Cardinality of Finite Set is Well-Defined

Theorem
Let $S$ be a finite set.

Then there is a unique natural number $n$ such that $S \sim \N_n$, where:
 * $\sim$ represents set equivalence

and:
 * $\N_n = \set {0, 1, \dotsc, n - 1}$ is the initial segment of $\N$ determined by $n$.

Proof
By the definition of finite set, there is an $n \in \N$ such that $S \sim \N_n$.

Suppose $m \in \N$ and $S \sim \N_m$.

It follows from Set Equivalence behaves like Equivalence Relation that $\N_n \sim \N_m$.

Thus by Equality of Natural Numbers, $n = m$.

Therefore the cardinality of a finite set is well-defined.