Proof of "if $x<y$, then $xz<yz$"

Question

I need help proving "For each $x,y,z$ in $\mathbb{N}$, if $x<y$, then $xz<yz$."

This must be done using Peano's axioms and the definitions of addition, multiplication, and ordering.

I have tried induction on z, and the base case works out but I run into an issue when it comes to the inductive part seeing as I am no longer dealing with equality.

Maybe induction is not the correct approach to this proof, and maybe a direct $P\implies Q$ proof would be more logical, however once I assume $x<y$, I don't have much to work with to deduce $xz<yz$, besides the definition of order.

Answer 1

The base case is $x<y$. Then $x+m=y$ where $m>0$.

Then assume $xk<yk$. Then $xk+n=yk$ and $n>0$.

Then $xk+x+m+n=y+yk$

or $x(k+1)+(m+n)=y(k+1)$

Thus $x(k+1)<y(k+1)$.

Thus $xz<yz$ for all $z\in\Bbb{N}$.