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enhance the proof section

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Mahler, Thomas 2020-04-17 16:37:03 +02:00
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functor-proof.md
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@ -1,8 +1,13 @@
# Proof of functor laws for Maybe
In this section I want to give a short example of how equational reasoning can be used to
proof certain properties of a given piece of code in Haskell.
**Facts:**
This is the `Functor` instance declaration of the type `Maybe`:
So without further ado let's begin:
## Known facts
The `Functor` instance declaration of the type `Maybe` is defined as:
```haskell
instance Functor Maybe where
@ -10,20 +15,21 @@ instance Functor Maybe where
fmap f (Just a) = Just (f a) -- (2)
```
This is the declaration of the composition operator `(.)`:
The composition operator `(.)` is defined as:
```haskell
(.) :: (b -> c) -> (a -> b) -> a -> c
f . g x = f (g x) -- (3)
f . g x = f (g x) -- (3)
```
This is the definition of the Identity function `id`:
The Identity function `id` is defined as:
```haskell
id :: a -> a
id x = x -- (4)
id x = x -- (4)
```
## Claim
Now we are asked to proof that `Maybe` fulfils the functor laws:
The claim is that `Maybe` fulfils the two functor laws:
```haskell
1.: fmap id = id
@ -65,7 +71,7 @@ Therefore, `fmap id m = id m` in all cases.∎
```haskell
fmap (f . g) m = fmap (f . g) Nothing -- by expansion of m
= Nothing -- by applying equation (1)
(fmap f . fmap g) m = fmap f (fmap g Nothing) -- by expansion of (.) and m
(fmap f . fmap g) m = fmap f (fmap g Nothing) -- by applying equation (4) and expanding m
= fmap f Nothing -- by applying equation (1)
= Nothing -- by applying equation (1)
```
@ -75,9 +81,15 @@ fmap (f . g) m = fmap (f . g) Nothing -- by expansion of m
```haskell
fmap (f . g) m = fmap (f . g) (Just a) -- by expansion of m
= Just ((f . g) a) -- by applying equation (2)
(fmap f . fmap g) m = fmap f (fmap g (Just a)) -- by expansion of (.), m
(fmap f . fmap g) m = fmap f (fmap g (Just a)) -- by applying equation (4) and expanding m
= fmap f (Just (g a)) -- by applying equation (2)
= Just (f (g a) -- by applying equation (2)
= Just ((f . g) a) -- by applying equation (3)
```
Therefore, `fmap (f . g) m = (fmap f . fmap g) m` in all cases. ∎
## Conclusion
You'll see this kind of reasoning quite a lot in Haskell documentation and online discussions.
The simple reason is: if you can prove something you don't have to test it.