This gives you a type like `Inv[A] with Inv[B] with Inv[C]`, and you get a subset test "for free", as it's equivalent to taking the supertype.
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You probably want to throw another type into the mix as well (IDK, pick one - it doesn't matter as long as you're consistent and it's got no subtype relationship with `Inv`) so that you have a good representation the empty set.
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