Proof of Arithmetic on [0,∞]: Bookkeeping

For (i), recall that for sets $u$ and $v$ the components of the pair $(u,v)$ are $(u,v)_{1}=u$ and $(u,v)_{2}=v$, by The Components of an Ordered Pair (i). Suppose first that $p\in[0,\infty]\times[0,\infty]$ and $\varphi(p_{1},p_{2})$. By The Cartesian Product of Two Classes (i) there are sets $u$ and $v$ with $u\in[0,\infty]$, $v\in[0,\infty]$ and $p=(u,v)$; so $p_{1}=u$ and $p_{2}=v$, and $\varphi(u,v)$ holds. Conversely, let $u,v$ be sets with $u,v\in[0,\infty]$, $p=(u,v)$ and $\varphi(u,v)$. Then $p\in[0,\infty]\times[0,\infty]$ by Membership in a Cartesian Product, and the Cartesian Product of Two Sets Is a Set (i), and again $p_{1}=u$ and $p_{2}=v$, so $\varphi(p_{1},p_{2})$ holds.

For (ii), let $x,y\in\mathbb{R}$ with $0\le x$ and $0\le y$. The sum $x+y$ is a real number. Since adding a common term preserves $\le$ in the ordered ring $\mathbb{R}$, from $0\le x$ we get $y=0+y\le x+y$, and with $0\le y$ transitivity gives $0\le x+y$. Hence $x+y\in[0,\infty]$ by The Extended Real Line, Its Order, and [0,∞] (ii).

For (iii), let $x,y\in\mathbb{R}$ with $0\le x$ and $0\le y$. The product $xy$ is a real number, and $0\le xy$ since $\mathbb{R}$ is an ordered ring. Hence $xy\in[0,\infty]$ by The Extended Real Line, Its Order, and [0,∞] (ii).