Let $\omega$ denote the minimally inductive set.
- $\omega = \aleph_0$
where $\aleph$ denotes the aleph mapping.
For all $n \in \omega$, $n \notin \NN'$ by the definition of the class of infinite cardinals.
Therefore, $\omega \le \aleph_0$.
Moreover, $\omega \in \NN'$ by Minimally Inductive Set is Infinite Cardinal.
Therefore, $\aleph_x = \omega$ for some ordinal $x$.
It follows that $\aleph_0 \le \aleph_x$ since $0 \le x$ and the definition of the aleph mapping, so $\aleph_0 \le \omega$.
Thus, $\aleph_0 = \omega$.