TITLE: A question on rank-to-rank embeddings
QUESTION [6 upvotes]: Consider a non-trivial elementary embedding $j:V_\lambda\to V_\lambda$ and, for each $A\subset V_\lambda$, set $j(A)=\bigcup_{\delta<\lambda}j(A\cap V_\delta)$.
In Implications between strong large cardinals, Annals of Pure and Applied Logic 90 (1997) 79-90, Laver says that $j:(V_\lambda, \in,A)\to (V_\lambda, \in, j(A))$ is an elementary embedding.
I think that this is equivalent to $j$ being $\Sigma_0^1$-elementary, i.e., it satisfies the definition of elementary embedding for formulas with second order variables but without second order quantifiers.
Could someone provide a hint or a reference for this fact? I think I could prove it from a particular case, namely, that $\forall x\in V_\lambda\exists y\in V_\lambda\ (x,y)\in A$ implies $\forall x\in V_\lambda\exists y\in V_\lambda\ (x,y)\in j(A)$.
Does this case follow easily from $j$ being elementary? I cannot see it.
Edited:
I have found a proof in chapter 2 of these notes (theorem 0.4), but I think that the proof has a gap, since it considers skolem functions $f_i$ and assumes tacitly that $j(f_i)$ are also total functions, but they could be just partial functions.

REPLY [7 votes]: Here is the author of the notes you quoted. Thank you very much for pointing this out, there is indeed a gap in the proof: I truly took for granted that if $f$ is total then $j^+(f)$ is total (more specifically, that if $f$ is a Skolem function, then $j^+(f)$ is a Skolem function, but once one has totality, the rest is easy).
Anyway, I think this is true. In fact, I think the following is true:
Lemma If $f:V_\lambda\to V_\lambda$ is a partial function, then $j^+(dom(f))=dom(j^+(f))$.
This is not immediate: by definition, $j^+(dom(f))=\bigcup_{n\in\omega}j(dom(f)\cap V_{\kappa_n})$; on the other hand, $dom(j^+(f))=\bigcup_{n\in\omega}j(dom(f\cap V_{\kappa_n}))$, and the single summands are not equal, for example if $f$ is the critical sequence then $j(dom(f\cap V_{\kappa_n}))=\{0,\dots,n-2\}$, while $j(dom(f)\cap V_{\kappa_n})$ is always $\omega$. 
The following is an attempt for proving this. I don't see any errors (but then again, I didn't in my notes). Sorry for the unpleasant wall of equations, that is just how I did it. Maybe there is another, more synthetic way to prove it.
Proof of Lemma. The trick is to uniformize the division in pieces, i.e., to consider $dom(f\cap V_{\kappa_n})\cap V_{\kappa_m}$. Then
$j^+(dom(f))=\bigcup_{m\in\omega}j(dom(f)\cap V_{\kappa_m})=\bigcup_{m\in\omega}j(\bigcup_{n\in\omega}dom(f\cap V_{\kappa_n})\cap V_{\kappa_m})$
because the dominion of the union of functions is the union of the dominions of the single functions. But now $\langle dom(f\cap V_{\kappa_n})\cap V_{\kappa_m}:n\in\omega\rangle$ is an $\omega$-sequence bounded in $V_{\kappa_m}$, therefore it is in $V_\lambda$, so we can carry the $j$ inside
$\bigcup_{m\in\omega}j(\bigcup_{n\in\omega}dom(f\cap V_{\kappa_n})\cap V_{\kappa_m})=\bigcup_{m\in\omega}\bigcup_{n\in\omega}j(dom(f\cap V_{\kappa_n})\cap V_{\kappa_m})=$
$=\bigcup_{n\in\omega}\bigcup_{m\in\omega}j(dom(f\cap V_{\kappa_n})\cap V_{\kappa_m})$.
$dom(f\cap V_{\kappa_n})$ is an element of $V_\lambda$, therefore there exists an $m$ such that it is in $V_{\kappa_m}$, so that is just a finite union.
$\bigcup_{m\in\omega}\bigcup_{n\in\omega}j(dom(f\cap V_{\kappa_n})\cap V_{\kappa_m})=\bigcup_{n\in\omega}j(dom(f\cap V_{\kappa_n}))=j^+(dom(f))$.