Equality of Natural Numbers

Theorem
Let $m, n \in \N$.

Then:
 * $\N_m \sim \N_n \iff m = n$

where $\sim$ denotes set equivalence and $\N_n$ denotes the set of all natural numbers less than $n$.

Proof
By Set Equivalence an Equivalence Relation, we have that:
 * $m = n \implies \N_m \sim \N_n$

It remains to show that:
 * $m \in \N_n \implies \N_m \nsim \N_n$

Let $S = \left\{{n \in \N: \forall m \in \N_n: \N_m \nsim \N_n}\right\}$.

That is, $S$ is the set of all the natural numbers $n$ such that $\N_m \nsim \N_n$ for all $m \in \N_n$.

We use mathematical induction to prove that $S = \N$.

It follows from Consecutive Subsets of $\N$ that $0 \in S$, as $\N_0 = \varnothing$. This is the basis for the induction.

Now, assume the induction hypothesis that $n \in S$.

We now complete the induction step, that is, to show that $n + 1 \in S$.

Let $m \in \N_{n+1}$.

If $m = 0$, then $\N_m \nsim \N_{n+1}$ because $\N_0 = \varnothing$ and $\N_{n+1} \ne \varnothing$.

Aiming for a contradiction, suppose that $m \ge 1$ and $\N_m \sim \N_{n+1}$.

Then, by Set Equivalence Less One Element, that means $\N_{m-1} \sim \N_n$.

But then $m - 1 \in \N_n$ which contradicts the induction hypothesis that $n \in S$.

Thus $n + 1 \in S$.

The result follows from the fact that Set Equivalence an Equivalence Relation, in particular the symmetry clause.