From @titsuki

See the following results.

$ perl6 -e 'use Test; my Set (%a, %b); %a{"abc"} = %a{"abc"} (|) set(0,1,2); %a{"ab"} = %a{"ab"} (|) set(0); %b{"abc"}
= %b{"abc"} (|) set(1,0,2); %b{"ab"} = %b{"ab"} (|) set(0); is %a, %b;'
not ok 1 -

# Failed test at -e line 1
# expected​: 'ab 0
# abc 1 0 2'
# got​: 'ab 0
# abc 0 1 2'

If "is-deeply" subroutine is used, it should return "not ok".
But in this case, "is" subroutine is used. So it should return "ok".

From @zoffixznet

The `is` sub uses `eq` operator for comparison, which coerces items to Str.

So this ticket comes down to whether or not .Str on sets should sort their elements. The problem with that is sets can contain different types of objects. You could further put the argument that .Strs (or .gists) of those objects should be used and sorted, but that still leaves the issue of two differing sets producing the same .Str result​:

$ 6 'class Bar { method gist { "Foo" } }; say set Bar; say set "Foo"'
set(Foo)
set(Foo)

Thus, I'm closing this ticket, as it appears `is` sub is simply not an appropriate way to compare two sets.

The RT System itself - Status changed from 'new' to 'open'

@zoffixznet - Status changed from 'open' to 'rejected'