CS 641 Lecture -*- Outline -*-

* The simply typed lambda calculus
	theory of application and substitution governed by types

** expressions
	ground type o, x in Variables
	---------------
	t ::= o | t -> t | (t)
	e ::= x | \x: t . e | e e | (e)
	---------------
		t is a type expression, e is a term expression (term)

*** 	parsing rules
	-------------
	\x:o. e e'  ==   (\x:o. (e e'))
	x y z       ==   (x y) z
	o -> o -> o ==   o -> (o -> o)
	-------------

***	free variables, Fv
	----------
	Fv(x) = {x}
	Fv(\x:t.e) = Fv(e) - {x}
	Fv(e e') = Fv(e) union Fv(e')
	----------

***	bound variables
	Def: variable is bound if it is not free
	bound variable names unimportant:
		treat raw terms as equivalence classes of terms under alpha-cnv
	------------
	(\x:t.x) == (\y:t.y)		syntactic identity
	------------

***	substitution
	------------
	[e/x]x         == e
	[e/x]y         == y		    if not(y == x)
	[e/x](e1 e2)   == ([e/x]e1)([e/x]e2)
	[e/x](\y:t.e') == \y:t . [e/x]e'    where not(y == x), not(y in Fv(e'))
	------------
	last restriction necessary to preserve lexical scoping
		e.g. [y/x](\y:o . x(y))  is not (\y:o . y(y))
			rather (\z:o. y(z))

** typing rules

***	type context
	a list x1:t1, ..., xn:tn
		where the xi are distinct
	usually written H,
		think of it as an environment

***	type judgements, has type relation
	"has type" is a ternary relation, written H |- e : t
		such that Fv(e) subset-of domain(H)
	the least relation satisfying
	-----------------
	[proj]		H, x:t, H' |- x : t
	[abs]		H, x:s |- e:t         ==>  H |- (\x:s.e) : s->t
	[appl]		H|-e:(s->t), H|-e':s  ==>  H |- (e e') : t
	-----------------

	by abuse of notation, identify the "has type" relation with
		the forms H |- x : t

***	type derivation
		demonstration of H |- e : t from the rules
		derivation tree: node is a judgement + rule name pair,
			branches (up) to other nodes on which the rule depends,
			branches of a node not ordered.

	exercise: show |- (\x:o \f:o->o . f(x)) : o -> (o->o) -> o
		how to proceed: write desired result at bottom of board
			look at form of expression in question, that gives rule

	only interested in terms e and env H such that H |- e:s for some s
		Def: e has a type (in H) if there is such a s

	exercise: show that if H |- e : s and H |- e : t, then s == t.
		  show that if a derivation of H |- e : t exists it is unique.

	HW exercise: if H |- e:s and x is not in domain(H), then H,x:t |- e:s.
		converse?

** equational rules, equational theory
	an equation is a 4-tuple H |- e = e' : s
		where terms e and e', a type env H, a type expr s
			such that H |- e : s  and H |- e' : s

	equations are typed,
		including H also tells what variables the equation ranges over.

	equational theory is generated by the following rules
	-------------------
	{refl}	H |- x = x : s
	{trans} H |- e = e' : s, H |- e' = e'' : s  ==>  H |- e = e'' : s
	{sym}   H |- e = e' : s  ==>  H |- e' = e : s

	        H |- e0 = e0' : s->t     H |- e1 = e1' : s
	{cong}	__________________________________________
			H |- e0(e1) = e0'(e1') : t

	{xi}	H,x:s |- e = e' : t  ==>  H |- (\x:s.e) = (\x:s.e') : s->t

	{beta}  H |- (\x:s.e)(e') = ([e'/x]e) : t 
	{eta}	H |- \x:s.e(x) = e : s->t, where x not in Fv(e)
	-------------------

	equivalence rules: [refl], [trans], [sym] used to prove that equality
		is an equivalence relation

	congruence rules: [cong], [xi] state that application and abstraction
		are congruences with respect to equality.

	basic axioms: [beta], [eta]

	a theory is a set of sequents (of form H |- e = e' : s).
		the above theory is consistent and decideable

	Def: the \-theory generated by a theory T
		is the least relation that includes T
		and satisfies the rules of the \-calculus

	HW exercise:
		show that equality is preserved under a renaming of free vars