What types of $\varphi(x)$ would satisfy the axiom of Regularity I just read in Bell Set Theory that the axiom of regularity can be written differently from the usual way as :
$$ \forall x [ (\forall y \in x  \text{ } \varphi(y)) \implies \varphi (x)] \implies \forall x \text{ } \varphi(x) \tag{1}$$
But for example :
Let $\varphi(x)$ mean that "x is an even number less than 11", then if x is a set of even numbers x := (n | $\varphi(n)$), then $\varphi(x)$ is not an even number. In addition isn't a set's meaning  defined entirely by its contents?
So what type of $\varphi(x)$ would satisfy (1) and is (1) equivalent to the usual form e.g. wiki (https://en.wikipedia.org/wiki/Axiom_of_regularity) which has no immediate references to a logical expression, but only to sets?
 A: I will refer to the axiom scheme you mentioned as the axiom scheme of foundation, and I will refer to the other as the axiom of regularity.
You are correct that $X$ is not an even number. Thus, we see that the hypothesis of the axiom is not satisfied - we do not have $\forall x (\forall y \in x \phi(y) \to \phi(x))$. Thus, we see that this instance of the axiom of foundation is vacuously true, since it is an implication whose hypothesis is false.
Suppose regularity, and suppose $\forall x (\forall y \in x \phi(y) \to \phi(x))$. Consider an arbitrary $x$, and let $T$ be the transitive closure of $\{x\}$. Let $W = \{y \in T \mid \neg \phi(y)\}$. Suppose $W$ is nonempty. Let $w$ be an $\in$-minimal element. Then for all $x \in w$, $x \in T \setminus W$, and thus $\phi(x)$. Therefore, $\phi(w)$. Then $w \notin W$; contradiction. Therefore, we know $W$ is empty. Then $x \in \{x\} \subseteq T$, and $x \notin W$. Thus, $\phi(x)$.
Now suppose foundation, and let $W$ be a nonempty set. Suppose $W$ has no $\in$-minimal element - that is, for all $y \in W$, there is $z \in y \cap W$. Let $\phi(x)$ be the statement $x \notin W$. Then consider an arbitrary $y$, and suppose for all $x \in y$, $x \notin W$. Then we see that $y \notin W$, since there is no $x \in y \cap W$. Apply set foundation to conclude $\forall x (\phi(x))$. That is, $W$ is empty. This is a contradiction.
