For all natural numbers $x$ and $y$, there is a natural number $z$ such that $x=y+z$ or $y=x+z$

Question

I am quite confussed because i can't proof the fallowing: $$\forall x,y \in \mathbb{N},\; \exists z \in N: x=y+z \;\text{ or }\; y=x+z.$$ I am allowed to use induction and definition of addition. Thanks for your support.

Answer 1

By induction on $x$:

For $x=0$ we have to show $$\forall y\in\mathbb N,\exists z\in\mathbb N\colon 0=y+z\lor y=0+z$$ This is clearly fulfilled by letting $z=y$.

Now assume we have for seom $x\in\mathbb N$ that $$\forall y\in\mathbb N,\exists z\in\mathbb N\colon x=y+z\lor y=x+z$$ We want to show $$\forall y\in\mathbb N,\exists z\in\mathbb N\colon Sx=y+z\lor y=Sx+z$$ So let $y\in\mathbb N$ be arbitrary. Then there is $z\in\mathbb N$ such that $x=y+z\lor y=x+z$. In the first case, we find $Sx=S(y+z)=y+Sz$. In the second case, either $z=0$ and then $y=x$, hence $Sx=y+1$, or $z=Sw$ for some $w\in\mathbb N$ and hence $y=x+z=x+Sw=S(x+w)=Sx+w$.