CS 641 Lecture -*- Outline -*-

* Boolean expressions (8.1)

  def: a Boolean expressions is a predicate,
       i.e., an expression of type S -> Bool

** pointwise extension

  Usually use the pointwise expended notation

  def: base term [[e]] associated with e is defined as:

       [[x: S -> G]] == x : G  , for a state attribute of type G
       [[y]]         == y      , for a logical variable (e.g. quantified)
       [[e1.e2]]     == [[e1]].[[e2]]
       [[(\forall i \in I :: b)]] == (\forall i \in I :: [[b]])
       [[(\exists i \in I :: b)]] == (\exists i \in I :: [[b]])

  Lemma 8.1. Let x1,...,xm be the attributes of an expression e.  Then
        e.s == [[e]][x1,...,xm := x1.s, ..., xm.s]

** reasoning about Boolean expressions (8.2)

   We want to be able to reduce reasoning about expressions to
   reasoning about the corresponding Boolean terms.

   Corollary 8.2 (reduction). Let x1,...,xm be the attributes of a Boolean
   expression b.  Then 

        e.s == (\exists x1,...,xm :: x1 == x1.s /\ ... /\ xm == xm.s /\ [[b]])

   This allows proofs to focus on the corresponding terms, avoiding
   extensionality. 

   Example: consider proving that x <= y \subseteq x - 10 < y

   Lemma 8.3 (alternative reduction). Let b and c be Boolean
   expressions, and xet x == x1,...,xm be the attributes that occur in
   these.  Then

    (a) |- (\forall x1...xm :: [[b]] ==> [[c]]) ==> b \subseteq c
    (b) var x1...xm |- (forall x1 ...xm :: [[b]] ==> [[c]]) <==> b \subseteq c

   Example: consider again proving that x <= y \subseteq x - 10 < y