Krdytkyu

Axiom of extensionality looks like this:

$forall{A,B}:(forall{X}:(X in A iff X in B) implies A=B)$

Now set $A= emptyset={forall{Y}:lnot(Y in A)}$ (axiom of the empty set)

Now look at all sentences produced by our Axiom of extensionality that uses the empty set.

$forall{B}:(forall{X}:(X in emptyset iff X in B) implies emptyset=B)$

But $X in emptyset $ is always $false$ for every $X$, it follows $X in B$ is $false$ (exception later)

That means $forall{X}:(X in emptyset iff X in B)$ is always $true$

Now the Implication $Aimplies B$ is $trueiff A=true land B=true$

and $false iff A=true land B=false$ (Rule of Implication)

We know that $forall{X}:(X in emptyset iff X in B)$ is $true$ and that implies that $emptyset=B$.

$forall{B}:(forall{X}:(X in emptyset iff X in B) implies emptyset=B)$ is $true$ when $B=emptyset$, but tricky to resolve when $B neq emptyset$, because imagine $B neq emptyset$, then we have:

$forall{B}:(forall{X}:(X in emptyset iff X in B) implies emptyset=B)$

We know for a fact $emptyset=B$ is false, but $forall{X}:(X in emptyset iff X in B)$ is tricky to resolve when $B neq emptyset$, because:

$X in emptyset$ is false for every $X$ but $X in B$ is at least true in some scenarios, because we choose $B neq emptyset$. In all those scenarious, the axiom

$forall{A,B}:(forall{X}:(X in A iff X in B) implies A=B)$

produce Implication of the form: $A implies B$, where A and B are false, but the whole sentence is true and A=true and B=false, so the whole sentence is false.

This proofs the inconsistency of the Axiom of extensionality with the axiom of the empty set

Example:

$A=emptyset$

$B={1,2,3}$

$forall{X}:(X in emptyset iff X in {1,2,3}) implies emptyset={1,2,3})$

$1 in emptyset iff 1 in {1,2,3})$ =wrong $implies emptyset={1,2,3})$ =wrong OR true, that means the whole sentence true (vacuous truth)

$2 in emptyset iff 2 in {1,2,3})$ same here

$3 in emptyset iff 3 in {1,2,3})$ same here

$4 in emptyset iff 4 in {1,2,3})$ =true $implies emptyset={1,2,3})$ =wrong OR false, that means the whole sentence is undecidable, but bexause the sentence is an axiom, it must be true and there fore if the first term is true it must be implied that $implies emptyset={1,2,3})$ is also true.

So the axiom of extensionality produces true sentences for some picks for A and B, especially if we pick $A=emptyset$ and $Bneq emptyset$, but it's not clear wether or not $emptyset={1,2,3})$

We could formulate it like this: If we ask the question: Is the sentence $emptyset={1,2,3})$ $true$ or $false$ from the perspective of the axioms. If you try the numbers 1,2,3 for X, then you get a sentence which is true, and the subsentence $emptyset={1,2,3})$ can be false or true. We can't know that.

But if we ask the question: Is the sentence $emptyset={1,2,3})$ $true$ or $false$ from the perspective of the axioms and we try every other element (except 1,2,3) for X, then the sentence is true or false, but because it's an axiom, it mus be true, therefore, because $4 in emptyset iff 4 in {1,2,3})$ is true the subsentence $emptyset={1,2,3})$ must be true.

But that contradicts the axiom of the empty set.

Where is the mistake?

The statement isn't

$forall x: [(xin A iff xin B)implies A = B]$

That would clearly fail frequently. Obviously $1in {1,2,3}iff 1in {1,4,5}$ but ${1,2,3} ne {1,4,5}$.

The statement is:

$[forall x: (xin A iff xin B)]implies A = B$ which is clearly true.

($[forall x: (x in {1,2,3} iff x in {1,4,5}]$ is obviously false so $[forall x: (xin {1,2,3} iff x in {1,4,5})]implies {1,2,3} = {1,4,5}$ is true.)

In the case $A =emptyset$

The $(x in emptyset iff xin B) = (xin emptyset implies x in B) land (x in B implies xin emptyset)$.

$(xin emptyset implies x in B)$ is always true. But $(x in B implies xin emptyset)$ is true if $x not in B$ but false if $x in B$.

So $(xin emptyset iff xin B)$ is true if and only if $x not in B$.

So if $Bne emptyset$ then $(xin emptyset iff xin B)$ is sometimes false and $[forall x: (xin emptyset iff xin B)]$ is false. In which case $[forall x: (xin emptyset iff xin B)]implies X$ is always true and $[forall x: (xin emptyset iff xin B)]implies B = emptyset$ is true.

If $B =emptyset$ then $(xin emptyset iff xin B)$ is always true. So $[forall x: (xin emptyset iff xin B)]$ is true. In which case $[forall x: (xin emptyset iff xin B)]implies X$ is only true if $X$ is true. And as $B = emptyset$ is true. $[forall x: (xin emptyset iff xin B)]implies B =emptyset$ is true.

So $[forall x: (xin emptyset iff xin B)]implies B = emptyset$ is always true.

$forall{B}:(forall{X}:(X in emptyset iff X in B) implies emptyset=B)$

But $X in emptyset $ is always $false$ for every $X$, it follows $X in B$ is $false$ (exception later)

This is where you went wrong. What this says is that $textbf{if}$ $Xin B$ is false $forall X$, then $varnothing=B$. This creates no contradiction.

Your axiom is
$$
forall{A,B}:(forall{X}:(X in A iff X in B) implies A=B)
$$
When you apply this to $A = emptyset$, you get
$$
forall{B}:(forall{X}:(X in emptyset iff X in B) implies emptyset=B)
$$

You've looked at $X in emptyset$, and observing that it's always false, concluded that the implication in the parens is true, because false implies anything.

What you seem to have failed to notice is that it's a double implication: for the main (one way) implication to hold for some particular set $B$ (so that you can conclude that $B$ is the empty set), you need to know that
$$
forall{X}:(X in emptyset iff X in B)
$$
holds; in particular, that both
$$
forall{X}:(X in emptyset implies X in B)
$$
and
$$
forall{X}:(X in B implies X in emptyset).
$$

You have not shown that second item (and indeed, it's not true for any set $B$ except for the empty set). In particular, if $B$ is the set ${1, 2}$, then $1 in B$, but $1 notin emptyset$, so that implication fails.