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Fair enough. On reflection, `\f xs -> []` is a better example of the point I was trying to make.



Yeah that's a better one (which I think is discussed (or a similar one) in Theorems for free!), but the wording of OP weasels itself out of that being a problem.

> any `b` in the result list _must_ have come from applying the function to some `a` in the input list.

Since there exists no a in [], the quoted statement holds! I find that really beautiful :)


That is indeed a weaselly but accurate statement. :)

I'll have to give Theorems for Free a read, thanks for the suggestion!




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