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Let $P \subset N$ be the set of primes. Is it true that the following statements are equivalent?

(1) There exists a natural number $N$ such that for all even natural numbers $n \geq N$, there exists two primes $p,q \in P$ such that $n=p+q$.

(2) There exists a finite set $S \subset N$ such that for all even $n \geq 4$ there exists $p,q \in P \cup S$ such that $n=p+q$.

(EDIT: One direction is easy. That is (1) implies (2). I have a feeling (2) implies (1) is possible to show. I'll admit I am somewhat confused how to show non-equivalence. I think that requires knowing the truth of the Goldbach conjecture as (2) implies (1) is only false if (2) is true and (1) is false. I'll accept a good argument why any attack using (2) to show (1) would likely fail in place of showing inequivalence if they are inequivalent.)

My motivation comes from thinking about sums of sets in general, that is $A+A$, and wondering just how important the distribution of the primes in the limit is or how much O(1) primes matter in making the Goldbach conjecture hard. My guess is that the limiting distribution is the most important and hence proving a statement like (2) is just as hard as proving the Goldbach conjecture.

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No. Suppose the Goldbach conjecture is false for only a finite set of numbers. Then S can be chosen to make the sum work for that set.

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  • $\begingroup$ So it is theoretically possible that GC is false yet (2) above is true, hence the statement "(2) $\implies$ GC" is false... $\endgroup$ – barak manos Oct 19 '16 at 15:38
  • $\begingroup$ @barakmanos, I think you mean the statement "(2) implies GC" could be false. Since GC is most likely true, in which case (2) is also true, "(2) implies GC" is most likely logically true. It's just there may not be a simple proof of the implication that doesn't depend on GC being true. $\endgroup$ – Barry Cipra Oct 19 '16 at 15:42
  • $\begingroup$ @BarryCipra: Well, no, I actually meant that even if GC is true, those two statements are not equivalent. In other words, (2) is weaker than GC... That is, if I understand this answer correctly... To be honest, I'm not sure how logic defines a statement when the value of some expression within that statement is unknown ("neither true nor false" - is that even possible?). $\endgroup$ – barak manos Oct 19 '16 at 15:44
  • $\begingroup$ @barakmanos, the only way a logical statement of the form $P\implies Q$ can be false is if $P$ is true and $Q$ is false. So to say that "(2) implies GC" is false is to say that (2) is true and GC is false. That's a very strong statement! $\endgroup$ – Barry Cipra Oct 19 '16 at 15:53
  • $\begingroup$ @BarryCipra: What about "true $\implies$ unknown"? $\endgroup$ – barak manos Oct 19 '16 at 15:56

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