Answer to this Haskell expression [closed]

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It works like a foreach loop, so

foreach (x in [1..3]) {
  foreach (y in [x .. x * 2]) {
    yield y;
  }
}

First x is 1, so y in [1 .. 2]

Then x is 2, so y in [2 .. 4]

Then x is 3, so y in [3 .. 6]

Concatenate these results together and you get the final result.

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