axiomatic sets 34 (regularity 3)

We shall give the easy proof in which Axiom of regularity
\[ \forall x[x\ne \phi\rightarrow \exists y[y\in x\wedge y\cap x=\phi]]  \]
means that there is not a infinite sequence $a_1\ni a_2\ni a_3\ni\cdots$ .

Suppose that there is a infinite sequence $a_1\ni a_2\ni a_3\ni\cdots$ .
Then, we are able to make the new set $A=\left\{ a_1,a_2,a_3,\cdots  \right\}$ .

If we accept axiom of regularity,
then there must be a $b$ such that $A\cap b=\phi$  in $a_1,a_2,a_3,\cdots $ .

Let $b$ be $a_i$ . However, as $\cdots\ni a_{i-1}\ni a_i\ni a_{i+1}\ni \cdots$ ,
$a_{i+1}\in b(=a_i)$ and $a_{i+1}\in A$ are true.

It is a contradiction, because $a_{i+1}\in(A\cap b)$ .
Therefore, there is not a infinite sequence $a_1\ni a_2\ni a_3\ni\cdots$ .

Conversely, we accept axiom of regularity.
For two arbitrary sets $A,a$ , even if $A\cap a=a_1$ ,$A\cap a_1=a_2$ ,$\cdots$ continue,
by axiom of regularity, the sequence will stop somewhere.

If $A\cap a_i=\phi$ , then by $a_i=b$ , axiom of regularity will be satisfied.