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I believe the following is true, however I'm stuck trying to prove.

If a is a set, then a ≠ {{a}}

Indeed I believe the result is true for any number of brackets, i.e a ≠ {{{a}}} or {{{{a}}}} etc.

The case a≠{a} follows from the axiom of foundation (think also known as axiom of regularity)

Am guessing the proof involves foundation and perhaps induction?

Thanks :)))

Asaf Karagila
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user1044791
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    Under the axiom of foundation it is also true that transitive closure $\overline{\in}$ is well founded (see here for instance). That might help. It demands that every non-empy set $a$ must have an element $b$ such that $a$ and the transitive closure of $b$ have empty intersection. Try that out in the case $a={{a}}$ and you will meet a contradiction. – drhab Apr 05 '22 at 12:51
  • Thank you both!!!!!! – user1044791 Apr 05 '22 at 14:27

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