Introduction
Type chasing in Haskell is about filling in the definition for a function given only its type signature.
In most cases, the actual definition of a parametrically polymorphic function is not only completely determined by the signature, but the types can actually drive the implementation toward the only reasonable, valid, solution (there may be a few valid but unreasonable solutions). All the components of the definition can be determined from the signature, and these will usually be the arguments declared on the righthand side of the equation. Presuming that no external (e.g. from Prelude) function is necessary for the implementation (which is most frequently the case), then you have all the necessary ingredients to write out the correct definition on the lefthand side of the equation. You just need to combine them somehow, using the reduced set of maneuvers: function application, pattern matching (occasionally).
An example
This is the exact signature for the function I kept failing to define correctly:
kapp :: (((a > b) > r) > r)  ContT r m (a > b)
> ((a > r) > r)  ContT r m a
> ((b > r) > r)  ContT r m b
The signature for kapp
is actually a standalone version of the Applicative's (<*>)
method, free of ContT
, as seen here.
 ContT monad transformer
newtype ContT r m a = ContT { runContT :: (a > m r) > m r }
 ContT's Applicative instance
instance Applicative (ContT r m) where
(<*>) :: ContT r m (a > b)
> ContT r m a
> ContT r m b
f <*> v = ContT $ \ c > runContT f $ \ g > runContT v (c . g)
Staring at the solution, i.e. at their definition of <*>
, I've found it particularly frustrating how they knew to introduce that lambda there (the one with the c
param), but to apply it way over there, near the end; and that other lambda (with the g
param) as well. Moreover, the composition, c . g
, implies the third lambda where that implicit parameter (x
) was declared, making that composition pointfull: c (g x)
.
And it was exactly these lambda introductions that have reminded me of the inference rule in propositional logic, the one for introducing an implication, prompting me to CurryHoward my way forward.
Logical derivation
The inference rule of implication introduction states that if you assume a p
and work your way to a q
, then you can conclude p => q
. Instead of coming up with an example for implication introduction, we better return to the problem at hand and demonstrate it there.
kapp :: (((a > b) > r) > r)  h
> ((a > r) > r)  k
> (b > r)  g
> r
kapp h k g =  how now brown cow?
Summary:
 we have 3 function arguments bound to parameters
h
,k
,g
h :: ((a > b) > r) > r
k :: (a > r) > r
g :: b > r
 we need to produce:

r
type

The 3 arguments are the 3 given propositions, and we need to come up with the conclusion r
. The form of this function is appropriate for conversion to the Fitchstyle proof derivation because everything needed to solve it is already present, i.e. (peeking at the solution) we know that no arbitrary (Prelude) function is involved, everything needed for a solution is available in the form of identifiers, we just need to rearrange them somehow. And since the types involved are function types, they translate directly into implications.
The Fitchstyle diagram:
┌──┬──────────────────────┬────────────────────────┐
│1 │ ((a > b) > r) > r │ proposition │
│2 │ (a > r) > r │ proposition │
│3 │ b > r │ proposition │
╞══╪══════════════════════╪════════════════════════╡
│4 │ ┌ (a > b)¹ │ assumption¹ │
│5 │ │ ┌ a² │ assumption² │
│6 │ │ │ b │ MP 4,5 │
│7 │ │ │ r │ MP 3,6 │
│8 │ │ a² > r │ >I 57 (discharging²) │
│9 │ │ r │ MP 2,8 │
│10│ (a > b)¹ > r │ >I 49 (discharging¹) │
│11│ r │ MP 1,10 │
└──┴──────────────────────┴────────────────────────┘
We need to come up with the conclusion r
. Lines 13 are given. So we start the solution on line 4.
 We assume
(a > b)
 We make another assumption
a
 Modus ponens: if we have
(a > b)
(4) and ana
(5), we can concludeb
 Modus ponens:
(b > r)
(3) andb
(6), we concluder
 Discharging assumption 2, we obtain
(a > r)
 Modus ponens:
(a > r) > r
(2) and(a > r)
(8), producer
 Discharging assumption 1, obtain
((a > b) > r)
 Modus ponens:
((a > b) > r) > r
(1) and above (10) giver
And that r
is the conclusion we sought. QED.
I was hoping this derivation form would reveal more so that the implementation of a function's definition would be compressed into a series of forced steps. But anyway, it seems as a helpful aid when you're stuck with an equation.
When doing a derivation, you can assume anything. Each assumption gets indented a level to the right as a reminder that you can only use the given propositions and the formulas from the same level as that assumption; also, it reminds that an assumption would have to be discharged.
When discharging a particular assumption, the form that is written on a line is always the same: first write that assumption, followed by an implication sign, followed by whatever formula was on the previous line (the line directly above the implication introduction). The justification given is called the implication introduction, abbreviated by >I
, along with the range indicating its scope.
Implication elimination, or >E
or MP
(modus ponens), is equivalent to function application: having an implication a > b
and an a
, you can conclude a b
.
Back to the function implementation
kapp :: (((a > b) > r) > r)  h
> ((a > r) > r)  k
> (b > r)  g
> r
kapp h k g = h (\f > k (\x > g (f x)))
 i.e.
kapp h k g = h \f > k \x > g $ f x
kapp h k g = h \f > k (g . f)
The derivation with code reference :
4 │ ┌(a > b)¹ assumption¹ . a > b :: f
5 │ │ ┌a² assumption² . . a :: x
6 │ │ │b MP 4,5 . . b :: f x
7 │ │ │r MP 3,6 . . r :: g (f x)
8 │ │ a² > r >I 57 (discharging²) . a > r :: c
9 │ │ r MP 2,8 . r :: k c
10│(a > b)¹ > r >I 49 (discharging¹) (a > b) > r :: d
11│r MP 1,10 r :: h d
It may not be crystal clear at first, but after some exercises, the steps behind many function implementations become easier to justify.
(hmm, what if the first assumption was a
?)
kapp :: (((a > b) > r) > r)
> ((a > r) > r)
> (b > r)
> r
kapp h k g = h $ \f > k $ \a > g $ f a
 vs
kapp h k g = k $ \a > h $ \f > g $ f a
Same signature, different derivation
┌──┬──────────────────────┬────────────────────────┐
│1 │ ((a > b) > r) > r │ proposition │
│2 │ (a > r) > r │ proposition │
│3 │ b > r │ proposition │
╞══╪══════════════════════╪════════════════════════╡
│4 │ ┌ a¹ │ assumption¹ │
│5 │ │ ┌ (a > b)² │ assumption² │
│6 │ │ │ b │ MP 5,4 │
│7 │ │ │ r │ MP 3,6 │
│8 │ │ (a > b)² > r │ >I 57 (discharging²) │
│9 │ │ r │ MP 1,8 │
│10│ a¹ > r │ >I 49 (discharging¹) │
│11│ r │ MP 2,10 │
└──┴──────────────────────┴────────────────────────┘
kapp h k g = k $ \a > h $ \f > g $ f a
Unlike the first one, this derivation matches the implementation in code, since a
is assumed first, while a > b
is assumed second, both, in the diagram and in the code... but something's gotta be wrong somewhere (!?)
Mapping continuations
Leaving the two ambiguous kapp
implementations above for now (lest interfere with the ongoing investigation), let's see the thing that should have been presented first  mapping a continuation by implementing the standalone function similar to fmap
but renamed to kmap
.
The signature of kmap
is more easily understood: the callback passed to a function, k
, is passed a function's wouldbereturned value, a
, along with a mapping function f
that will be applied to a
before passing it on to k
. So, a function's original return value is caught right before being passed to the continuation and mapped (and then passed to k
).
First version
Like before, the signature below can be simplified by removing the extra parenthesis on the right, so we get 3 proper arguments. It is actually equivalent whether we declare 3 parameters on the LHS of an equation, or we catch each argument with a lambda on the RHS, or any distribution in between. But for the first version of kmap
, all 3 arguments will have a corresponding proper, i.e. LHSdeclared parameter.
 v.1
kmap :: (a > b) > ((a > r) > r) > (b > r) > r
 just to make it more readable
kmap :: (a > b)  f
> ((a > r) > r)  k
> (b > r)  g
> r
kmap f k g = k $ \a > g (f a)
The implementation now follows smoothly from the corresponding Fitch diagram:
a > b, (a > r) > r, b > r ⊢ r
1 │ a > b proposition  f
2 │ (a > r) > r proposition  k
3 │ b > r proposition  g

4 │ ┌ a¹ assumption¹  \a > (
5 │ │ b MP 1,4  f a
6 │ │ r MP 3,5  g (f a)
7 │ (a¹ > r) =>I 46 (discharging¹)  )
8 │ r MP 2,7  k $ \a > g (f a)
Second version
Another version of kmap
, but this time the third argument doesn't have a corresponding parameter on the LHS, rather it is bound by introducing an extra lambda. This is often the case when dealing with the continuations (and other types as well) wrapped inside a newtype, because then the data constructor gets in the way, forcing the introduction of a lambda to handle it (usually the last, often the third, argument). Or maybe because you'd just prefer to infix the function you're defining, in which case 2 parameters on the LHS are just right (e.g. f 'kmap' k = \g > …
instead of kmap f k g = …
).
Now we have 2 premises instead of 3, and the consequent we're looking for is not just r
but (b > r) > r
. However, the entire effect of that is the introduction of another assumption, which, translated to the code, means introduction of an extra lambda (whose body will wrap the implementation from the first version), as shown below.
a > b, (a > r) > r ⊢ (b > r) > r
1 │ a > b proposition  f
2 │ (a > r) > r proposition  k

3 │ ┌ (b > r)¹ assumption¹  \g > (
4 │ │ ┌ a² assumption²  \a > (
5 │ │ │ b MP 1,4  f a
6 │ │ │ r MP 3,5  g (f a)
7 │ │ (a² > r) =>I 46 (discharging²)  )
8 │ │ r MP 2,7  k $ \a > g (f a)
9  (b > r)¹ > r =>I 38 (discharging¹)  )
The diagram now translates to the following implementation:
 v.2
kmap :: (a > b) > ((a > r) > r) > ((b > r) > r)
 making the dig more readable
kmap :: (a > b)  f
> ((a > r) > r)  k
> ((b > r) > r)
kmap f k = \g > (k $ \a > g (f a))
Chaining continuations
To complete the trio, the definition of the bind
like function for chaining continuations follows.
kbind :: ((a > r) > r) > (a > ((b > r) > r)) > ((b > r) > r)
 rearranged sig
kbind :: ((a > r) > r)
> (a > ((b > r) > r))
> ((b > r) > r)
 removing redundant parens
kbind :: ((a > r) > r)  k
> (a > (b > r) > r)  h
> b > r  g
> r
kbind k h g = k \a > (h a g)
This implementation follows from the diagram smoothly:
(a > r) > r, a > (b > r) > r, b > r ⊢ r
1  (a > r) > r proposition  k
2  a > (b > r) > r proposition  h
3  b > r proposition  g

4  ┌ a¹ assumption¹  \a > (
5  │ (b > r) > r MP 2,4  h a
6  │ r MP 5,3  (h a) g
7  a¹ > r =>I 46 (dis¹)  )
8  r MP 1,7  k (\a > h a g)
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