Is this proof that the sum of two natural number is a natural number? Introduction
I saw a video of a guy talking about proving that a even number squared is still even. It was something like this:

Let n be a natural/whole number, and 2n defines all even numbers:
2n * 2n = 4n² = 2(2n²)
So, 2(2n²) is an even number.

When I saw it, I thought: assuming that n² and 2n are natural numbers, right? People told me that it's unnecessary to prove that those are natural numbers. It goes out of the scope of the question, and it's very easy to see that n² and 2n are natural numbers.
It left me thinking. How to prove it? I recently asked a question about rigour of a proof, and someone showed me the proof of infinite many primes. It has "...". I always thought that "..." was prohibited in proving,  like a lazy way to write. But it apparently ain't. Then I thought "it makes things easier", so I decided to give it a try. Here's my proof that the result of a sum of two natural numbers is a natural number:
My try
Definitions
It'll be shown that a number x is a natural number using the notation N(x). [I've already seen people using it, so I ain't inventing notation].
Natural Number:

*

*N(0)

*~[N(-1)]

*[N(x)] → [N(x+1)]

"Which I think it means: -1 ain't a natural number, 0 is a natural number, and if a number is a natural number, this number plus one is also a natural number".
"A → B → C" means "A implies B, which implies C". This part is because that's not right/common notation, I think.
Proof
Since N(0) and [N(x)] → [N(x+1)]:
N(0) → N(0+1) → N(0+2) ... → N(0+a)
This chain reaction only works because the first number is a natural number, so we can generalize it:
N(b) → N(b+1) → N(b+2) ... → N(b+a)
We know that N(a) because N(0+a), which can be simplified to N(a).
So it can be said that [N(a) ^ N(b)] → N(a+b). Which hopefully means: if a and b are natural numbers, their sum is also a natural number.
Ending
How accurate or how right is my proof?
And thanks for reading this Bible-long text.
P.S.: I wrote this down at 0500 and in my cellphone, so I am 99% sure I made grammar and spelling mistakes. Also, if this text is too long or not focusing on the matter, please, tell me or, if you want, edit it.
 A: Your definition of natural numbers is problematic. For example this: $[N(x)]\to[N(x+1)]$. What is $x+1$ anyway? What is $1$? And in this: $\sim [N(-1)]$, what is $-1$? You have circular references here, you define naturals using naturals (or even integers) and operations on naturals.
So that won't do. In mathematics we typically treat naturals axiomatically via Peano axioms. With that we have an axiomatic $0$ and the successor function $S(n)$. One of the axioms is that $0\in\mathbb{N}$, the other is that for any $n\in\mathbb{N}$ we have $S(n)\in\mathbb{N}$. We also have that $S:\mathbb{N}\to\mathbb{N}$ is injective and that there is no $n\in\mathbb{N}$ with $S(n)=0$.
In that setup we first define some constants: $1:=S(0)$, $2:=S(1)$, $3:=S(2)$, etc. Then we define addition recursively via
$$a+0:=a$$
$$a+S(b):=S(a+b)$$
analogously we define multiplication:
$$a\cdot 0:=0$$
$$a\cdot S(b):=a+(a\cdot b)$$
and then exponentiation:
$$a^1:=a$$
$$a^{S(b)}:=a\cdot (a^b)\text{ for }b\neq 0$$
Note that addition, multiplication and exponentiation produce new naturals by definition.
So all of that guarantees that:

*

*$2\in\mathbb{N}$

*$2\cdot n\in\mathbb{N}$

*$n^2\in\mathbb{N}$
