval $n$ up to $8n$ still by using the fact that Bertrand–Chebyshev Theorem guarantee a prime number in the interval $4n$ to $8n$.

ad infinitum.

Roughly Goldbach's Conjecture is again proved.