Closed Form for Triangular Numbers/Proof using Cardinality of Set

Proof
Let $\N_n^* = \left\{ {1, 2, 3, \cdots, n}\right\}$ be the initial segment of natural numbers.

Let $A = \left\{ {\left({a, b}\right): a \le b, a, b \in \N_n^*}\right\}$

Let $B = \left\{ {\left({a, b}\right): a \ge b, a, b, \in \N_n^*}\right\}$

Let $\phi: A \to B$ be the mapping:
 * $\phi \left({x, y}\right) = \left({y, x}\right)$

By definition of dual ordering, $\phi$ is a bijection:
 * $(1): \quad \left\vert{A}\right\vert = \left\vert{B}\right\vert$

We have:

Thus:

Combined with $\left({1}\right)$ this yields:
 * $\left\vert{A}\right\vert = \dfrac {n^2 + n} 2 = \dfrac {n \left({n + 1}\right)} 2$

It remains to prove that:
 * $T_n = \left\vert{A}\right\vert$