> AFAICS it is a silly bug in 'subset?'.  If we have Set(X) and
> X is ordered (like Integer) then w use one algorithm.  In
> unorderd case we use:
> 
>    subset?(s, t)   ==
>        #s < #t and every?((x : S) : Boolean +-> member?(x, t), parts s)
> 
> from FiniteSetAggregate.  Of course we should have '#s <= #t'.

Wow! That this bug stayed was undetected for sooo many years...

Will you commit a fix? I go to bed now.

Ralf

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