Finding the statement of n = (2 – y), for all positive natural numbers of ℕ.

(∀n ϵ ℕ) (∃y ϵ ℕ) (n = 2 – y).

(∀n ϵ ℕ).

n/2 + y/2 = 1

n/2 = 1 – y/2

n = 2(1 – y/2)

n = 2–2 y/2

n = 2–2y/2

n = 2 – y

Finding the k, element of ℕ, where n = 2k, for all even numbers.

(∀n ϵ ℕ) (∃k ϵ ℕ) (n = 2k) for some number (k ϵ ℕ).

For all even numbers.

n = 2k = 2 – y

Where;

k = (2 – y)/2

= 1– y/2

For all even numbers.

Finding the k, element of ℕ, where n = 2k + 1, for all prime numbers.

(∀n ϵ ℕ) (∃k ϵ ℕ) written as n = 2k + 1, for all prime numbers (k ϵ ℕ).

For all prime numbers.

n = 2k + 1 = 2 – y

Where;

2k + 1 = 2 – y

2k = 2 – y – 1

2k = 1 – y

k = 1/2 – y/2

For all prime numbers.

Adding two prime numbers together will result in a (y ϵ k) value for the resulting even number.

2k + 1 + 2k + 1 = 0

2(1/2 – y/2) + 1 + 2(1/2 – y/2) + 1 = 0

(1–2/2y) + 1 + (1–2/2y) + 1 = 0

– y + 2 + 1 – y + 1 = 0

–2y + 4 = 0

–2y = – 4

y = – 4/– 2

y = 2

Substituting this (y ϵ k) from the addition of two primes as the (y ϵ k) for the resulting even number gives an (n ϵ k).

n = 2k

= 2(1– y/2)

= 2(1–2/2)

= 2 (0)

= 0

Substituting the (n ϵ k) of the resulting even number from the addition of two primes into the (n ϵ ℕ) gives a (y ϵ ℕ).

n = 2 – y

0 = 2–0

0–2 = – y

– 2 = – y

2 = y

When (n ϵ k) equals zero, (y ϵ ℕ) equals two.

Demonstrating the proof for the first even number two where the (n ϵ ℕ) = 2 and produces a (y ϵ ℕ).

n = 2 – y

n – 2 = – y

2–2 = – y

0 = – y

Substituting this (y ϵ ℕ) into the (n ϵ ℕ) equation n = 2 – y, proves the conjecture.

n = 2 – y

= 2 – – y