Open mwhipple opened 6 years ago
The current workaround would be a combination of including
(which amounts to a subset test) and length which taken together would prove set equality though the pigeonhole principle.
The specifics around doing this in a decently expressive way are TBD.
An additional consideration would be that there is the distinction between comparing sets and comparing unordered sequences in that the latter could include duplicated values...currently that only inspires me to procrastinate on this ticket.
currently equality will also test ordering, an additional assertion should exist to do comparison of the values without ordering. Maybe:
an equivalent set of