Natural numbers proof via Peano's axioms (not trichotomy)

Question

Prove that for each $x,y$ an element of the natural numbers ($\mathbb{N}$), $x<y$ or $x=y$ or $x>y$. So at least one is true. I have the definition of order to work with and the basic algebra of the natural numbers to work with (i.e. commutativity of add./mult., associativity of add./mult., etc.). I figured it best to begin by doing this by cases. So, for case 1: assume $x\nless y$ and $x\neq y$, show $x>y$. I'm not sure, however, what $x\nless y$ and $x\neq y$ allows me to conclude. Thanx for the help and my apologies for the formatting.

Answer 1

Proof by induction on $x$.

Base case: $x=1$. Then either $y=1$ (and $x=y$) or $y\neq 1$, and then $x<y$ because each natural number is a successor of $1=x$.

Inductive case: $x\neq 1$. This breaks into two parts:
1. $y=1$. Then $x>y$, since each natural number except $1$ is a successor of $1=y$.
2. $y\neq 1$. Then both $x,y$ are successors, so we apply the inductive hypothesis to $x-1, y-1$. Either $x-1<y-1$, $x-1=y-1$, or $x-1>y-1$. In the first case, $x-1+k=y-1$, so $x+k=y$ and hence $x<y$. The last two cases are similar.