CS 641 Lecture -*- Outline -*-

* Conditional Expressions (8.3)
** conditionals

------------------------------------------
                CONDITIONALS
             ^
cond.t.t1.t2 = 
  (\choose x :: (t ==> x == t1)
                /\ (!t ==> x == t2))
  * assuming x is not free in t, t1 or t2

Sugar:                   ^
 if t then t1 else t2 fi = cond.t.t1.t2

Inference Rules:

  Phi |- t
  ___________________________(if true rule)
  Phi |- if t then t1 else t2 fi == t1

  Phi |- !t
  __________________________(if false rule)
  Phi |- if t then t1 else t2 fi == t2

------------------------------------------
        Q: what does the definition of cond mean?

        Proving the inference rules is exercise 8.2.

  Proof of (if true rule): Suppose Phi |- t. Suppose x is a variable
  that is not free in t, t1, and t2.  We calculate as follows.

  Phi
  |- if t then t1 else t2 fi
  ==  { definition of if }
     cond.t.t1.t2
  ==  { definition of cond }
     (\choose x :: ("t" ==> x == t1) /\ (!"t" ==> x == t2))
  ==  { by asumption t <==> T, twice }
     (\choose x :: "(T ==> x == t1) /\ (!T ==> x == t2)")
  ==  { predicate calculus }
     (\choose x :: x == t1)
  ==  { by "one point rule for \choose", x is not free in t1 }
     t1
  QED

  The one point rule for \choose is proved in
           ../predicates-and-sets/selection-individuals.txt

  Q: can you fill in the details of the predicate calculus step?

  The proof of the (if false rule) is similar.  

  Q: How would you  prove the other inference rule:
        _____________________________________(if same rule)
        Phi |- if t then t' else t' fi == t'
     from the previous two rules?

------------------------------------------
          FOCUSING RULES

Let ~ be a reflexive relation.

Phi, t |- t1 ~ t1'
___________________________(if focus true)
Phi |- if t then t1 else t2 fi
       ~ if t then t1' else t2 fi

Phi, !t |- t2 ~ t2'
__________________________(if focus false)
Phi |- if t then t1 else t2 fi
       ~ if t then t1 else t2' fi
------------------------------------------

        Q: how would you prove these?
        Q: Why must ~ be reflexive?

** conditional expressions and state transformers
   these are just the pointwise extension

   Def: let b be a predicate s and let f: S -> G, g: S -> G 
   be two state functions.  Then the *conditional state function*
                             ^
   (if b then f else g fi).s = if b.s then f.s else g.s fi.

   This is a *conditional expression* if f and g are expresions.
   This is a *conditional state transformer* if f and g are state transformers.

   look at the theorems on page 145-146

** proving properties about conditional state transformers
------------------------------------------
        ADDING CONDITIONALS
     TO STRAIGHT-LINE PROGRAMS

  f ::= id | x := e | f1; f2
      | if b then f1 else f2 fi

where
 x is a list of distinct state attributes,
 e is a corresponding expression list,
 b is a Boolean expression     
------------------------------------------

        Q: how would you prove that
             var x : Nat |- (x := x + 2); (x := x - 2) == id  ?

        note the use of functionality of assignment
             e == e' ==> (x := e) == (x := e').
        we need a similar kind of functionality for functions over states

        Def: A function f: S -> G is pointwise functional iff
             (\forall s : e.s == e'.s : f.e.s == f.e'.s) .
        That is, iff (e == e') \subseteq (f.e == f.e').

        Q: are pointwise extended functions pointwise functional?
        Q: are expressions in general pointwise functional?
           yes, lemma 8.6.

        Sequential composition is pointwise functional in its
        first argument, but not the second
              (f == f') \subseteq (f;g == f';g)
        however, while
              (g == g').(f.s) ==> (f;g == f;g').s    { by Liebnitz }
        we don't have the converse, because, for example,
            (y := 2; x:= 3 == y:= 2; x := y+1).s
            but x := 3 != x := y + 1

** Example proof
   Q: can you prove that
       var x, y |- (x := x max y) == if x <= y then x := y else id fi ?

   Q: what does "generalization" mean in the middle (big) step of the
   proof on p. 148?