Proof of Arithmetic of Multiplication on Omega: Recursion Rules, Distributivity, Associativity, Commutativity, No Zero Divisors, Cancellation and Compatibility with the Order

Throughout, sums and products of elements of $\omega$ lie in $\omega$ by Addition on Omega §addition and Multiplication on Omega §multiplication, and $0\in\omega$ and $S(n)\in\omega$ for every $n\in\omega$ by Omega Is the Least Inductive Class: It Is a Set, Induction from Zero, the Peano Properties, and Transitivity §inductive. Let $\le$ be the order on $\omega$.

Recursion equations. Let $m\in\omega$, and let $f:\omega\to\omega$ be the map given in Multiplication on Omega §multiplication by The Recursion Theorem on Omega §recursion with $a=\omega$, $c=0$ and $g=\tau_{m}$, so that $f(0)=0$ and $f(S(k))=\tau_{m}((k,f(k)))=f(k)+m$ for every $k\in\omega$. By Multiplication on Omega §multiplication, $f$ is a set witnessing the formula defining $m\cdot n$ with $z=f(n)$, so $m\cdot n=f(n)$ for every $n\in\omega$. Hence, for all $m,n\in\omega$,

$$m\cdot0=0\qquad\text{and}\qquad m\cdot S(n)=m\cdot n+m.\qquad(\ast)$$

Induction classes. Each induction below is run with Omega Is the Least Inductive Class: It Is a Set, Induction from Zero, the Peano Properties, and Transitivity §induction on a class $A=\{n\in\omega:\varphi\}$ formed by class abstraction with set parameters among $k,m$ and the class parameter $\omega$. Each $\varphi$ is an equation between terms built from $0$ (the empty set, by The Class Omega of Natural Numbers with Zero §zero), $S(\cdot)$ (by The Successor of a Set §successor), sums (by Addition on Omega §addition) and products (by Multiplication on Omega §multiplication), all defined set symbols whose arguments lie in $\omega$ by the conjunct $n\in\omega$ and the remarks above; so $\varphi$ quantifies over set variables only and is predicative as Class Theory NBG: the Axioms, Standing Conventions and Basic Notation §comprehension requires. By Class Abstraction: the Class of All Sets Satisfying a Predicative Formula §abstraction, $n\in\omega$ lies in $A$ if and only if $\varphi$ holds of $n$. For each class we check $0\in A$ and, for $n\in\omega$, that $n\in A$ implies $S(n)\in A$; then $\omega\subseteq A$, which is the claim for every element of $\omega$.

Order facts. By Omega Is the Least Inductive Class: It Is a Set, Induction from Zero, the Peano Properties, and Transitivity §set, $\omega$ is a set; by The Order on Omega Is a Well-Order with Membership as Its Strict Order, and Nothing Lies between n and Its Successor §well-order and Well-Orders on a Set §well-order, $\le$ is a total order on $\omega$, and by The Order on Omega Is a Well-Order with Membership as Its Strict Order, and Nothing Lies between n and Its Successor §strict, $<$ is its associated strict relation. So Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders applies with $a=\omega$ and these $\le$ and $<$.

Zero. $m\cdot0=0$ is $(\ast)$. For $0\cdot m=0$, let $A=\{n\in\omega:0\cdot n=0\}$. By $(\ast)$, $0\cdot0=0$; and if $0\cdot n=0$, then $0\cdot S(n)=0\cdot n+0=0+0=0$ by $(\ast)$ and Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §zero.

Successor. $m\cdot S(n)=m\cdot n+m$ is $(\ast)$. For the second equation fix $m$ and let $A=\{n\in\omega:S(m)\cdot n=m\cdot n+n\}$. By $(\ast)$ and Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §zero, $S(m)\cdot0=0=0+0=m\cdot0+0$. If $S(m)\cdot n=m\cdot n+n$, then by $(\ast)$, Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §associative, Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §successor and Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §commutative,

$$S(m)\cdot S(n)=S(m)\cdot n+S(m)=(m\cdot n+n)+S(m)=m\cdot n+S(n+m)=m\cdot n+S(m+n)=(m\cdot n+m)+S(n)=m\cdot S(n)+S(n),$$

where the third and fifth equalities use associativity together with $n+S(m)=S(n+m)$ and $m+S(n)=S(m+n)$.

One. By $(\ast)$ and Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §zero, $m\cdot S(0)=m\cdot0+m=0+m=m$. By the clauses successor and zero above and Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §zero, $S(0)\cdot m=0\cdot m+m=0+m=m$.

Commutative. Fix $m$ and let $A=\{n\in\omega:m\cdot n=n\cdot m\}$. By the clause zero above, $m\cdot0=0=0\cdot m$. If $m\cdot n=n\cdot m$, then by $(\ast)$ and the clause successor above (its second equation with $n$ and $m$ interchanged), $m\cdot S(n)=m\cdot n+m=n\cdot m+m=S(n)\cdot m$.

Distributive. Fix $k,m$ and let $A=\{n\in\omega:k\cdot(m+n)=k\cdot m+k\cdot n\}$. By Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §zero and $(\ast)$, $k\cdot(m+0)=k\cdot m=k\cdot m+0=k\cdot m+k\cdot0$. If $k\cdot(m+n)=k\cdot m+k\cdot n$, then by Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §successor, $(\ast)$ and Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §associative,

$$k\cdot(m+S(n))=k\cdot S(m+n)=k\cdot(m+n)+k=(k\cdot m+k\cdot n)+k=k\cdot m+(k\cdot n+k)=k\cdot m+k\cdot S(n).$$

So $k\cdot(m+n)=k\cdot m+k\cdot n$ for all $k,m,n\in\omega$. With the clause commutative above, $(m+n)\cdot k=k\cdot(m+n)=k\cdot m+k\cdot n=m\cdot k+n\cdot k$.

Associative. Fix $k,m$ and let $A=\{n\in\omega:(k\cdot m)\cdot n=k\cdot(m\cdot n)\}$. By $(\ast)$, $(k\cdot m)\cdot0=0=k\cdot0=k\cdot(m\cdot0)$. If $(k\cdot m)\cdot n=k\cdot(m\cdot n)$, then by $(\ast)$ and the clause distributive above,

$$(k\cdot m)\cdot S(n)=(k\cdot m)\cdot n+k\cdot m=k\cdot(m\cdot n)+k\cdot m=k\cdot(m\cdot n+m)=k\cdot(m\cdot S(n)).$$

No-zero-divisors. Let $m\cdot n=0$ and $n\neq0$. By Omega Is the Least Inductive Class: It Is a Set, Induction from Zero, the Peano Properties, and Transitivity §cases, $n=S(n')$ for some $n'\in\omega$, so $m\cdot n'+m=m\cdot n=0$ by $(\ast)$, and $m=0$ by Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §zero-sum.

Order. Let $k\neq0$. Suppose $m<n$. By The Order on Omega Is a Well-Order with Membership as Its Strict Order, and Nothing Lies between n and Its Successor §successor-below, $S(m)\le n$, so by Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §difference there is $j\in\omega$ with $n=S(m)+j$, and $S(m)+j=S(m+j)=m+S(j)$ by Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §successor. By the clause distributive above, $k\cdot n=k\cdot m+k\cdot S(j)$. Since $S(j)\neq0$ by Omega Is the Least Inductive Class: It Is a Set, Induction from Zero, the Peano Properties, and Transitivity §successor-nonzero and $k\neq0$, the clause no-zero-divisors above gives $k\cdot S(j)\neq0$; with $0\le k\cdot S(j)$ by The Order on Omega Is a Well-Order with Membership as Its Strict Order, and Nothing Lies between n and Its Successor §zero-least, The Order on Omega Is a Well-Order with Membership as Its Strict Order, and Nothing Lies between n and Its Successor §strict gives $0<k\cdot S(j)$. By Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §order and Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §zero, $k\cdot m=k\cdot m+0<k\cdot m+k\cdot S(j)=k\cdot n$.

Conversely suppose $k\cdot m<k\cdot n$. By Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §trichotomy, $m<n$, $m=n$ or $n<m$. If $m=n$, then $k\cdot m<k\cdot m$, contradicting Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-irreflexive. If $n<m$, then $k\cdot n<k\cdot m$ by the forward direction, and with $k\cdot m<k\cdot n$, Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-transitive gives $k\cdot m<k\cdot m$, again a contradiction. Hence $m<n$.

Cancellation. Let $k\neq0$ and $k\cdot m=k\cdot n$. If $m<n$, the clause order above gives $k\cdot m<k\cdot n=k\cdot m$, and if $n<m$ it gives $k\cdot m=k\cdot n<k\cdot m$; both contradict Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-irreflexive. So $m=n$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §trichotomy.