Originally Posted by

**chopet** I perfectly understand your explanation, except this part where the so-called "first element" is singled out.

Is the "first element" arbitrary?

Could it well be the "second element"?

Ok. Let me attempt to write a proof with my own Axioms.(please bear with me...I'm but a beginner)

__Proof__

**Axiom of Order:**A set can either be ORDERED () or UNORDERED {}.

The relationship between them is:

{a,b} = (a,b) OR (b,a)

Trivial:

{a} = a

An analogy is with $\displaystyle \sqrt9 = -3 \ or \ 3$

**Axiom of Inequality:**

A set with 1 element is DIFFERENT from a set with 2 elements.

This axiom is actually a corollary from :

(a,b) = (c,d) iff a=c and b=d

So we dissect the expression {{a,b}, {a}} using the 1st axiom (converting {} to () ).

{ {a,b},{a} }

= { {a,b} , a }

= ( {a,b} , a ) OR ( a, {a,b} )

= ( (a,b), a ) OR ( (b,a),a) OR ( a, (a,b) ) OR ( a, (b,a) )

Using the 2nd axiom of inequality:

(a,b) is none of the above.

So the definition :

(a,b) = { {a,b},{a} }

is absurb.