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Solution you can compile
Problem 1
untag : Tagged tag value -> value untag (Tagged value) = value
Problem 2
map : (oldValue -> newValue) -> Tagged tag oldValue -> Tagged tag newValue map f (Tagged value) = Tagged <| f value
Problem 3
map2 : (a -> b -> c) -> Tagged tag a -> Tagged tag b -> Tagged tag c map2 f (Tagged one) (Tagged two) = Tagged <| f one two
With three values, I thought a -> b -> c was as descriptive as longer identifiers and easier to skim.
a -> b -> c