Counting: Intervals, Empty Sets and Singletons, Injective Images, Subsets, Unions and Products

In the setting of The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion, let finite sets be as in Finite Sets §finite, $\#$ as in The Number of Elements of a Finite Set §cardinality, and intervals and $[n]$ as in Intervals of Natural Numbers §interval and Intervals of Natural Numbers §segment. Intervals, $\emptyset$ and singletons are finite, and so are subsets, unions, products and images of finite sets, by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice §naturals, Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice §small, Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice §subset, Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice §union and Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice §image. Let $A$ and $B$ be finite sets and $m,n\in\mathbb{N}_{0}$.

¶ $\#[n]=n$, and if $m\le n+1$, then $\#\{m,\dots,n\}=(n+1)-m$, the difference in $\mathbb{N}_{0}$.

¶ $\#A=0$ if and only if $A=\emptyset$, and $\#\{x\}=1$ for every set $x$.

¶ For every map $f:A\to C$ to a set $C$, $\#f(A)\le\#A$, with equality if $f$ is injective.

¶ If $B\subseteq A$, then $\#B\le\#A$, and $\#B=\#A$ only if $B=A$.

¶ $\#(A\cup B)+\#(A\cap B)=\#A+\#B$; in particular $\#(A\cup B)=\#A+\#B$ if $A$ and $B$ are disjoint.

¶ $\#(A\times B)=\#A\cdot\#B$.