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If we use $H_1(X) = H_0(X) \oplus firstnbits(X)$, this would seem to be trivial. EDIT: As C├ędric Van Rompay pointed out, this is only a counterexample if $H_1$ winds up being preimage-resistant. This may not be a necessary consequence of $H_0$ being preimage-resistant, but I really only need one case where it is.

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