Let $a,b$ be natural numbers. Then $a < b$ if and only if $a\texttt{+}\texttt{+}\leq b$.
MY ATTEMPT
If $a < b$, we may suppose by contradiction that $a\texttt{+}\texttt{+} > b$. Thus we conclude that $a < b < a\texttt{+}\texttt{+}$, which is impossible, because there is no natural number between $a$ and $a\texttt{+}\texttt{+}$.
Conversely, if $a\texttt{+}\texttt{+}\leq b$, one has that $a < a\texttt{+}\texttt{+}\leq b$, and $a < b$. That is because $a\texttt{+}\texttt{+} = a + 1 > a$ by definition of order, where it has been used the fact that $a\texttt{+}\texttt{+} = (a+0)\texttt{+}\texttt{+} = a + 0\texttt{+}\texttt{+} = a + 1$.
My question is: how do we prove there is no natural number between $a$ and $a\texttt{+}\texttt{+}$?
Let me use $a^+$ for the successor, or the argument would be very cumbersome to write.
Saying that $a<b$ is the same as saying that $b=a+c$ for some $c>0$. Since $c>0$, we have $c=d^{+}$ and so $$ b=a+d^{+}=a^++d $$ Therefore $a^+\le b$.
Conversely, if $a^+\le b$, then $b=a^++c$ for some $c$. Hence $b=a+c^+$ and so $a<b$, because $c^+>0$.
This relies only on $x+y^+=x^++y$, that you can prove by induction.