## Not quite a monad
-To demonstrate that this is indeed the continuation monad's ¢
-operator:
+The unit and the combination rule work very much like we are used to
+from the various monads we've studied.
+
+In particular, we can easily see that the ¢ operator defined just
+above is exactly the same as the ¢ operator from the continuation monad:
¢ (\k.g[kf]) (\k.h[kx])
= (\MNk.M(\m.N(\n.k(mn)))) (\k.g[kf]) (\k.h[kx])
== ------
fx
-However, these continuations do not form an official monad. The
-reason is that (see Wadler's paper `Composable continuations' for details).
+However, these continuations do not form an official monad.
+One way to see this is to consider the type of the bind operator.
+The type of the bind operator in a genuine monad is
+
+ mbind : Mα -> (α -> Mβ) -> Mβ
+
+In particular, the result type of the second argument (Mβ) must be
+identical to the result type of the bind operator overall. For the
+continuation monad, this means that mbind has the following type:
+
+ ((α -> γ) -> ρ)
+ -> α -> ((β -> δ) -> ρ)
+ -> ((β -> δ) -> ρ)
+
+But the type of the bind operator in our generalized continuation
+system (call it "kbind") is
+
+ kbind :
+ ((α -> γ) -> ρ)
+ -> α -> ((β -> δ) -> γ)
+ -> ((β -> δ) -> ρ)
+
+Note that `(β -> δ) -> γ` is not the same as `(β -> δ) -> ρ`.
+
+ kbind u f = \k. u (\x. f x k )
+ β->δ (α->γ)->ρ α α->(β->δ)->γ α β->δ
+ -----------------
+ (β->δ)->γ
+ ------------------------
+ γ
+ -------------------
+ α->γ
+ ---------------------
+ ρ
+ --------------
+ (β->δ)->ρ
+
+See Wadler's paper `Composable continuations' for discussion.
+
+Neverthless, it's easy to see that the generalized continuation system
+obeys the monad laws. We haven't spent much time proving monad laws,
+so this seems like a worthy occasion on which to give some details.
+Since we're working with bind here, we'll use the version of the monad
+laws that are expressed in terms of bind:
+
+ Prove u >>= ⇧ == u:
+
+ u >>= (\ak.ka)
+ == (\ufk.u(\x.fxk)) u (\ak.ka)
+ ~~> \k.u(\x.(\ak.ka)xk)
+ ~~> \k.u(\x.kx)
+ ~~> \k.uk
+ ~~> u
+
+The last two steps are eta reductions.
+
+ Prove ⇧a >>= f == f a:
+
+ ⇧a >>= f
+ == (\ak.ka)a >>= f
+ ~~> (\k.ka) >>= f
+ == (\ufk.u(\x.fxk)) (\k.ka) f
+ ~~> \k.(\k.ka)(\x.fxk)
+ ~~> \k.fak
+ ~~> fa
+
+ Prove u >>= (\a.fa >>= g) == (u >>= f) >>= g:
+
+ u >>= (\a.fa >>= g)
+ == u >>= (\a.(\k.fa(\x.gxk)))
+ == (\ufk.u(\x.fxk)) u (\a.(\k.fa(\x.gxk)))
+ == \k.u(\x.(\a.(\k.fa(\x.gxk)))xk)
+ ~~> \k.u(\x.fx(\y.gyk))
+
+ (u >>= f) >>= g
+ == (\ufk.u(\x.fxk)) u f >>= g
+ ~~> (\k.u(\x.fxk)) >>= g
+ == (\ufk.u(\x.fxk)) (\k.u(\x.fxk)) g
+ ~~> \k.(\k.u(\x.fxk))(\y.gyk)
+ ~~> \k.u(\x.fx(\y.gyk))
+
+## Syntactic refinements: subtypes of implication
+
+Because natural language allows the functor to be on the left or on
+the right, we replace the type arrow -> with a left-leaning version \
+and a right-leaning version, as follows:
+
+ α/β β = α
+ β β\α = α
-Neverthless, obeys the monad laws.
+This means (without adding some fancy footwork, as in Charlow 2014) we
+need two versions of ¢ too, one for each direction for the unmonadized types.
-Oh, one more thing: because natural language allows the functor to be
-on the left or on the right, we replace the type arrow -> with a
-left-leaning version \ and a right-leaning version, as follows:
+\\ //
- α/β β = α
- β β\α = α
-This means we need two versions of ¢ too (see Barker and Shan 2014
-chapter 1 for full details).
This is (almost) all we need to get some significant linguistic work
done.