09.Answer - Fpinscala

Let's rewrite the following in terms of flatMap:

scala

compose(compose(f, g), h) == compose(f, compose(g, h))

a => compose(f, g)(a).flatMap(h) == a => f(a).flatMap(compose(g, h))
a => ((b => flatMap(f(b))(g))(a)).flatMap(h) == a => f(a).flatMap(b => g(b).flatMap(h))

So far we have just expanded the definition of compose. Equals substituted for equals. Let's simplify the left side a little:

scala

a => f(a).flatMap(g).flatMap(h) == a => f(a).flatMap(b => g(b).flatMap(h))

Let's simplify again by eliminating the a argument and substituting a hypothetical value x for f(a):

scala

x.flatMap(g).flatMap(h) == x.flatMap(b => g(b).flatMap(h))

This now looks exactly like the monad law stated in terms of flatMap, just with different names:

scala

x.flatMap(f).flatMap(g) == x.flatMap(a => f(a).flatMap(g))

Q.E.D.