User:Dfeuer/Natural Number does not Contain Itself

Theorem
Let $n$ be a natural number.

Then $n \notin n$.

Proof
Then proof proceeds by induction.

Let $a$ be the set of all natural numbers $n$ such that $n \notin n$.

By the definition of the empty class, $\varnothing \notin \varnothing$.

Suppose that $n \in a$.

Suppose for the sake of contradiction that $n^+ \in n^+$.

Then $n^+ \in n \cup \{n\}$.

Thus $n^+ \in n$ or $n^+ = n$.

That is, $n \cup \{n\} \in n$ or $n \cup \{n\} = n$.

Then $n \cup \{n\} = n$.

Then $n \in n$, contradicting the fact that $n \notin a$.

Suppose instead that $n \cup \{n\} \in n$.

Then since $n$ is transitive by User:Dfeuer/Natural Number is Transitive:


 * $n \cup \{n\} \subseteq n$

Thus $n \in n$, another contradiction.

Since assuming $n^+ \in n^+$ leads to a contradiction, we conclude that:


 * $n^+ \notin n^+$