# Lambda calculus exercises

## Slide 12

### Round 1

Using square bracket notation for redexes:

   [(\x.x) z]
-> z
   
   (\xy.x) z
=  [(\x.(\y.x)) z]
-> \y.z
   
   (\xy.x) z u
=  ([(\x.(\y.x)) z]) u
-> [(\y.z) u]
-> z
   
Using underline notation for redexes:

   (\x.x) z
   --------
-> z

   (\xy.x) z
=  (\x.(\y.x)) z
   -------------
-> \y.z

   (\xy.x) z u
=  ((\x.(\y.x)) z) u
    -------------
-> (\y.z) u
   --------
-> z


### Round 2

   (\x.x x) (\y.y)
   ---------------
-> (\y.y) (\y.y)
   -------------
-> (\y.y)

   \x.(\y.y) z
      --------
-> \x.z

   (\x.(x (\y.x))) z
   -----------------
-> z (\y.z)


## Examples of free and bound variables

(\x.x y) z

Free: y and z
Bound: x

(\x.x y) x
Free: x and y
Bound: x (this is a different x!)

(\x.x y) z
Within the body of the abstraction (x y):
Free: x and y


## Slide 25

Applicative-order reduction of Example 1:

   (λx.x x) ((λxy.y x) z (λx.x))
 = (\x.x x) (((\x.(\y.y x)) z) (\x.x))
              ---------------
-> (\x.x x) ((\y.y z) (\x.x))
             ---------------
-> (\x.x x) ((\x.x) z)
             --------
-> (\x.x x) z
   ----------
-> z z

Normal-order reduction of Example 1:

   (\x.x x) (((\x.(\y.y x)) z) (\x.x))
   -----------------------------------
-> (((\x.(\y.y x)) z) (\x.x)) (((\x.(\y.y x)) z) (\x.x))
...
-> z z


Applicative-order reduction of Example 2:
   
   (λxyz.x z) (λz.z) ((λy.y) (λz.z)) x
 = (((\x.\y.\z.x z) (\z.z)) ((\y.y) (\z.z))) x
     ---------------------
-> (((\y.\z'.(\z.z) z')) ((\y.y) (\z.z))) x
             ---------
-> ((\y.\z'.z') ((\y.y) (\z.z))) x
                 -------------
-> ((\y.\z'.z') (\z.z)) x
    ------------------
-> (\z'.z') x
   ----------
-> x


## Church encoding practice

   succ two
 = (\nfx.f (n f x)) (\fx.f (f x))
   ------------------------------
-> \fx.f ((\fx.f (f x)) f x)
          ---------------
-> \fx.f ((\x.f (f x)) x)
          --------------
-> \fx.f (f (f x))