Arithmetic and Order of the Natural Numbers

In the setting of Class Theory NBG: the Axioms, Standing Conventions and Basic Notation, let $\mathbb{N}$ and $1$ be as in The Set of Natural Numbers and the Number One §naturals and The Set of Natural Numbers and the Number One §one, let $+$ and $\cdot$ be the addition and the multiplication on the set $\omega$ of The Class Omega of Natural Numbers with Zero §omega, of which $\mathbb{N}$ is a subset, products being formed before sums as in Multiplication on Omega §precedence, and let $\le$ and $<$ be the order and the strict order there. By Natural Numbers Are the Successors in Omega: One Is Least and Not a Successor of a Natural Number, the Successor Is Injective, and N Is Closed under Addition and Multiplication §closed, $a+b\in\mathbb{N}$ and $a\cdot b\in\mathbb{N}$ for all $a,b\in\mathbb{N}$. Let $a,b,c\in\mathbb{N}$.

¶ $a+b=b+a$ and $a\cdot b=b\cdot a$.

¶ $(a+b)+c=a+(b+c)$ and $(a\cdot b)\cdot c=a\cdot(b\cdot c)$.

¶ $a\cdot(b+c)=a\cdot b+a\cdot c$.

¶ $a\cdot1=a$ and $1\cdot a=a$.

¶ If $a+c=b+c$, then $a=b$; if $a\cdot c=b\cdot c$, then $a=b$.

¶ $a\le a$; if $a\le b$ and $b\le a$, then $a=b$; if $a\le b$ and $b\le c$, then $a\le c$; $a\le b$ if and only if $a<b$ or $a=b$; and if $a<b$ and $b<c$, then $a<c$.

¶ Exactly one of $a<b$, $a=b$ and $b<a$ holds.

¶ $a<b$ if and only if there is $d\in\mathbb{N}$ with $b=a+d$.

¶ $a<b$ if and only if $a+c<b+c$, and $a<b$ if and only if $a\cdot c<b\cdot c$; likewise $a\le b$ if and only if $a+c\le b+c$, and $a\le b$ if and only if $a\cdot c\le b\cdot c$.

¶ $1\le a$.

¶ $a<a+1$, $a+1\neq1$, and there is no $d\in\mathbb{N}$ with $a<d$ and $d<a+1$.

¶ If $a\neq1$, there is $d\in\mathbb{N}$ with $a=d+1$.

¶ Let $A$ be a class with $A\subseteq\mathbb{N}$ such that $1\in A$ and $n+1\in A$ for every $n\in A$. Then $A=\mathbb{N}$.

¶ The digits $2,\dots,9$ and the numeral $10$ of Digits and Decimal Numerals §numerals are natural numbers, and $2=1+1$, $3=2+1$, $4=3+1$, $5=4+1$, $6=5+1$, $7=6+1$, $8=7+1$, $9=8+1$ and $10=9+1$.

¶ Every subset $s$ of $\mathbb{N}$ with at least one element has an element $m\in s$ with $m\le v$ for every $v\in s$.