Is there always a membership chain between an element and the tuple it's in? I've just started learning about ZFC set theory, and I'm looking into how relations over sets are defined in this context. The definition I found is that a relation $R$ over two sets $E$ and $F$ can either be implemented as:  
1) A subset of $E \times F$, or
2) As the triplet $(E,F,G)$ where $G$ is a subset of $E \times F$.
I was thinking about relations over sets of relations, and was wondering wether such a relation $R$ could compare itself. (i.e. could $R \space R \space S$ or $S \space R \space R$ (infix notation), where $S$ is another relation, be a valid statement?)
My reasoning was that, if it were possible, there would necessarily be a cyclic membership chain like $R \in \dotsb \in R$ if we're going with definition 1), or $R \in \dotsb \in G \in \dotsb \in R$ with definition 2). And since the axiom of foundation prohibits it, it is not possible for a relation to compare itself with another.
But while it is pretty simple to proove that this is correct if ordered pairs are implemented as $(a,b)=\{\{a\},\{a,b\}\}$ and n-tuples as nested ordered pairs or as functions, my real question is if this would also be correct regardless of the implementation of those objects.  
Since any definition for n-tuples must validate their caracteristic property for equality,
$$ \forall(a_1,\dotsb,a_n)\forall(b_1,\dotsb,b_n),\space(a_1,\dotsb,a_n)=(b_1,\dotsb,b_n)\Longleftrightarrow\bigwedge_{i=1}^n (a_i=b_i)$$
can this be used to show that, regardless of the details of how they are defined, there will allways be a membership chain from $x_i$ to $(x_1,\dotsb,x_i,\dotsb,x_n)$?
I didn't find any answer on the internet of in the math stack exchange, unsurprisingly since to me this seems like a pretty tough question to answer...
Does anybody know if there is an answer to this question?
P.S.: I haven't yet looked into class theory, category theory, or really any extentions of set theory, but if there is such a proof that requires knowlege in those fields I'd still be happy to hear them. I just want to know if there is an answer, even if it is beyond my comprehension.
 A: There are ways of implementing $(a,b)$ such that $a\not\in(a,b)$ (and there is no longer $\in$-chain connecting $a$ to $(a,b)$ either), perhaps surprisingly there are even useful ways to to do such a thing!
Among the many implementations of ordered pairs available there is the so called Quine-Rosser pair, defined as follows. Let $\sigma$ be the class function defined by $\sigma(x)=x+1$ if $x\in\Bbb N$ and $\sigma(x)=x$ otherwise.
Given two sets $A$ and $B$ consider $\sigma[A]=\{\sigma(a)\mid a\in\ A\}$, note that no element of $\sigma[A]$ contains $0$, so if we now consider $C=\{\sigma(b)\cup\{0\}\mid b\in B\}$ we can define $(A,B)=\sigma[A]\cup C$.
This is an honest definition of pair, given $(A,B)$ you can recover $A$ by looking at $\{a\in (A,B)\mid 0\not\in a\}$ and undoing $\sigma$ (shifting back integers by one), while $B$ can be recovered in a similar way by looking at the elements of $(A,B)$ that do contain $0$.
Why is this useful? Given any set $x$ let $\mathrm{rank}(x)$ denote the least ordinal $\xi$ such that $x\subseteq V_\xi$. Note that with the standard implementation of ordered pairs $\mathrm{rank}((a,b))>\max\{\mathrm{rank}(a),\mathrm{rank}(b)\}$ while the Quine-Rosser definition does not increase rank (as long as one of $A$ and $B$ has infinite rank) and the existence of such a flat pairing function is useful occasionally. 
A: You could artificially redefine tuples to make this work:
Take a set $a$ and consider $R = \{(n, a) : n \in \mathbb N\} \subset \mathbb N \times \{a\}$.
When $n \in \mathbb N$, define $(n, a)' := (n+1, a)$. Define also $(R, a)' := (0, a)$. For all other pairs of sets $x, y$, define $(x, y)' = (x, y)$. This is a good notion of tuple, and we have accordingly a notion of product $E \times' F$, and a new notion of relation.
We have $R \subset (\mathbb N \cup \{R\}) \times' \{a\}$, and with the new notion of relation we have $R \; R \; a$ because $(R, a)' = (0, a) \in R$.
