
Let me show you an example to prove it.
The example is limited to composition of unary functions defined on int
u::Int->Int
v::Int->Int
o ::(Int->Int)->(Int->Int)->(Int->Int)
o u v= \x->u(v(x))
#include
Let me show you an example to prove it.
The example is limited to composition of unary functions defined on int
u::Int->Int
v::Int->Int
o ::(Int->Int)->(Int->Int)->(Int->Int)
o u v= \x->u(v(x))
#include