CS 641 meeting -*- Outline -*-

* the syntactic algebra (4.7)

	need to define choice and composition syntax for A(.)

	Why is this a problem?
	A(.) is semantics, not syntax.

	Consider the following
		(c1 [] (c2 ; c3 [] c4 ; c5)) ; c6
	This isn't a part of A(.), even though c2;c3 and c4;c5 are,
	because it isn't flattened as required by A(.).

	Q: how could you translate the above into an element of A(.)?
		{c1;c6, c2;c3;c6, c4;c5;c6}

	so we want syntactic operators [] and ; that are given meaning by
		just such a translation.

	Q: What's the translation for []?
		Let I be nonempty
		([] i \in I :: q.i) =def= (\cup i \in I :: q.i)

	Q: And for ;?
		s \in q;r equiv (exists t \in q, u \in r :: s = t;u).
		(that last ; is composition as strings in A*.)

----------------------
   SEMANTICS OF CHOICE
   AND COMPOSITION

Let A([],;) be set of commands obtained
  from A(.) by using ; and [].

Let q, q.i \in A([],;)
Let r,s,t \in A(.)

meaning : A([],;) -> A(.)
meaning.r = r
meaning.([] i \in I :: q.i)
  = (\cup i \in I :: meaning.(q.i))
meaning.(q;r)
  = {s | t \in meaning.q, u \in meaning.r,
         and s = t;u}
-----------------------
	that last ; is composition as strings in A*.

	Hesselink doesn't use meaning or A([],;),
		he thinks of [] and ; as operators "on A(.)"
		and thus *identifies* the two examples above

	Use S(.) to denote the subset of A(.) that consists of nonempty sets
		of strings over S.  (as in section 2.6)

	Thm (distributivity): For nonempty sets C and D of A(.),
		  ([] c \in C :: c) ; ([] d \in D :: d)
	        = ([] c \in C, d \in D :: c;d)

	Note that this is equality instead of semantic equivalence.
		(compare chapter 3)

	Q: can you prove this?