CS 641 meeting -*- Outline -*-

* syntax for commands with unbounded choice

	a necessary step in the definition of wp and wlp for recursive procs

	the point is to define the syntax of commands so can do induction

** layers
	this will be done in 3 layers
-------------------------
      LAYERS OF SYNTAX

Let S be a set of simple commands.
Let H be a set of procedure names.

                             examples
       (.)            
     A     (choices)   ([] i \in I :: c.i)
    ________________
      *
     A  (compositions)     x:=x-1; h
    ______________
     A = S \cup H          x:=x-1  ,  h
----------------------------------

** definitions
----------------------------------
     SYNTAX SETS (LANGUAGES)

Let A be a set of symbols not containing
\epsilon or `;'.

      *
def: A  = set of strings over A
    with \epsilon the empty string,
    and concatenation written with `;'.

      (.)             *
def: A    = PowerSet(A ) \ {}.
    elements of this set are *commands*.
----------------------------------
	Can regard A as subset of A* and A* as subset of A(.).
	An element s \in A* is sequential composition of its terms.
	An element r \in A(.) is choice of its element strings.
		Q: so why is A(.) non-empty?

---------------------------
      LANGUAGE EXTENSIONS

def: Let v: A -> PT.
The *string extension* of v,
 v*: A* -> PT
is defined inductively as
   v* .\epsilon = identity
   v*.(a;s) = v.a o v*.s

def: Let v: A -> PT.
The *language extension* of v,
 v(.): A(.) -> PT
is defined as
   v(.).r = (inf s \in r :: v*.s)

Lemma: for any predicate p \in |P,
   v(.).r.p = (all s \in r :: v*.s.p)
---------------------------

	Lemma: (a) Let V be an inf-closed subset of PT that contains
		   the identity function and is closed under composition.
		   Then v(.) : A(.) -> V for every v : A -> V.
	       (b) The subsets MT, MC, MP, MU of PT satisfy the condition
			in part (a).
	Q: how would you prove this?