Let bigset(X) and rel(X,Y) be otherwise arbitrary formulas in the language of second-order arithmetic with the indicated variables free, and thmemberof(Z,x,X) be the formula asserting that X is the xth member of the sequence of sets coded by Z. Does it follow that second-order arithmetic proves


$((\exists X)(bigset(X)) \; \land \; (\forall X)(bigset(X) \implies (\exists Y)(bigset(Y) \; \land \; rel(X,Y)))) \implies$  

$(\exists Z)(\forall x)(\exists X)(\exists Y)($  

$bigset(X) \; \land \; bigset(Y) \; \land \; thmemberof(Z,x,X) \; \land \; thmemberof(Z,\operatorname{S}(x),Y) \; \land \; rel(X,Y))$


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By second-order arithmetic, I mean Robinson arithmetic + full comprehension + the second-order induction axiom.



Does second-order arithmetic prove every expressible instance of Dependent Choice?
Carl has pointed out that my previous answer missed a clause in the theorem I cited.


Simpson's book, Subsystems of Second Order Arithmetic, does address this in section VII.6. He shows that dependent choice for $\Sigma^1\_2$ formulas is equivalent to $\Delta^1\_2$ comprehension plus $\Sigma^1\_2$ induction (Theorem VII.6.9).


However even regular (non-dependent) $\Sigma^1\_3$ choice is independent of full comprehension; he attributes this result to Feferman and Levy, and cites Theorem 8 of Levy's "Definability in axiomatic set theory, II". 


The result I mentioned before, that $\Sigma^1\_k$ dependent choice is equivalent to $\Delta^1\_k$ comprehension plus $\Sigma^1\_k$ induction for $k\geq 2$, holds for $k\geq 3$ requires the additional assumption that the universe is constructible from from some set of integers.


Your statement of dependent choice is a bit more complicated than necessary; you can fold bigset into rel (take $rel'(X,Y)$ to hold if either $\neg bigset(X)$ and $bigset(Y)$, or if $bigset(X)$, $bigset(Y)$, and $rel(X,Y)$). Conversely, it's a bit simpler than the version Simpson uses (which I believe is standard), in which $rel$ can depend on the parameter $x$ as well.


