Throughout the post $\mathcal{L}_\text{ZF}$ is the language of $\text{ZF}$, while $\text{fld}(R)$ is the field of the relation $R$, and $\text{wf}(R)$ means the relation $R$ is well-founded i.e.
- $\forall x\in\text{fld}(R)\exists y\left(x\in y\land R^{-1}[y]\subseteq y\right)$, and
- $\forall x\subseteq\text{fld}\left(x\neq 0\implies \exists y\in x\forall z\in x\left(\sim z R y\right)\right).$
I do not understand the meaning of Remark 1.3.3 c), which is an application of Theorem 1.3.2, in Holz's Introduction to Cardinal Arithmetic:
Theorem 1.3.2 (Transfinite Induction Principle) If $R$ is well-founded and if $\phi$ is a formula of $\mathcal{L}_\text{ZF}$, then $$\text{ZF} \vdash \text{wf}(R)\land \forall x\in\text{fld}(R)\Big(\big[\forall y\in\widehat{x}\ \phi(y)\big]\implies \phi(x)\Big) \implies \forall x\in\text{fld}(R)\ \phi(x).$$
Remark 1.3.3 Important special cases of Theorem 1.3.2 are: $$[\ldots]$$ $$ \text{c) $R = \in$: From the axiom of regularity and Lemma $1.3.8$ (see below) we can conclude that the relation $\in$ is well-founded. Thus we can use in ZF the so-called }\in\textbf{-Induction Principle:} $$ $$ \text{ZF}\vdash\forall x \Big(\big[\forall y < x\ \phi(y)\big]\implies \phi(x)\Big) \implies \forall x\ \phi(x). $$
As the $\in$-Induction Principle is a special case of Theorem $1.3.2$ we have that $R$ must be a relation, and $\in$ by itself is not (right?). Some options are $R=\in\cap(\text{ON}\times\text{ON})$ or $R=\in\cap(A\times A)$ for some arbitrary set $A$. If it is the latter, the principle could be restated as
Result: given any set $A$, and any formula $\phi(x)$, we have that $$ \text{ZF}\vdash\forall x\in A \Big(\big[\forall y\in A \text{ such that } y \in x\ \phi(y)\big]\implies \phi(x)\Big) \implies \forall x\in A\phi(x). $$
So, What exactly is the statement of the $\in$-Induction Principle? What is $R$?
$\in$-induction is written exactly as stated. However, you are correct that it is not literally an instance of TIP since $\in$ is "too large" to be a relation. There are two ways to think about the relation between TIP and $\in$-induction:
Sweep details under the rug and "schematize" TIP. For each pair of formulas $\varphi, \psi$, $\mathsf{ZFC}$ proves "If $\varphi$ defines a well-founded class-relation and $\psi$ defines a subclass of the field of $\varphi$ then $\psi$ has a $\varphi$-minimal element." This is a "theorem scheme," and taking $\varphi(x,y)\equiv x\in y$ yields TIP.
Don't treat it as a literal instance. While $\in$ itself is not a relation, $\in\upharpoonright V_\alpha$ is for every ordinal $\alpha$. And thinking about how TIP applies to the $\in\upharpoonright V_\alpha$s will give a quick proof of $\in$-induction.
Generally this sort of detail is often ignored in natural-language expositions, but you are right that - technically - $\in$-induction is not a genuine instance of TIP as written.