# Thread: Axiom of Regularity Problem

1. ## Axiom of Regularity Problem

In conjunction with my other post on checking the regularity of a database, I am interested in solving Problem 1 on page 56 of Suppes' Axiomatic Set Theory:

Prove that for all sets $A$, $B$, and $C$ it is not the case that

$A\in B\land B\in C\land C\in A.$

Now, the proof of Theorem 2.106 on page 54 is bound to be similar. Here's the theorem statement:

$\neg(A\in B\land B\in A).$

Proof: (Quoting directly from Suppes) Suppose that $A\in B\land B\in A.$ Then

$(1)\quad A\in\{A,B\}\cap B\quad\text{and}\quad B\in\{A,B\}\cap A.$

By the axiom of regularity there is an $x$ in $\{A,B\}$ such that

$\{A,B\}\cap x=0$

and by Theorem 2.43 [Ackbeet: on page 31. It states that $z\in\{x,y\}\iff z=x\vee z=y.$]

$x=A\quad\text{or}\quad x=B.$

Hence

$\{A,B\}\cap A=0\quad\text{or}\quad\{A,B\}\cap B=0,$

So here's my attempt at a proof of the non-existence of the 3-cycle:

Suppose the assertion were false. Then the following hold:

$(1)\quad A\in\{A,B,C\}\cap B,$
$(2)\quad B\in\{A,B,C\}\cap C,$
$(3)\quad C\in\{A,B,C\}\cap A.$

By the axiom of regularity there is an $x\in\{A,B,C\}$ such that

$\{A,B,C\}\cap x=0.$

By the obvious extension of Theorem 43,

$x=A\quad\text{or}\quad x=B\quad\text{or}\quad x=C.$

Hence, at least one of $\{A,B,C\}\cap A$ or $\{A,B,C\}\cap B$ or $\{A,B,C\}\cap C$ is zero, contradicting (1), (2), or (3). QED.

Is this a valid proof?

2. Originally Posted by Ackbeet
Is this a valid proof?
Took me a little while to decipher, but I say yes, it is valid.