Oh I totally misunderstood it.
isOn should be true if there's an odd number of trues, right? (Conversely, true if there is not an even number of trues)
Couldn't that be expressed as just a counting operation?
isOn = !switches
.lazy
.filter(\.self)
.count
.isMultiple(of: 2)
This is where a count(where:) would come in really useful, to remove the need for .lazy.filter(). An isOdd and isEven function would be cool too. If you add extensions for those, you get this down to:
isOn = switches.count(where: \.self).isOdd