Proof of Multiplication of Dedekind Cuts: Associativity, Commutativity, Unit, Distributivity, Reciprocals and Nonnegative Products

Throughout, $<$ is the strict relation of the order $\le$ of $\mathbb{Q}$. By The Real Numbers §operations and The Real Numbers §constants, $x+y=x\oplus y$, $0_{\mathbb{R}}=0^{*}$ and $1_{\mathbb{R}}=1^{*}$; $x\le_{C}y$ means $x\subseteq y$ (Construction of the Dedekind Cuts: They Form a Set Totally Ordered by Inclusion, Closed under Sums, Negatives and Products of Nonnegative Cuts); and $\oplus$, $\ominus$ and $\odot$ are as in Construction of the Dedekind Cuts: They Form a Set Totally Ordered by Inclusion, Closed under Sums, Negatives and Products of Nonnegative Cuts. Two classes with the same elements are equal by Axiom of Extensionality for Classes; every equality of cuts below that is not a computation is proved by comparing elements. Let $P=\{x\in C:0^{*}\subseteq x\}$, the class of nonnegative cuts.

Step 0 (facts about $\mathbb{Q}$). By The Rational Numbers Form an Archimedean Ordered Field Containing the Integers §ordered-field, $\mathbb{Q}$ is an ordered field and $\le$ is a total order on the set $\mathbb{Q}$; the field laws (associativity, commutativity, distributivity, $1\cdot c=c$, and $c\cdot c^{-1}=1$ for $c\neq0$ by Negatives, Differences, Reciprocals and Quotients §reciprocal) are used without further mention. Let $a,b,c,d\in\mathbb{Q}$.

(Q1) If $a\le b$ and $b<c$, or $a<b$ and $b\le c$, then $a<c$, by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §weak-strict and Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-transitive. Moreover $a\le b$ fails if and only if $b<a$ (Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §total-negation), and $a<a$ fails (Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-irreflexive).

(Q2) If $a\le b$ and $0\le c$, then $a\cdot c\le b\cdot c$; if $a<b$ and $0<c$, then $a\cdot c<b\cdot c$. Indeed, if $c=0$ both sides of the first claim are $0$ by Rules of Arithmetic and Order in an Ordered Field §zero; otherwise $0<c$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §weak-strict, and both claims are Rules of Arithmetic and Order in an Ordered Field §order-product. Taking $a=0$: if $0\le b$ and $0\le c$, then $0\le b\cdot c$.

(Q3) We use Rules of Arithmetic and Order in an Ordered Field §order-sum ($a<b$ if and only if $a+c<b+c$; $a\le b$ and $c\le d$ give $a+c\le b+d$), the sign rules Rules of Arithmetic and Order in an Ordered Field §signs, $0<1$ from Rules of Arithmetic and Order in an Ordered Field §squares, $0<c^{-1}$ whenever $0<c$ from Rules of Arithmetic and Order in an Ordered Field §positive-reciprocal, and midpoints from Rules of Arithmetic and Order in an Ordered Field §midpoint. In particular, if $0<d$ then $-d<0$ (add $-d$ to both sides), and if $a<1$ then $0<1-a$ (add $-a$).

Step 1 (facts about cuts). Let $x\in C$. By the definition of $C$ in Construction of the Dedekind Cuts: They Form a Set Totally Ordered by Inclusion, Closed under Sums, Negatives and Products of Nonnegative Cuts, $x\subseteq\mathbb{Q}$, $x\neq\emptyset$, $x\neq\mathbb{Q}$, every $v\in\mathbb{Q}$ with $v<u$ for some $u\in x$ lies in $x$ (downward closure), and every $u\in x$ has some $v\in x$ with $u<v$ (openness).

(C1) If $c\in x$, $a\in\mathbb{Q}$ and $a\le c$, then $a\in x$: by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §weak-strict, $a<c$ or $a=c$.

(C2) If $c\in x$ and $a\in\mathbb{Q}\setminus x$, then $c<a$: otherwise $a\le c$ by (Q1), and $a\in x$ by (C1).

(C3) If $0\in x$, then $x\in P$ and $x\neq0^{*}$: every $v<0$ lies in $x$ by downward closure, and $0\notin0^{*}$ by (Q1). Conversely, if $x\in P$ and $x\neq0^{*}$, then $0\in x$, there is $a\in x$ with $0<a$, and $0<q$ for every $q\in\mathbb{Q}\setminus x$. Indeed, since $0^{*}\subseteq x$ and $x\neq0^{*}$, some $c\in x$ is not in $0^{*}$, so $0\le c$ by (Q1) and $0\in x$ by (C1); $a$ exists by openness; and $0<q$ by (C2).

(C4) For every $d\in\mathbb{Q}$ with $0<d$ there are $c\in x$ and $q\in\mathbb{Q}\setminus x$ with $q<c+d$. Indeed, $\ominus x\in C$ by Construction of the Dedekind Cuts: They Form a Set Totally Ordered by Inclusion, Closed under Sums, Negatives and Products of Nonnegative Cuts §negative and $x\oplus\ominus x=0^{*}$ by Addition and Order of Dedekind Cuts: an Ordered Abelian Group in Which Every Nonempty Set Bounded Above Has a Supremum §group; as $-d<0$ (Q3), $-d\in x\oplus\ominus x$, so $-d=c+v$ with $c\in x$ and $v\in\ominus x$. By the definition of $\ominus x$ in Construction of the Dedekind Cuts: They Form a Set Totally Ordered by Inclusion, Closed under Sums, Negatives and Products of Nonnegative Cuts there is $r\in\mathbb{Q}$ with $0<r$ and $q:=-v-r\notin x$. Then $q+r=-v=c+d$, and $q=q+0<q+r$ by (Q3), so $q<c+d$.

(C5) For $x,y\in C$ and $w\in\mathbb{Q}$: $w\in x\odot y$ if and only if $w<0$ or $w=u\cdot v$ for some $u\in x$ and $v\in y$ with $0\le u$ and $0\le v$ (definition of $\odot$ in Construction of the Dedekind Cuts: They Form a Set Totally Ordered by Inclusion, Closed under Sums, Negatives and Products of Nonnegative Cuts). Hence $x\odot y=y\odot x$; $x\odot0^{*}=0^{*}$, since no $u\in0^{*}$ has $0\le u$ (Q1); and if $y\subseteq y'$ with $y'\in C$, then $x\odot y\subseteq x\odot y'$. If $x,y\in P$, then $x\odot y\in P$ by Construction of the Dedekind Cuts: They Form a Set Totally Ordered by Inclusion, Closed under Sums, Negatives and Products of Nonnegative Cuts §product, so (C1) applies to $x\odot y$. Finally $1^{*}\in P$: $1^{*}\in C$ by Construction of the Dedekind Cuts: They Form a Set Totally Ordered by Inclusion, Closed under Sums, Negatives and Products of Nonnegative Cuts §rational, and every $v<0$ has $v<1$ by $0<1$ and (Q1); likewise $0^{*}\in P$.

Step 2 (group facts). For $a,b\in C$, $a+b\in C$ and $\ominus a\in C$ by Construction of the Dedekind Cuts: They Form a Set Totally Ordered by Inclusion, Closed under Sums, Negatives and Products of Nonnegative Cuts §sum and Construction of the Dedekind Cuts: They Form a Set Totally Ordered by Inclusion, Closed under Sums, Negatives and Products of Nonnegative Cuts §negative, and by Addition and Order of Dedekind Cuts: an Ordered Abelian Group in Which Every Nonempty Set Bounded Above Has a Supremum §group the operation $+$ is associative and commutative with $a+0^{*}=a$ and $a+\ominus a=0^{*}$. Let $a,b,c\in C$.

(G1) If $a+b=0^{*}$ and $a+c=0^{*}$, then $b=c$: $b=b+(a+c)=(b+a)+c=(a+b)+c=0^{*}+c=c$.

(G2) $\ominus\ominus a=a$, $\ominus0^{*}=0^{*}$, $\ominus(a+b)=\ominus a+\ominus b$, and $(a+b)+\ominus b=a=(a+\ominus b)+b$. Indeed $\ominus a+a=0^{*}=\ominus a+\ominus\ominus a$; $0^{*}+0^{*}=0^{*}=0^{*}+\ominus0^{*}$; and $(a+b)+(\ominus a+\ominus b)=(a+\ominus a)+(b+\ominus b)=0^{*}=(a+b)+\ominus(a+b)$ by associativity and commutativity; apply (G1) in each case. The last two identities follow from associativity, $b+\ominus b=\ominus b+b=0^{*}$ and $a+0^{*}=a$.

(G3) If $b\in P$, then $a\subseteq a+b$. Indeed $0^{*}\le_{C}b$ gives $0^{*}+a\le_{C}b+a$ by Addition and Order of Dedekind Cuts: an Ordered Abelian Group in Which Every Nonempty Set Bounded Above Has a Supremum §order, where $0^{*}+a=a$ and $b+a=a+b$. Hence if $a,b\in P$, then $0^{*}\subseteq a\subseteq a+b$, so $a+b\in P$, and also $b\subseteq a+b$.

(G4) If $a\in P$ and $\ominus a\in P$, then $a=0^{*}$: by (G3), $a\subseteq a+\ominus a=0^{*}\subseteq a$. Consequently, by Construction of the Dedekind Cuts: They Form a Set Totally Ordered by Inclusion, Closed under Sums, Negatives and Products of Nonnegative Cuts §absolute, if $a\neq0^{*}$ then exactly one of $a\in P$ and $\ominus a\in P$ holds.

Step 3 (the product $\odot$ on $P$). Let $x,y,z\in P$. By (C5), $x\odot y=y\odot x\in P$ and $x\odot0^{*}=0^{*}$. Each class compared below is an element of $P$ or of the form $s\odot t$, so contains $0^{*}$; hence for each inclusion only elements $w$ with $0\le w$ need checking (Q1), and for such $w$, $w\in s\odot t$ means $w=u\cdot v$ with $u\in s$, $v\in t$, $0\le u$ and $0\le v$.

(3a) $(x\odot y)\odot z=x\odot(y\odot z)$. Let $0\le w\in(x\odot y)\odot z$: $w=e\cdot c$ with $e\in x\odot y$, $c\in z$, $0\le e$, $0\le c$. As $0\le e$, $e=u\cdot v$ with $u\in x$, $v\in y$, $0\le u$, $0\le v$. Then $w=u\cdot(v\cdot c)$ with $v\cdot c\in y\odot z$ and $0\le v\cdot c$ (Q2), so $w\in x\odot(y\odot z)$. Conversely, if $0\le w=u\cdot e$ with $u\in x$, $e\in y\odot z$, $0\le u$, $0\le e$, then $e=v\cdot c$ with $v\in y$, $c\in z$, both nonnegative, and $w=(u\cdot v)\cdot c$ with $0\le u\cdot v\in x\odot y$, so $w\in(x\odot y)\odot z$.

(3b) $x\odot1^{*}=x$. If $0\le w\in x\odot1^{*}$, then $w=u\cdot v$ with $u\in x$, $0\le u$ and $0\le v<1$, so $w\le1\cdot u=u$ by (Q2) and $w\in x$ by (C1). If $0\le w\in x$, choose $u\in x$ with $w<u$ (openness); then $0<u$ by (Q1). Let $v=w\cdot u^{-1}$. Then $0\le v$ by (Q2) and (Q3), $v<u\cdot u^{-1}=1$ by (Q2), and $u\cdot v=w$; so $v\in1^{*}$ and $w\in x\odot1^{*}$.

(3c) $x\odot(y+z)=x\odot y+x\odot z$. Here $y+z\in P$ by (G3), and both sides lie in $P$ by (C5) and (G3). If $y=0^{*}$, then $y+z=z$ and $x\odot y+x\odot z=0^{*}+x\odot z=x\odot z$, so both sides are $x\odot z$; the case $z=0^{*}$ is symmetric. So let $y\neq0^{*}\neq z$; then $0\in y$ and $0\in z$ by (C3).

For $\subseteq$, let $0\le w\in x\odot(y+z)$: $w=u\cdot s$ with $u\in x$, $s\in y+z$, $0\le u$, $0\le s$, and $s=e+f$ with $e\in y$, $f\in z$. We may assume $0\le e$ and $0\le f$: if $e<0$, then $s=e+f<0+f=f$ by (Q3), so $s\in z$ by (C1), and we replace $(e,f)$ by $(0,s)$, using $0\in y$; if $f<0$, symmetrically replace $(e,f)$ by $(s,0)$. Then $w=u\cdot e+u\cdot f$ with $u\cdot e\in x\odot y$ and $u\cdot f\in x\odot z$, so $w\in x\odot y+x\odot z$.

For $\supseteq$, let $0\le w\in x\odot y+x\odot z$: $w=p+q$ with $p\in x\odot y$ and $q\in x\odot z$. If $p<0$, then $w<q$ by (Q3); as $z\subseteq y+z$ by (G3), (C5) gives $q\in x\odot z\subseteq x\odot(y+z)$, and $w\in x\odot(y+z)$ by (C1). If $q<0$, the same argument applies with $y\subseteq y+z$. Otherwise $0\le p$ and $0\le q$, so $p=u_{1}\cdot e$ and $q=u_{2}\cdot f$ with $u_{1},u_{2}\in x$, $e\in y$, $f\in z$, all nonnegative. Let $u$ be the larger of $u_{1}$ and $u_{2}$ for the total order $\le$; then $u\in x$, and $p\le u\cdot e$, $q\le u\cdot f$ by (Q2), so $w\le u\cdot e+u\cdot f=u\cdot(e+f)$ by (Q3). Since $e+f\in y+z$ and $0\le e+f$ (Q3), $u\cdot(e+f)\in x\odot(y+z)$, and $w\in x\odot(y+z)$ by (C1).

Step 4 (reciprocals in $P$). Let $x\in P$ with $x\neq0^{*}$. By (C3), $0\in x$, there is $a\in x$ with $0<a$, and $0<q$ for every $q\in\mathbb{Q}\setminus x$; since $x\neq\mathbb{Q}$, fix $q_{0}\in\mathbb{Q}\setminus x$. Let

$$w=\{p\in\mathbb{Q}:\exists q\,(q\in\mathbb{Q}\wedge q\notin x\wedge p\cdot q<1)\}.$$

As a subclass of the set $\mathbb{Q}$, $w$ is a set by Subclasses of Sets Are Sets, the Union and Power Set of a Set Exist Uniquely, Binary Unions of Sets Are Sets, and the Universal Class Is Proper §subclass, hence an element of $\mathcal{P}(\mathbb{Q})$ by The Union Set and the Power Set of a Set §power.

(4a) $w\in P$ and $w\neq0^{*}$. If $p\le0$, then $p\cdot q_{0}\le0\cdot q_{0}=0<1$ by (Q2) and (Q1), so $p\in w$; in particular $0\in w$ and $w\neq\emptyset$. Also $a^{-1}\notin w$, so $w\neq\mathbb{Q}$: if $a^{-1}\cdot q<1$ with $q\notin x$, then $q=(a^{-1}\cdot q)\cdot a<1\cdot a=a$ by (Q2), so $q\in x$ by downward closure, a contradiction. Downward closure: if $p\in w$ with witness $q$ and $p'<p$, then $p'\cdot q<p\cdot q<1$ by (Q2) and $0<q$, so $p'\in w$. Openness: if $p\in w$ with witness $q$, then $p=(p\cdot q)\cdot q^{-1}<1\cdot q^{-1}=q^{-1}$ by (Q2) and (Q3); take $p'$ with $p<p'<q^{-1}$ (midpoint, (Q3)); then $p'\cdot q<q^{-1}\cdot q=1$ by (Q2), so $p'\in w$. Thus $w\in C$, and $w\in P$, $w\neq0^{*}$ by (C3) since $0\in w$.

(4b) $x\odot w\subseteq1^{*}$. Elements $t<0$ lie in $1^{*}$ by (Q1). If $0\le t\in x\odot w$, then $t=u\cdot p$ with $u\in x$, $p\in w$, $0\le u$, $0\le p$, and $p\cdot q<1$ for some $q\in\mathbb{Q}\setminus x$. By (C2), $u<q$, so $t=u\cdot p\le q\cdot p<1$ by (Q2) and (Q1), i.e. $t\in1^{*}$.

(4c) $1^{*}\subseteq x\odot w$. Let $t\in1^{*}$, so $t<1$; if $t<0$, then $t\in x\odot w$. Let $0\le t$. Then $0<1-t$ (Q3), and $d:=a\cdot(1-t)$ satisfies $0<d$ by (Q2). By (C4) choose $c\in x$ and $q\in\mathbb{Q}\setminus x$ with $q<c+d$, and let $u$ be the larger of $c$ and $a$. Then $u\in x$, $0<a\le u$ (so $0<u$ by (Q1)), and $q<c+d\le u+d$ by (Q3) and (Q1). We claim $t\cdot q<u$. If $t=0$, then $t\cdot q=0<u$. If $0<t$, then $t\cdot q<t\cdot(u+d)=t\cdot u+t\cdot d$ by (Q2); moreover $t\cdot a\le1\cdot a=a\le u$ by (Q2), so $t\cdot d=(t\cdot a)\cdot(1-t)\le u\cdot(1-t)$ by (Q2), and therefore $t\cdot q<t\cdot u+u\cdot(1-t)=u$ by (Q3) and (Q1). Now let $p=t\cdot u^{-1}$. Then $0\le p$ by (Q2) and (Q3), $u\cdot p=t$, and $p\cdot q=(t\cdot q)\cdot u^{-1}<u\cdot u^{-1}=1$ by (Q2), so $p\in w$ with witness $q$. Hence $t=u\cdot p\in x\odot w$.

By (4b) and (4c), $x\odot w=1^{*}$.

Step 5 (signs). For $x\in C$, $|x|_{C}\in P$ by The Real Numbers §operations. Write $\mathrm{id}$ for the identity map of $C$, and let $\varepsilon,\eta,\zeta$ range over $\{\mathrm{id},\ominus\}$, unary operations on $C$, given by Maps and Relations Given by Formulas §map since $\ominus x\in C$ for every $x\in C$ by Construction of the Dedekind Cuts: They Form a Set Totally Ordered by Inclusion, Closed under Sums, Negatives and Products of Nonnegative Cuts §negative. Since $\ominus\ominus a=a$ (G2), the composite $\varepsilon\eta$ of two of them again lies in $\{\mathrm{id},\ominus\}$, and composition is associative. Let $x,y\in C$.

(5a) Let $\varepsilon_{x}=\mathrm{id}$ if $x\in P$ and $\varepsilon_{x}=\ominus$ otherwise. Then $x=\varepsilon_{x}(|x|_{C})$: if $x\notin P$, then $|x|_{C}=\ominus x$ and $\ominus|x|_{C}=\ominus\ominus x=x$ by (G2).

(5b) $|\ominus x|_{C}=|x|_{C}$, and if $x\neq0^{*}$, then $\ominus x\in P$ if and only if $x\notin P$. For $x=0^{*}$ we have $\ominus x=0^{*}$ by (G2). For $x\neq0^{*}$, exactly one of $x$ and $\ominus x$ lies in $P$ by (G4); if $x\in P$, then $|\ominus x|_{C}=\ominus\ominus x=x=|x|_{C}$, and if $x\notin P$, then $|\ominus x|_{C}=\ominus x=|x|_{C}$.

(5c) $x\cdot y=y\cdot x$: the case distinction in The Real Numbers §operations is symmetric in $x$ and $y$, and $|x|_{C}\odot|y|_{C}=|y|_{C}\odot|x|_{C}$ by (C5).

(5d) If $x=0^{*}$ or $y=0^{*}$, then $x\cdot y=0^{*}$: since $0^{*}\in P$, $|0^{*}|_{C}=0^{*}$, so $|x|_{C}\odot|y|_{C}=0^{*}$ by (C5), and $\ominus0^{*}=0^{*}$ by (G2).

(5e) If $x,y\in P$, then $x\cdot y=x\odot y\in P$, by The Real Numbers §operations and (C5).

(5f) $(\ominus x)\cdot y=\ominus(x\cdot y)$. If $x=0^{*}$, both sides are $0^{*}$ by (5d) and (G2). If $x\neq0^{*}$, then by (5b) the case of The Real Numbers §operations that applies to the pair $(\ominus x,y)$ is the other one than for $(x,y)$, while $|\ominus x|_{C}\odot|y|_{C}=|x|_{C}\odot|y|_{C}=:m$. So one of $x\cdot y$ and $(\ominus x)\cdot y$ is $m$ and the other is $\ominus m$, and since $\ominus\ominus m=m$ (G2), $(\ominus x)\cdot y=\ominus(x\cdot y)$ in either case. By (5c), $x\cdot(\ominus y)=(\ominus y)\cdot x=\ominus(y\cdot x)=\ominus(x\cdot y)$, and then $(\ominus x)\cdot(\ominus y)=\ominus\ominus(x\cdot y)=x\cdot y$. Hence $(\varepsilon x)\cdot(\eta y)=\varepsilon\eta(x\cdot y)$ for all $\varepsilon,\eta$.

Step 6 (lem:dedekind-cut-multiplication-nbg-2026a#ring). Let $x,y,z\in C$, $\bar x=|x|_{C}$, $\bar y=|y|_{C}$, $\bar z=|z|_{C}$ (elements of $P$), and $\varepsilon=\varepsilon_{x}$, $\eta=\varepsilon_{y}$, $\zeta=\varepsilon_{z}$, so that $x=\varepsilon\bar x$, $y=\eta\bar y$, $z=\zeta\bar z$ by (5a).

Commutativity is (5c). Unit: $1_{\mathbb{R}}=1^{*}\in P$ by (C5), so by (5f), (5e) and (3b), $x\cdot1^{*}=\varepsilon(\bar x\cdot1^{*})=\varepsilon(\bar x\odot1^{*})=\varepsilon\bar x=x$.

Associativity: $\bar x\odot\bar y$ and $\bar y\odot\bar z$ lie in $P$ by (C5), so by (5f) and (5e),

$$(x\cdot y)\cdot z=(\varepsilon\eta(\bar x\odot\bar y))\cdot(\zeta\bar z)=\varepsilon\eta\zeta((\bar x\odot\bar y)\odot\bar z),\qquad x\cdot(y\cdot z)=(\varepsilon\bar x)\cdot(\eta\zeta(\bar y\odot\bar z))=\varepsilon\eta\zeta(\bar x\odot(\bar y\odot\bar z)),$$

and these agree by (3a).

Distributivity: by (5f), $x\cdot(y+z)=\varepsilon(\bar x\cdot(y+z))$ and $x\cdot y+x\cdot z=\varepsilon(\bar x\cdot y)+\varepsilon(\bar x\cdot z)$; when $\varepsilon=\ominus$, $\ominus(\bar x\cdot y)+\ominus(\bar x\cdot z)=\ominus(\bar x\cdot y+\bar x\cdot z)$ by (G2). So it suffices to prove, for all $y,z\in C$, the identity $D(y,z)$: $\bar x\cdot(y+z)=\bar x\cdot y+\bar x\cdot z$. By commutativity of $+$, $D(y,z)$ holds if and only if $D(z,y)$ does.

(6i) $D(y,z)$ implies $D(\ominus y,\ominus z)$: by (G2), (5f), $D(y,z)$, (G2) and (5f) in turn, $\bar x\cdot(\ominus y+\ominus z)=\bar x\cdot\ominus(y+z)=\ominus(\bar x\cdot(y+z))=\ominus(\bar x\cdot y+\bar x\cdot z)=\ominus(\bar x\cdot y)+\ominus(\bar x\cdot z)=\bar x\cdot(\ominus y)+\bar x\cdot(\ominus z)$.

(6ii) If $y,z\in P$, then $D(y,z)$: $y+z\in P$ by (G3), so by (5e) and (3c), $\bar x\cdot(y+z)=\bar x\odot(y+z)=\bar x\odot y+\bar x\odot z=\bar x\cdot y+\bar x\cdot z$.

(6iii) If $y\in P$, $z\notin P$ and $y+z\in P$, then $D(y,z)$. Indeed $\ominus z\in P$ by Construction of the Dedekind Cuts: They Form a Set Totally Ordered by Inclusion, Closed under Sums, Negatives and Products of Nonnegative Cuts §absolute, and $y=(y+z)+\ominus z$ by (G2). By (6ii) for the pair $(y+z,\ominus z)$ and by (5f), $\bar x\cdot y=\bar x\cdot(y+z)+\bar x\cdot(\ominus z)=\bar x\cdot(y+z)+\ominus(\bar x\cdot z)$. Adding $\bar x\cdot z$ and using (G2), $\bar x\cdot y+\bar x\cdot z=(\bar x\cdot(y+z)+\ominus(\bar x\cdot z))+\bar x\cdot z=\bar x\cdot(y+z)$.

(6iv) Let $y,z\in C$ be arbitrary. If $y,z\in P$, (6ii) applies. If $\ominus y,\ominus z\in P$, then (6ii) gives $D(\ominus y,\ominus z)$ and (6i) gives $D(\ominus\ominus y,\ominus\ominus z)$, which is $D(y,z)$ by (G2). Otherwise, since at least one of $y,\ominus y$ and at least one of $z,\ominus z$ lies in $P$ (Construction of the Dedekind Cuts: They Form a Set Totally Ordered by Inclusion, Closed under Sums, Negatives and Products of Nonnegative Cuts §absolute), exactly one of $y,z$ lies in $P$ and exactly one of $\ominus y,\ominus z$ lies in $P$ (if $y\in P$ then $z\notin P$, so $\ominus z\in P$ and $\ominus y\notin P$; if $y\notin P$ then $\ominus y\in P$, so $\ominus z\notin P$ and $z\in P$). By Construction of the Dedekind Cuts: They Form a Set Totally Ordered by Inclusion, Closed under Sums, Negatives and Products of Nonnegative Cuts §absolute applied to $y+z$, and (G2), $y+z\in P$ or $\ominus y+\ominus z=\ominus(y+z)\in P$. In the first case (6iii), applied to $(y,z)$ or to $(z,y)$, gives $D(y,z)$. In the second, (6iii) likewise gives $D(\ominus y,\ominus z)$, and (6i) with (G2) gives $D(y,z)$.

Step 7 (lem:dedekind-cut-multiplication-nbg-2026a#nontrivial, lem:dedekind-cut-multiplication-nbg-2026a#reciprocal, lem:dedekind-cut-multiplication-nbg-2026a#nonnegative). Nontriviality: $0\in1^{*}$ since $0<1$, and $0\notin0^{*}$ by (Q1); so $0_{\mathbb{R}}=0^{*}\neq1^{*}=1_{\mathbb{R}}$.

Reciprocals: let $x\in C$ with $x\neq0^{*}$. If $x\in P$, Step 4 gives $w\in P$ with $x\odot w=1^{*}$, and $x\cdot w=x\odot w=1_{\mathbb{R}}$ by (5e). If $x\notin P$, then $a:=\ominus x\in P$, and $a\neq0^{*}$ since otherwise $x=\ominus\ominus x=\ominus0^{*}=0^{*}$ by (G2). Step 4 gives $w'\in P$ with $a\odot w'=1^{*}$; let $w=\ominus w'\in C$. Then $x=\ominus a$ by (G2), and by (5f) and (5e), $x\cdot w=(\ominus a)\cdot(\ominus w')=a\cdot w'=a\odot w'=1_{\mathbb{R}}$.

Nonnegative products: $0_{\mathbb{R}}\le_{C}x$ and $0_{\mathbb{R}}\le_{C}y$ mean $0^{*}\subseteq x$ and $0^{*}\subseteq y$, i.e. $x,y\in P$. By (5e), $x\cdot y=x\odot y\in P$, i.e. $0_{\mathbb{R}}\le_{C}x\cdot y$.

Step 8 (lem:dedekind-cut-multiplication-nbg-2026a#rational).

(8a) If $u,v\in\mathbb{Q}$ with $0\le u$ and $0\le v$, then $u^{*},v^{*}\in P$ and $u^{*}\cdot v^{*}=(u\cdot v)^{*}$. Indeed $u^{*}\in C$ by Construction of the Dedekind Cuts: They Form a Set Totally Ordered by Inclusion, Closed under Sums, Negatives and Products of Nonnegative Cuts §rational and $0^{*}\le_{C}u^{*}$ by Addition and Order of Dedekind Cuts: an Ordered Abelian Group in Which Every Nonempty Set Bounded Above Has a Supremum §rational-order, so $u^{*}\in P$, and likewise $v^{*}\in P$; by (5e), $u^{*}\cdot v^{*}=u^{*}\odot v^{*}$. Both $u^{*}\odot v^{*}$ and $(u\cdot v)^{*}$ contain $0^{*}$ (the latter since $0\le u\cdot v$ by (Q2), using (Q1)), so only elements $w$ with $0\le w$ need checking. If $0\le w\in u^{*}\odot v^{*}$, then $w=e\cdot f$ with $0\le e<u$ and $0\le f<v$; so $0<u$ by (Q1), and $w=e\cdot f\le u\cdot f<u\cdot v$ by (Q2) and (Q1), i.e. $w\in(u\cdot v)^{*}$. Conversely let $0\le w<u\cdot v$. Then $u\cdot v\neq0$ by (Q1), so $u\neq0$ by Rules of Arithmetic and Order in an Ordered Field §zero and $0<u$. By (Q2) and (Q3), $0\le w\cdot u^{-1}<u\cdot v\cdot u^{-1}=v$; take $f$ with $w\cdot u^{-1}<f<v$ (midpoint), so $0<f$ by (Q1), and let $e=w\cdot f^{-1}$. Then $0\le e$ by (Q2) and (Q3), $e\cdot f=w$, and $w=(w\cdot u^{-1})\cdot u<f\cdot u$ by (Q2), whence $e=w\cdot f^{-1}<u\cdot f\cdot f^{-1}=u$ by (Q2) and (Q3). So $e\in u^{*}$, $f\in v^{*}$ and $w=e\cdot f\in u^{*}\odot v^{*}$.

(8b) For $u\in\mathbb{Q}$, $(-u)^{*}=\ominus(u^{*})$: by Addition and Order of Dedekind Cuts: an Ordered Abelian Group in Which Every Nonempty Set Bounded Above Has a Supremum §rational-sum and The Rational Numbers Form an Archimedean Ordered Field Containing the Integers §ordered-field, $u^{*}+(-u)^{*}=(u+(-u))^{*}=0^{*}$, while $u^{*}+\ominus(u^{*})=0^{*}$ by Addition and Order of Dedekind Cuts: an Ordered Abelian Group in Which Every Nonempty Set Bounded Above Has a Supremum §group; apply (G1).

(8c) Let $u,v\in\mathbb{Q}$, and let $s=|u|$ and $t=|v|$, so $0\le s$ and $0\le t$ by Rules of Arithmetic and Order in an Ordered Field §absolute-value. By Absolute Value in an Ordered Field §absolute-value, if $0\le u$ then $u=s$, and we set $\varepsilon=\mathrm{id}$; otherwise $s=-u$, so $u=-s$ by Rules of Arithmetic and Order in an Ordered Field §signs, and we set $\varepsilon=\ominus$. In both cases $u^{*}=\varepsilon(s^{*})$, by (8b) in the second. Define $\eta$ from $v$ and $t$ in the same way, so $v^{*}=\eta(t^{*})$. By Rules of Arithmetic and Order in an Ordered Field §signs and commutativity, $u\cdot v=s\cdot t$ if $\varepsilon=\eta$, and $u\cdot v=-(s\cdot t)$ if $\varepsilon\neq\eta$; in the first case $\varepsilon\eta=\mathrm{id}$ by (G2), in the second $\varepsilon\eta=\ominus$, so in both cases $(u\cdot v)^{*}=\varepsilon\eta((s\cdot t)^{*})$, using (8b) in the second. On the other hand, by (5f) and (8a), $u^{*}\cdot v^{*}=(\varepsilon(s^{*}))\cdot(\eta(t^{*}))=\varepsilon\eta(s^{*}\cdot t^{*})=\varepsilon\eta((s\cdot t)^{*})$. Hence $(u\cdot v)^{*}=u^{*}\cdot v^{*}$.