For $I=\{\mathrm{\varnothing },\{\mathrm{\varnothing }\}\}$ with $A_{\mathrm{\varnothing }}=A$ and $A_{\{\mathrm{\varnothing }\}}=B$, the map $f\mapsto (f(\mathrm{\varnothing }),f(\{\mathrm{\varnothing }\}))$ is a bijection ${\prod }_{i\in I}{A}_i\to A\times B$

Given: sets $A$ and $B$, the index set $I:=\{\mathrm{\varnothing },\{\mathrm{\varnothing }\}\}$, the family $(A_i{)}_{i\in I}$ above, and $P:={\prod }_{i\in I}{A}_i$.

an element of ${\prod }_{i\in I}{A}_i$ is a function with domain $I$ that takes its value at each index inside the member carried by that index (The product ${\prod }_{i\in I}{A}_i:=\{f:I\to {\bigcup }_{i\in I}{A}_i\mid f(i)\in A_i\text{ for every }i\in I\}$).

$z\in A\times B$ holds if and only if $z=(a,b)$ for some $a\in A$ and some $b\in B$ (The Cartesian product $A\times B:=\{z\in \mathcal{P}(\mathcal{P}(A\cup B)):\mathrm{\exists }a\in A\mathrm{\exists }b\in Bz=(a,b)\}$).

$(a,b)=(c,d)$ if and only if $a=c$ and $b=d$ ($(a,b)=(c,d)$ if and only if $a=c$ and $b=d$).

$f$ is injective (one-to-one) if $f(x)=f(y)$ implies $x=y$, for all $x,y\in A$ (Injection, surjection, bijection).

$f$ is surjective (onto) if for every $b\in B$ there is some $x\in A$ with $f(x)=b$ (Injection, surjection, bijection).

$f=g$ if and only if $\mathrm{dom}f=\mathrm{dom}g$ and $f(x)=g(x)$ for every $x\in \mathrm{dom}f$ (Functions $f$ and $g$ are equal if and only if $\mathrm{dom}f=\mathrm{dom}g$ and $f(x)=g(x)$ for every $x$ in that common domain).

$\{x,y\}$ is the set whose elements are exactly $x$ and $y$, and $\{x\}:=\{x,x\}$ (The unordered pair $\{x,y\}$ and the singleton $\{x\}=\{x,x\}$).

There is exactly one set with no elements, written $\mathrm{\varnothing }$ (There is exactly one set with no elements, written $\mathrm{\varnothing }$).

A function is a relation $f$ such that $(a,b)\in f$ and $(a,c)\in f$ imply $b=c$ (A function is a relation $f$ with $(a,b)\in f$ and $(a,c)\in f$ implying $b=c$; $f:A\to B$, the value $f(a)$, domain and codomain).

An indexed family with index set $I$ is a function $A$ with $\mathrm{dom}A=I$ (An indexed family $(A_i{)}_{i\in I}$ is a function with domain $I$; $\{A_i:i\in I\}$ is its range).

$\mathrm{dom}R:=\{a:\mathrm{\exists }b(a,b)\in R\},\mathrm{ran}R:=\{b:\mathrm{\exists }a(a,b)\in R\}$ (Relation, $\mathrm{dom}R$, $\mathrm{ran}R$, $\mathrm{fld}R$, and the specialisations "relation from $A$ to $B$" and "relation on $A$").

For any parameters $\bar{p}$ and any set $x$, there is a set $y$ whose elements are exactly the elements $z$ of $x$ for which $\phi (z,\bar{p})$ holds (The Axiom Schema of Separation: for each formula $\phi$, $\mathrm{\forall }\bar{p}\mathrm{\forall }x\mathrm{\exists }y\mathrm{\forall }z(z\in y\leftrightarrow (z\in x\wedge \phi (z,\bar{p})))$).

$z\in {\bigcup }_{i\in I}{A}_i$ holds if and only if $z\in A_i$ for some $i\in I$ (${\bigcup }_{i\in I}{A}_i:=\bigcup \{A_i:i\in I\}$, and ${\bigcap }_{i\in I}{A}_i:=\bigcap \{A_i:i\in I\}$ for $I\mathrm{\ne }\mathrm{\varnothing }$).

$z\in a\cup b$ holds if and only if $z\in a$ or $z\in b$ ($\bigcup \mathrm{\varnothing }=\mathrm{\varnothing }$, $\bigcup \{a\}=a$, $\bigcup \{a,b\}=a\cup b$, $\bigcap \{a\}=a$, and $\bigcap \{a,b\}=a\cap b$).