CS 641 Lecture -*- Outline -*-

* inference rules for complete lattices (7.3)

  Now that we have sets, we can talk about meeting joins over
  arbitrary sets of elements.  That is, we can talk about complete
  lattices.

  Q: are the Booleans a complete lattice?  The predicates?
     Yes, and yes

** general meet and join
------------------------------------------
         ASCII NOTATION

  ``<='' means the partial order
              (latex's \sqsubseteq)
  ``>='' means the reverse partial order
              (latex's \sqsupseteq)

  ``\bot'' means the least element
  ``\bot'' means greatest element

  ``==>'' means logical implication

  ``(\glb i \in I :: t.i)''
       means meet (greatest lower bound)
       over the set { t.i | i \in I }

  ``(\lub i \in I :: t.i)''
       means join (least upper bound)
       over the set { t.i | i \in I }

                             ^
  (\glb i \in I : b.i : t.i) =
      (\glb i \in { i \in I | b.i} :: t.i)

                             ^
  (\lub i \in I : b.i : t.i) =
      (\lub i \in { i \in I | b.i} :: t.i)
------------------------------------------

        Can think of (\glb i \in I :: t.i)
                  as (\glb i : i \in I : t.i)
        and (\glb i \in I : b.i : t.i)
         as (\glb i : i \in I /\ b.i : t.i)

        But the chosen notation is closer to the book...

        Q: What do these mean for the booleans?
        \glb is \forall, and \lub is \exists

*** inference rules
------------------------------------------
             INFERENCE RULES

general \meet introduction:

     Phi, i \in I |- s <= t
 _________________________________
 Phi |- s <= (\glb i \in I :: t)
 * i not free in s, Phi, or I


general \meet elimination:

 _____________________________________
 t' \in I
   |- (\glb i \in I :: t) <= t[i:= t']
 * t' is free for i in t


general \join introduction:

 _______________________________________
 t' \in I
   |- t[i:= t'] <= (\lub i \in I :: t) 
 * t' is free for i in t


general \join elimination:

     Phi, i \in I |- t <= s
 _________________________________
 Phi |- (\lub i \in I :: t) <= s
 * i not free in s, Phi, or I

------------------------------------------
        
        Q: What do these mean?
        Q: Why the side conditions?

        Q: How can you get rid of the assumptions that t' \in I in the
        middle two rules?
               use the discharge rule

        Q: Is alpha conversion valid for these?

        Q: Can you formulate introduction and elimination
        rules for the bounded forms of the index meet and join?

        I think...

        general bounded \meet introduction:

             Phi, i \in I, b |- s <= t
         __________________________________
         Phi |- s <= (\glb i \in I : b : t)
         * i not free in s, Phi, or I

        general \meet elimination:

         __________________________________________
         t' \in I, b[i := t']
            |- (\glb i \in I : b : t) <= t[i:= t']
         * t' is free for i in t


        general \join introduction:

                
         __________________________________________
         t' \in I, b[i := t']
             |- t[i:= t'] <= (\lub i \in I : b : t) 
         * t' is free for i in t

        general \join elimination:

             Phi, i \in I, b |- t <= s
         _________________________________
         Phi |- (\lub i \in I : b: t) <= s
         * i not free in s, Phi, or I

*** index range homomorphisms

------------------------------------------
        INDEX RANGE HOMOMORPHISMS

Motivation: 

  (\glb i \in I :: a.i)
==   { notation }
  \glb { a.i | i \in I }
==   { definition of image }
  \glb (im.a.I)
==   { beta conversion }
  (\ I . \glb (im.a.I)).I

So as for fixed a: S -> G,
where and G is a complete lattice
  (\glb i \in I :: a.i)
is the following function (of I):
  (\ I . \glb (im.a.I))
------------------------------------------

------------------------------------------
     HOMOMORPHISM PROPERTIES

Range monotonicity:
  I \subseteq J
   ==> (\glb i \in I :: a.i)
       >= (\glb i \in J :: a.i)

  I \subseteq J
   ==> (\lub i \in I :: a.i)
       <= (\lub i \in J :: a.i)

Empty range:
  (\glb i \in {} :: a.i) == \top
  (\lub i \in {} :: a.i) == \bot

Range split rules:
(\glb i \in (I \union J) :: a.i)
  == (\glb i \in I :: a.i)
     \meet (\glb i \in J :: a.i)

(\lub i \in (I \union J) :: a.i)
  == (\lub i \in I :: a.i)
     \join (\lub i \in J :: a.i)
------------------------------------------
        Q: What do these mean for the booleans?
        interpret
        see the file ../truth-values/cohen-predicates.txt 

*** index function homomorphisms
------------------------------------------
       INDEX FUNCTION HOMOMORPHISMS

Suppose we fix the index set I.

If a: S -> G,
then (\ I . \glb (im.a.I)) : (S -> G) -> G

Image monotonicity:
  a \sqsubseteq b
   ==> (\glb i \in I :: a.i)
       <= (\glb i \in I :: b.i)

  a \sqsubseteq b
   ==> (\lub i \in I :: a.i)
       <= (\lub i \in I :: b.i)

Bottom and top homomorphic:
i != {}
 ==> (\glb i \in I :: \bot.i) == \bot

(\lub i \in I :: \bot.i) == \bot

(\glb i \in I :: \top.i) == \top

i != {}
 ==> (\lub i \in I :: \top.i) == \top

Meet and join homomorphic:
(\glb i \in I :: (a \meet b).i)
  == (\glb i \in I :: a.i)
     \meet (\glb i \in I :: b.i)

(\lub i \in I :: (a \join b).i)
  == (\lub i \in I :: a.i)
     \join (\lub i \in I :: b.i)

------------------------------------------

        Q: What is \bot in S -> G?  What is \top in S -> G?

        Q: is indexed meet universally meet homomorphic?
           yes, because it's meet homomorphic on a complete lattice

** bounded quantification (7.4)
------------------------------------------
        BOUNDED QUANTIFICATION
              NOTATION

                       ^
(\forall i \in I :: t) =
   (\forall i : i \in I : t)

                       ^
(\exists i \in I :: t) =
   (\exists i : i \in I : t)

Thus, 
  (\forall i \in I : b : t)
     == (\forall i :: i \in I /\ b ==> t)

------------------------------------------

------------------------------------------
         BOUNDED QUANTIFICATION
        AS INDEXED MEET AND JOIN

(\forall i \in I :: t)
    == (/\ i \in I :: t)

(\exists i \in I :: t)
    == (\/ i \in I :: t)

So what are the rules for...

Range monotonicity:
  I \subseteq J
   ==> ((\forall i \in I :: a.i)
            (\forall i \in J :: a.i))

  I \subseteq J
   ==> ((\exists i \in I :: a.i)
           (\exists i \in J :: a.i))

Empty range:
  (\forall i \in {} :: a.i) == 
  (\exists i \in {} :: a.i) == 

Range split rules:
(\forall i \in (I \union J) :: a.i)
  == (\forall i \in I :: a.i)
         (\forall i \in J :: a.i)

(\exists i \in (I \union J) :: a.i)
  == (\exists i \in I :: a.i)
         (\exists i \in J :: a.i)

------------------------------------------
        ... fill these in, remembering that \top == T, \bot == F,
            <= == ==>, \meet == /\, \join == \/

        The index function rules are exercise 7.4

*** general conjunction and disjunction for predicates
------------------------------------------
   GENERAL CONJUNCTION AND DISJUNCTION
           FOR PREDICATES

General conjunction:

  (\intersect i \in I :: p.i).s
      <==> (\forall i \in I :: p.i.s)

General disjunction:
  (\union i \in I :: p.i).s
      <==> (\exists i \in I :: p.i.s)
------------------------------------------

        Q: What inference rules would we have for these?
        the ones for general meets and joins, specialized to predicates