For each $x,y,z∈\mathbb N$, if $xFor each $x,y,z∈\mathbb{N}$, if $x<y$ then $x+z<y+z$
I know how to solve this if it was $x=y$ then $x+z=y+z$ (which my professor said I need to do)
Is there a definition that says I can change the $<$ into a $=$ and then change it back to $<$ at the end?
I have a paper full of Peano's axioms that I'm able to build off of but I don't know what one I should be using.
Would For each $x,y∈\mathbb{N}$, $x<y$ or $x=y$ or $x>y$ 
be able to be used to change the $<$ to $=$?
 A: Typically, Peano axioms include, that $N$ contains $0$.
You said in your comment that you define $x<y$ as $(\exists k\in N) y=x+k$. If we include zero, this would mean $x<x$. So I assume that you meant
$$
\begin{align*}
x\le y &\Leftrightarrow (\exists k\in N) x+k=y\\
x<y &\Leftrightarrow (\exists k\in N) x+S(k)=y
\end{align*}
$$
If you have $x\le y$, that means that $x+k=y$ for some $k$. From this you get:
$$(x+z)+k = x+(z+k) = x+(k+z) = (x+k)+z = y+z$$
which means $x+z\le y+z$. (Note that we have used associativity and commutativity; so this works if you have shown these properties before.)
The proof that $x<y$ implies $x+z<y+z$ is similar. (Just replace $k$ by $S(k)$.)
This is basically just more detailed version of the approach suggested in Hagen von Eitzen's comment. 
A: If it is peano it is fairly simple, let $\sigma$ denote our "next" function such that by definition we have $a+1=\sigma(a)$. By definition of $x<y$ we have that $y=\sigma^n(x)$ for some $n$ where it denotes iterative application of $\sigma$. Then we have that $y+z=\sigma^{m}(y)=\sigma^{m+n}(x)=\sigma^n(x+z)$ which implies that $y+z>x+z$, by definition of how the next-function works and addition.
