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.
