Full Idea
Ante rem structuralism, eliminative structuralism formulated over a sufficiently large domain of abstract objects, and modal eliminative structuralism are all definitionally equivalent. Neither is to be ontologically preferred, but the first is clearer.
Gist of Idea
The main versions of structuralism are all definitionally equivalent
Source
Stewart Shapiro (Philosophy of Mathematics [1997], 7.5)
Book Reference
Shapiro,Stewart: 'Philosophy of Mathematics:structure and ontology' [OUP 1997], p.242
A Reaction
Since Shapiro's ontology is platonist, I would have thought there were pretty obvious grounds for making a choice between that and eliminativm, even if the grounds are intuitive rather than formal.