# Multiple functions applied to a single value

Graham Klyne GK at ninebynine.org
Mon Dec 8 16:33:53 EST 2003

```A little while ago, I sent a long rambling message  containing a key
question in response to an earlier helpful response.  I guess my ramblings
were quite reasonably passed over as too much waffle, and the question was
lost.  Here, I isolate the question.

From the earlier response, I understand that:

((->) r)

is an instance of Monad, but I haven't found a definition of the Monad
instance functions.  I think these do the job, and they satisfy the monad
laws :

[[
return a  =  const a
g1 >>= f  =  \e -> f (g1 e) e
]]

Can anyone confirm, or point me at an actual definition?

#g
--

 Checking the Monad laws for the above definitions

(A)
return a >>= k
= [g1/const a][f/k] \e -> f (g1 e) e
= \e -> k (const a e) e
= \e -> k a e
= k a    -- as required.

(B)
m >>= return
= [g1/m][f/const]  \e -> f (g1 e) e
= \e -> const (m e) e
= \e -> (m e)
= m      -- as required.

(C)
m >>= (\x -> k x >>= h)
= [g1/m][f/(\x -> k x >>= h)] \e -> f (g1 e) e
= \e -> (\x -> k x >>= h) (m e) e
= \e -> (k (m e) >>= h) e
= \e -> ([g1/k (m e)][f/h] \e1 -> f (g1 e1) e1) e
= \e -> (\e1 -> h (k (m e) e1) e1) e
= \e -> h (k (m e) e) e

(m >>= k) >>= h
= ([g1/m][f/k] \e -> f (g1 e) e) >>= h
= (\e -> k (m e) e) >>= h
= [g1/(\e -> k (m e) e)][f/h] \e1 -> f (g1 e1) e1
= \e1 -> h ((\e -> k (m e) e) e1) e1
= \e1 -> h (k (m e1) e1) e1
= \e -> h (k (m e) e) e

So both expressions are equivalent, as required.

------------
Graham Klyne
For email:
http://www.ninebynine.org/#Contact

```