If $x,y \in \mathbb{N}$ then $x+y=0 \iff x=y=0$

Question

Let $x,y \in \mathbb{N}$. The operation $(+)$ is defined by:

$$x+0=x$$ $$ x+(y+1)=(x+y)+1$$

Then prove that $x+y=0 \iff x=y=0$.

The second implication $x=y=0 \implies x+y=0 $ is simple and I have already done it. I imagine the other implication is also simple but I'm missing something obvious.

I tried approaching by assuming the opposite and splitting it into 3 cases

$1)\ x=0, y \neq 0$

$2)\ y=0, \ x \neq 0$

$3)\ x \neq 0, \ y \neq 0$

For the first two I used the definition $x+0=x$ and commutativity (which I proved earlier).

I really don't know how to approach the third case. Any other method is also appreciated.

Answer 1

Suppose, wlog, that $y \neq 0$. Then $y = n + 1$ for some $n \in \mathbb{N}$. Then, for any $x \in \mathbb{N}$, we have: $$x + y = x + (n + 1) = (x + n) + 1$$ Because this is the successor of some natural natural, this sum can't be $0$. This proves $x + y = 0 \Rightarrow x = y = 0$. The other implication shouldn't be a problem.