Comment by JadeNB

> I wasn't just claiming Turing completeness of Haskell. I was pointing out that every language construct, every subexpression in Haskell, directly represents a corresponding lambda term, with corresponding semantics (e.g. laziness).

I was referring to the Turing completeness of the lambda calculus, not of Haskell. But, again, I think that trying to work directly with lambda expressions everywhere, even if it is possible and, as you say, straightforward for "vanilla" Haskell, quickly shows why we put some semantic sugar over it. That is to say, it's certainly true that, in an obvious sense, the layer of semantic sugar is thinner for Haskell than for C, but it's still "just" semantic sugar, and still just as conceptually important, in both cases.