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In the end of a discussion with a friend, we arrived at the following condition:

$$A\in A^C$$

where $A^C$ denotes the complement of $A$ (say, to a referencial set $R$ such that $A\subseteq R$ and $A\in R$). I wonder: is that condition ill-posed in ZF set theory?

Edit: As Eclipe Sun points out, you can construct a rather simple example using the empty set, I wonder if you can construct more examples after imposing the condition $A\neq\varnothing$.

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Set $A=\emptyset$ and $R=\{\emptyset\}$.

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    I should add $A\neq\varnothing$. Do you mind? – 2017-02-01
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    No. If you know that for any natural number we can write it as a set: $0=\emptyset$, $1=\{\emptyset\}$, $2=1\cup\{1\}$,..., then you can consider $A=n$ and $R=n'(=n+1)$. – 2017-02-01
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    Brilliant, many thanks. – 2017-02-01
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In ZF, as long as $A\in R$ the answer is yes, since no set is an element of itself. So if $A\in R$ and $A\notin A$, then $A\in A^c$.

Without the axiom of foundation, though, it is consistent that $A\in A$, so we cannot prove that anymore.