Sum of odd prime and odd semiprime as sum of two odd primes?

How to prove that each sum of odd prime and odd semiprime can be written as sum of two odd primes $(p_1+p_2p_3=p_4+p_5)$ ? Since we know that each prime number greater than $3$ is of the form $6k\pm 1$ we can take one of the combinations and write:

$p_1=6k_1+1 ; p_2=6k_2+1 ; p_3=6k_3+1 \Rightarrow p_1+p_2p_3=(6k_1+1)+(6k_2+1)(6k_3+1)=$

$=6k_1+1+6k_2+6k_3+36k_2k_3+1=6(k_1+k_2+k_3)+1+6(6k_2k_3)+1=$

$=(6k_4+1)+(6k_5+1)$ , Similarly we can show that for each combination of the $p_1 , p_2 ,p_3$ forms we get $(6k_4\pm 1)+(6k_5\pm 1)$ form. So we have necessary but not sufficient condition for $p_4$ and $p_5$ to be primes. Any idea how to find a sufficient condition for this proof ?

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In short, what you are stating should be true but it's not proven yet. By Chen's theorem, every (sufficiently large) even number is either a sum of two primes or a prime and semiprime. If we could show your question, then every (sufficiently large) even number is a sum of 2 primes. That is, we would have more-or-less proved the Goldbach conjecture... –  Srivatsan Sep 24 '11 at 8:32
@SrivatsanNarayanan,In my opinion this statement is weaker than Goldbach conjecture –  pedja Sep 24 '11 at 8:35
Strictly speaking, it is weaker because you will prove it only for sufficiently large even numbers. –  Srivatsan Sep 24 '11 at 8:41
+1 great question –  draks ... Mar 22 '12 at 22:27