Lemma
For $n>5$, every even number $2n$ can be expressed as a sum of four primes $p_0 + p_1 + p_2 + p_3$.
Proof
Let $p_0$ be an odd prime and $m = 2n - p_0$ an odd number.
Applying the Weak Goldbach Conjecture (H. Helfgott), $m = p_1 + p_2 + p_3$.
Rearranging, $2n = p_0 + p_1 + p_2 + p_3 \blacksquare$
Question
From the lemma, $2n = p_0 + (p_1 + p_2 + p_3)$ and Goldbach's conjecture states $2n = q_0+q_1$.
Equating the two expressions $q_0+q_1 = p_0 + p_1 + p_2 + p_3$
Doesn't this impose constraints on the form of the $q_0, q_1$ primes?
Example
If $q_0=p_0$ then $q_1=p_1+p_2+p_3$ is constrained to the sum of $3$ primes.
For example, let $n=6$, by the lemma, $12=5+(3+2+2)$ and by Goldbach, $12=5+7$.
Equating yields $5+7=5+(3+2+2)$ where $q_0=p_0=5$ and $q_1=p_1+p_2+p_3=3+2+2=7$.