The continuity of prime numbers can lead to even continuity(Goldbach conjecture)

Two prime numbers{ 3,5 }.{ 3,5 }→{ 3 + 3 = 6，3 + 5 = 8，5 + 5 = 10 } →{ 6,8,10 }. { { 10 }→( 5 + 5 =10 = 3 + 7 ) → 7} Increased by 7 →{ 3, 5, 7 }.{ 3,5,7 }→{ 3 + 3 = 6, 3 + 5 = 8, 5 + 5 = 10, 5 + 7 = 12, 7 + 7 = 14 }→{ 6,8,10,12,14 }. { { 14 }→( 7 + 7 =14 = 3 + 11 ) → 11} Increased by 11 →{ 3, 5, 7, 11 }.Note changes: { 3,5,7 }→{ 6,8,10,12,14 }→ { 3, 5, 7,11 }.The same logic would be:{ 3,5,7,11 }→{ 6,8,10,12,14,16 }→ { 3, 5, 7,11,13 }.The same logic would be:{ 3,5,7,11,13 }→{ 6,8,10,12,14,16,18,20 }→ { 3, 5, 7,11,13,17 }.If you expand infinitely in the above specified mode: { 3, 5, 7, 11, 13, 17 ,... Get: { 6,8,10,12,14,16,18,20,22,... The above is: the continuity of prime numbers can lead to even continuity.Get: Goldbach conjecture holds.If it is mandatory: Authenticity stops at an even number 2n. { { 3, 5, 7, 11 ,...,p 1 } →{ 6,8,10,12,...,2n}. { 3, 5, 7, 11, ...,p 1 } ↛ ( 2n+2). ∀p+∀p≠ 2n+2 } It can be proved that: It violates the "Bertrand Chebyshev" theorem.∴ { 3, 5, 7, 11, 13, ... →{ 6,8,10,12,14,... Get: Goldbach conjecture holds.