Trying to prove the claim:
$$\forall x \; \forall y \; \lnot (x \in y \in x)$$
I know we should apply the axiom of regularity to the set $\{x, y\}$.
Can anyone help?
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${x,y}$ is non empty: try intersecting it with its only two elements and check whether these intersections are empty or not... if none of them is, you found that this example contradicts the regularity axiom – TheMadcapLaughs Dec 07 '16 at 16:41
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I edited your question to improve the math formatting. If you want to see waht I did, click "edit". – Caleb Stanford Dec 07 '16 at 17:07
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See also: Are there sets $A$ and $B$ such that $A \in B$ and $B \in A$? – Martin Sleziak Mar 01 '20 at 10:42
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Since you want to prove a NOT statement, i.e. for all $x$ and $y$, it is NOT the case that both $x \in y$ and $y \in x$, it makes sense to proceed by contradiction. Assume -- towards a contradiction -- that we have some $x$ and $y$ such that $x \in y$ and $y \in x$.
Now, applying regularity to $\{x,y\}$, we must have that $\{x,y\}$ contains an element disjoint from itself. So either $\{x,y\}$ is disjoint from $x$, or $\{x,y\}$ is disjoint from $y$.
Case 1: $\{x,y\}$ is disjoint from $x$.
But $y \in \{x,y\}$, and $y \in x$. So they are not disjoint. Contradiction.
Case 2: $\{x,y\}$ is disjoint from $y$.
Can you do this case?
Caleb Stanford
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