Two hash models is functionally equivalent to a particular type of one double-sized hash model. So it shouldn't be any harder to recompute against a 2nd hash, i

Two hash models is functionally equivalent to a particular type of one double-sized hash model. So it shouldn't be any harder to recompute against a 2nd hash, if that 2nd hash were public. Of course, it won't be public (and if it ever became public they'd replace it with a different secret hash).