im like poly, i'm in prog

inprog/idiots-guide-lambda.md

@@ -21,4 +21,64 @@ what you would need is....
 
 ## a way to describe computability
 
+and one way of doing that is with lambda calculus. to define lambda calculus we need just a few rules - variables, abstraction, application, reduction.
 
+these concepts aren't too difficult, so awayyyy we go.
+
+### variables
+
+variables are simply a name, most of the time we use letters. there are two catches here, if you're used to variables in maths or other languages:
+
+* unlike pretty much every other language, lambda calc variables don't have a value, only a name. this is because we don't have any concept of what a value is.
+
+* unlike any language worth using, variables in lambda calculus do not have a *type*. again this is because we don't have a concept of what type is.
+
+this guide is gonna use lowercase letters for the most part, so the examples are `a b c d e f` etc etc
+
+### abstraction
+
+abstractions are the functions of lambda calc, and take the form off:
+
+```
+(λx.M)
+```
+
+where `λ` denotes that this is a function  
+`x` is an argument name (that is local to this function)  
+and `M` is a list of variables that will form the function body
+
+a quick example of an abstraction is the identity function:
+
+```
+(λx.x)
+```
+
+we'll get onto more to do with fuctions in a bit, but next
+
+### application
+
+application is what we pretentious people call "actually using a function" and it takes the form of:
+
+```
+(M N)
+```
+
+where `M` is an abstraction, and `N` is any lambda term at all. i will refer to `M` as an abstraction and `N` as the application body
+
+to do the application we just use `N` as the argument for `M`, and then do a find and replace based on the abstraction definition.
+
+for example, lets do an easy one using the identity function, which is a function that returns its input unchanged:
+
+if we have the abstraction `(λx.x)` and the variable `y` to use for the application, and then lay the application out as so:
+
+```
+( (λx.x) y )
+```
+
+we can then begin applying.
+
+the identity function has `x` as its argument, so we take the application body, in this case `y`, and every time we see `x` in the abstraction body (the bit past `.`) we replace it with our argument, `y`.
+
+so our abstraction body is `x`, and `x` is equal to our argument `y`, so therefore our function will return `y`, which is the identity.
+
+a sligtly more interesting example is something like