Here is a proof by induction, using relatively heavy machinery. It clearly holds for by assumption. Suppose you want to write as a sum of distinct primes, and that you have managed to write all numbers up to as a sum of distinct primes. By Bertrand's postulate, there is a prime . Then . Now write as a sum of distinct primes, say . Clearly none of the 's are equal to because and . Thus is a representation of as a sum of distinct primes.
Aside from your base case, you didn't use the fact that was considered a prime. Therefore if we started at we could have the same theorem without considering as a prime. But is the only way to write as a sum of distinct primes.
With that said, could you explain where else you used as a prime in this proof?
"Prove that every positive integer can be written as a sum of one or more distinct primes"
Consider the even numbers. Goldbach's conjecture says that every even number can be written as the sum of primes. Problem here is that no one has yet proven Goldbach's conjecture.
Implicitly, Goldbach's Conjecture does consider distinctiveness because, besides 2, every even number 4 or greater when divided by 2, yields only even numbers which can't be prime so the only other choice for possible prime numbers must be distinct numbers.
Goldbach"s Conjecture says every positive integer above 1 (counting 1 as a prime number) can be written as the sum of two prime numbers while Keityo puts forth the problem as "Prove that every positive integer can be written as a sum of one or more distinct primes." This lets out Goldbach's Conjecture as applying to this problem (I get onto the internet from my library which has a time restriction so I didn't get a chance to carefully review what I've written - sorry about that).