Context: self-study from Devlin's "The Joy of Sets" -- theorem 1.7.11.
Let $(X, \le)$ be a woset such that for each $a \in X$, $X_a$ is isomorphic to an ordinal. Then $X$ is isomorphic to an ordinal.
His proof goes as follows:
For each $a \in X$, let $g_a: X_a \cong Z(a)$ be an isomorphism of $X_a$ onto an ordinal $Z(a)$. By Theorems 1.7.10 and 1.7.3, both $Z(a)$ and $g_a$ are unique. Hence this defines a function $Z$ on $X$. Let $W$ be its range. That is. $$W = \lbrace Z(a) | a \in X \rbrace $$ Define $f: X \to W$ by $$f(a) = Z(a)$$
Claim: If $x, y \in X$, then $x < y \to Z(x) \subset Z(y)$.
Proof of claim: Let $x, y \in X$, $x < y$. Then
$$(1): \quad g_x: X_x \cong Z(x)$$
Also, since:
$X_x = \lbrace x \in X | x < z\rbrace$
$= \lbrace x \in X | z < y \land z < z\rbrace$
$= \lbrace z \in X_y | z < x\rbrace$
$= (X_y)_x$
we have: $$(2) \quad (g_y X_x) : X_x \cong (Z(y))_{g_y (x)} = F$$
Now, $Z(y)$ is an ordinal, so by Theorem 1.7.6, $F$ is an ordinal. But by $(1)$ and $(2)$, $Z(x) \cong F$. Hence by Theorem 1.7.10: $$(3) \quad Z(x) = (Z(y))_{g_y(x)}$$ Thus, in particular, $Z(x) \subset Z(y)$.
The claim is proved. $\blacksquare$
The proof continues, but the rest of it is straightforward, proving that $W$ is in fact an ordinal.
The question I have is: what does the notation mean on $(2)$?
Specifically: $$(2) \quad (g_y X_x) : X_x \cong (Z(y))_{g_y (x)}$$
the $(g_y X_x)$ symbol.
The suggestion is that it may be the restriction of $g_y$ to $X_x$, that is, $g_x$.
But if it meant that, Devlin would have written $g_y \restriction X_x$, would be not?
I confess I'm all at sea over this proof, I still haven't understood it fully.