Evaluation Contexts
-------------------

Motivation:  A concise notation for communicating operational rules.

The language with simplified operational semantics:

t ::= bool | t1 -> t2
e ::= x | v | if e1 then e2 else e3 | e1 e2
v ::= true | false | \x:t.e
E ::= [] | E e | v E | if E then e1 else e2

E is an evaluation context where E[e] means the context where [] is replaced
with e. 

Eval rules:

---------------------
(\x:t.e) v -> e[v/x]

-----------------------------   ------------------------------
if true then e1 else e2 -> e1	if false then e1 else e2 -> e2

   e -> e'
--------------  (E-context)
E[e] -> E[e']

The above definition of E eliminates the need to include a number of
"search rules" from the original semantics and hence makes the semantics
more compact. For example,

E e corresponds to

   e1 -> e1'
---------------
e1 e2 -> e1' e2

v E corresponds to

   e2 -> e2'
---------------
 v e2 -> v e2'

if E then e2 else e3 corresponds to

      e1 -> e1'
------------------------
if e1 then e2 else e3 ->
if e1' then e2 else e3

Suppose we have an expression:

if (\x:bool.x) ((\x:bool.x) true) then false else true

Let E1 = if (\x:bool.x) []  then false else true
and e1 = ((\x:bool.x) true)

We have according to (E-context):

   e1 -> true
----------------
 E[e1] -> E[true]

Let E2 = (\.x:bool.x) [], e2 = (\x:bool.x) true

Let E3 = if (\x:bool.x) [] then false else true

   e2 -> true
--------------------
 E3[e2] -> E3[true]

We define judgement e ->(toplevel) e', then

       e -> e'
------------------------
E[e] ->(toplevel) E[e']


Preservation Lemma:
-------------------
If |- e : t and e -> e', then |- e' : t

Proof: By induction on the derivation of e -> e'

Case:
     e -> e'
-----------------
  E[e] -> E[e']

Given that |- E[e] : t, 
must prove that |- E[e'] : t

How do we do that?  We some additional lemmas.


Typing Evaluation Contexts
--------------------------

We define a typing rule for eval context:

G, x : t1 |- E[x] : t2
-----------------------		(T-context)
  G |- E : t1 ==> t2

Lemma 1. If |- E[e] : t2 then there exists a type t1, such that
x: t1 |- E[x] : t2 and |- e : t1

Proof: By induction on the structure of E

Case: E = []

|- [][e] : t2	(given)
same as: |- e : t2
Let t1 = t2, must prove |- e : t1
True since t1 = t2.

x: t1 |- [][x] : t2
same as x : t1 |- x : t1 	(by variable rule and t1 = t2)

Case: E = E e
|- (E1 e1)[e] : t2
Same as |- (E1[e]) e1 : t2

(1) |- (E1[e]) : t_a -> t2	(by inversion of typing rule)
(2) |- e1 : t_a			(by inversion of typing rule)
(3) x: t1 |- e1 : t_a		(by weakening form lemma)
(4) x: t1 |- E1[x] : t_a -> t2	(by I.H.)
and |- e : t1			(by I.H.)
x: t1 | (E1[x]) e1 : t2		(by (3) and (4) and typing rule for application)

----------

Corollary 1. If |- E[e] : t2 then there exists a type t1, such that
|- E: t1 ==> t2 and |- e : t1

Proof: follows immediately from Lemma 1 and T-context rule.