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Clarify proof of 2nd Functor law
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quchen committed Feb 27, 2016
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Expand Up @@ -57,7 +57,8 @@ Similarly, you can generate the free theorem for `fmap`, which reads
In other words, this says that whenever functions compose, fmapping all of them
still composes.

Now choose `p = id` and `q = f . g` in (2),
Now choose `p = id` and `q = f . g`. (1) clearly holds in this case, so we can
derive

```haskell
fmap f . fmap g
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