The lambda operator creates an anonymous function.
λx • x × x
Apply a function to its argument by setting them next to each other.
(λx • x × x) 2
variable | ( MN ) | (λ variable • M )
Bound and free: Bound variables are parameters; free are what's left.
Substitution M[x ← N].
Generally, just substitute N for all
free occurrences of x in M.
First, though, if N's un-bound variables are bound in M
(this is, if they are used as parameters), rename them to remove
conflicts, so the substitution leaves the free variables of
N free. (Obviously, parameter names are indifferent.)
(zx)[x ← y] = (zy)
(zx)[z ← (λz • x + z)] = ((λz • x + z)x)
(λx • x × x)[x ← (λz • x + z)] = (λx • x × x)
((λx • M) N) ⇒ M[x ← N]
Values are always functions, since there's nothing else.
Applied lamdba calculus adds number and such.
Lisp variations langauges approximate an applied lamda calculus.
Notice that define is really a kind of assignment. No one has
figured out how to dispense with the impurity.