CS 641 meeting -*- Outline -*-

* quantified expressions (Cohen, Ch. 3)

	Context: Now we study universal and existential quantification.

	Plan: Cohen makes an interesting abstraction first,
		generalized expressions, and then specializes to
		universal and existential quantification

	In the following, we ``all'' for the universal quantifier,
	and ``exists'' for the existential quantifier.

	Note: all the theorems are here from Cohen's book,
		but it's not useful to just list them.
		Refer the students to the book or this file,
			and just work a few of their proofs.

** quantified expressions (3.1, can omit)
	Contrary to Cohen, we'll indicate dependence explicitly,
		so when we write R.i, you can see it depends on i,
		but Z does not.

	generalization of *symmetric and associative* binary operators
		doesn't work for, say subtraction or division.
----------------------
QUANIFIED EXPRESSIONS
	(OP i : R.i : T.i)
	 --

where op is symmetric and associative

Examples:
    (PLUS i : 1 <= i < 4 : i )
	= 1 plus 2 plus 3
	= 6
    (MAX i : 1 <= i < 9 : i+7) = 15
----------------------
	definitions:
		i is a list of dummies,
		R.i is the range,
		T.i is the term (an expression)

	ranges of true are omitted

	def: a range is empty if it's false everywhere

	Q: Why doesn't this work for subtraction?	
	in what follows, the underlining will be left off, due to limitations
		of ASCII, we'll just capitalize whatever.

	Cannonical examples:
		PLUS (summation), TIMES (product),
		AND (universal quantifier), OR (existential quantifier)
		other examples would be EQUIV, DIFFS
	but not implication or consequence

*** laws
----------------
LAWS OF QUANTIFIED EXPRESSIONS
----------------
	For each law, give concrete example, then law, and apply to
		universal and existential quantification
----------------
  (MAX i : 1<=i<4 : (i+8) max (i+5))
=	<def>
  (1+8) max (1+5) max (2+8) max (2+5)
    max (2+8) max (2+5)
=	<associtivity>
  (1+8) max (2+8) max (2+8)
   max (1+5) max (2+5) max (2+5)
= 	<def>
  (MAX i : 1<=i<4 : i+8)
   max (MAX i : 1<=i<4 : i+5)
----------------
	so this generalizes to the
----------------
term rule:
   (OP i : R.i : P.i op Q.i)
 = (OP i : R.i : P.i) op (OP i : R.i : Q.i)
-----------------
	in hints, left to right is called "splitting the term"
		 and right to left is called "joining the term".
	Q: why is this correct (in terms of the semantics)?
	example:
		(AND i : 0<=i<5 : i<7 /\ i<8)
		equiv (AND i : 0<=i<5 : i<7) /\ (AND i : 0<=i<5 : i<8)

------------------
  (MAX i : 1<=i<4 \/ 3<=i<5 : i+6)
=	<def, 1,2,3,4 satisfy the range>
  (1+6) max (2+6) max (3+6) max (4+6)
=	<idempotence>
  (1+6) max (2+6) max (3+6)
   max (3+6) max (4+6)
=	<def>
  (MAX i : 1<=i<4 : i+6)
   max (MAX i: 3<=i<5 : i+6)
------------------
	but what about non-idempotent operators?
------------------
  (PLUS i : 1<=i<4 \/ 7<=i<9 : i)
=	<def>
  1 + 2 + 3 + 7 + 8
=	<def>
  (PLUS i : 1<=i<4 : i)
   + (PLUS i: 7<=i<9 : i)

but
  (PLUS i : i<=i<4 \/ 3<=i<5 : i)
=	<def>
  1 + 2 + 3 + 4
?
  1 + 2 + 3 + 3 + 4
=	<def>
  (PLUS i : i<=i<4 : i)
  + (PLUS i : 3<=i<5 : i)?
------------------
	so we have...
------------------
range rule:
 if op is idempotent (i.e., for all P, P op P = P),
 or P.i diffs Q.i for all i,
   (OP i : P.i \/ Q.i : T.i)
 = (OP i : P.i : T.i) op (OP i : Q.i : T.i)
------------------
	in hints, "splitting the range", "joining the range"
	Q: Are conjuction and disjunction idempotent?
		So example: 	(AND i : P.i \/ Q.i : T.i)
			  equiv (AND i : P.i : T.i) /\ (AND i : Q.i : T.i)
				because conjunction is idempotent
	But only works for + if P.i diffs Q.i for all i

------------------
  (MAX i : 1<=i<4 : 5*i)
=	<def>
  5*1 max 5*2 max 5*3
=	<* distributes over max>
  5 * (1 max 2 max 3)
=	<def>
  5 * (MAX i : 1<=i<4 : i)

distributivity:
 if * distributes over op,
 and the range is non-empty,
   (OP i : R.i : Z * T.i)
 = Z * (OP i : R.i : T.i)
------------------
	Recall, can move Z out because it doesn't depend on i
	Q: does conjunction distribute over disjunction?  vice versa?
		yes, so for example
		(AND i : R.i : Z \/ Q.i)
		= Z \/ (AND i : R.i : Q.i)

-----------------
(PLUS i : false : i+7) = 0
(TIMES i : false : i+7) = 1
(MAX i : false : i+7) = -infinity

empty range rule:
 if op has a unit, 1_op,
   (OP i : false : T.i) = 1_op
-----------------
	Recall that a unit is x such that P op x = P
		(as for multiplication)
	Q: What's the unit of conjunction?
		true, so (AND i : false : T.i) = true
	   for disjunction the unit is false,
		so (OR i : false : T.i) = false

----------------
(MAX i: 1<=i<4 : 641) = 641

constant term rule:
 if op is idempotent,
 and the range is non-empty,
   (OP i: R.i : Z) = Z
-----------------
	Q: So what is (AND i : 1 <= i < 27 : true)?
	Q: how about (OR i : 1 <= i < 27 : false)?
		note this latter works even if the range is empty

	A trick question...
	Q: How about (EQUIV i : 1 <= i < 5 : false)?
		false equiv false equiv false equiv false
	equiv     < P equiv P equiv true, twice >
		true equiv true
	equiv	  < P equiv P equiv true>
		true
	why did that happen?

---------------
  (MAX i : 1<=i<4 : i)
=	<def>
  1 max 2 max 3
=	<def>
  (MAX i : 0<=j /\ j<3 : j+1)
=	<arithmetic>
  (MAX i : 1<=j+1 /\ j+1<4 : j+1)
=	<notation>
  (MAX i : 1<=j+1<4 : j+1)

dummy transformation rule:
 if f is invertible
        (i.e., i neq j equiv f.i neq f.j),
   (OP i : R.i : T.i)
    = (OP j : R.(f.j) : T.(f.j))
----------------
	Note: assumes by the notation, j does not appear in R.i or T.i
	invertability of f is needed to convert the one into the other.

	Remark, this allows renaming, see below.

--------------
nesting rule:
 (OP i : P.i : (OP j : Q.i.j : T.i.j))
= (OP i,j : P.i /\ Q.i.j : T.i.j)
--------------
	from left to right this is called "nesting",
	from right to left this is called "unnesting".

-------------
(MAX i : i = 3 : i+4) = 3+4

1-point rule:
 if T.Z is defined
   (OP i : i = Z : T.i) = T.Z
--------------
	Recall that by the notation, Z does not depend on i

	Q: why does T.Z have to be defined?

*** theorems
	Thm (renaming the dummy):
		if j is not free in R or T,
		(OP i : R.i : T.i) = (OP j : R.j : T.j)
	Thm (generalized distributivity):
		if * distributes over op,
		and either the range, R.i, is non-empty
		    or the unit of op is the same as the unit of *,
		then (OP i : R.i : Z * T.i) = Z * (OP i : R.i : T.i)

** Universal quantification (3.2)
	this is generalized conjuction

	The only additional thing that doesn't follow from the general theory
		is "trading" below.
	Also note "instantiation".

----------------
UNIVERSAL QUANTIFICATION

(all i : R.i : T.i)

term rule:
       (all i : R.i : P.i /\ Q.i)
 equiv (all i : R.i : P.i) /\ (all i : R.i : Q.i)
-----------------
	in hints, left to right is called "splitting the term"
		 and right to left is called "joining the term".
	(Dijkstra calls this rule "all distributes over /\")
	Q: why is this correct?

------------------
range rule:
       (all i : P.i \/ Q.i : T.i)
 equiv (all i : P.i : T.i) /\ (all i : Q.i : T.i)
------------------
	in hints, left to right is "splitting the range",
		right to left is "joining"

------------------
conjunction distributes over universal quantifier
with non-empty range:
 if the range is non-empty,
       (all i : R.i : Z /\ T.i)
 equiv Z /\ (all i : R.i : T.i)

disjunction distributes over universal quantifier:
       (all i : R.i : Z \/ T.i)
 equiv Z \/ (all i : R.i : T.i)
------------------
	the latter follows from generalized distributivity (why?)

-----------------
empty range rule:
   (all i : false : T.i) equiv true
-----------------
	Q: What's the unit of conjunction?
		true, so (AND i : false : T.i) = true
----------------
constant term rule:
 if the range is non-empty,
   (all i : R.i : Z) equiv Z
-----------------
	recall that conjunction is idempotent
---------------
dummy transformation rule:
 if f is invertible (i.e., i neq j equiv f.i neq f.j),
   (all i : R.i : T.i) = (all j : R.(f.j) : T.(f.j))
----------------
	Note: assumes by the notation, j does not appear in R.i or T.i
	invertability of f is needed to convert the one into the other.

--------------
nesting rule:
 (all i : P.i : (all j : Q.i.j : T.i.j))
= (all i,j : P.i /\ Q.i.j : T.i.j)
--------------
	from left to right this is called nesting,
	from right to left this is called unnesting.

-------------
1-point rule:
 if T.Z is defined,
   (all i : i = Z : T.i) = T.Z
--------------
	Recall that by the notation, Z does not depend on i

	Q: why does T.Z have to be defined?

	Thm: implication distributes over universal quantification.
		(all i : R.i : Z ==> T.i) equiv Z ==> (all i : R.i : T.i)

---------------
trading:
       (all i : R.i : T.i)
 equiv (all i :: ~R.i \/ T.i)
---------------
	this needs to be postulated, as it doesn't follow from above.
		Gives meaning to the range.

	Thm (trading):        (all i : R.i /\ S.i : T.i)
		        equiv (all i : R.i : S.i ==> T.i)

	Thm (instantiation): (all i :: f.i) ==> f.x

** existential quantification (3.3)
	generalized disjunction

	The only additional thing that doesn't follow from the general theory
		is "trading" below.
	Also note "instantiation".

---------------
EXISTENTIAL QUANTIFICATION

(exists i : R.i : T.i)

postulate (Generalized De Morgan):
       (exists i : R.i : T.i)
 equiv ~(all i : R.i : ~T.i)
---------------
	From this postulate, rules similar to the above follow.

	Thm (term rule):
		       (exists i : R.i : P.i \/ Q.i)
		 equiv (exists i : R.i : P.i) \/ (exists i : R.i : Q.i)
	Thm (range rule):
		      (exists i : P.i \/ Q.i : T.i)
		equiv (exists i : P.i : T.i) \/ (exists i : Q.i : T.i)
	Thm (disjunction distributes over existentials with non-empty range):
		If the range is non-empty, then
		      (exists i : R.i : Z \/ T.i)
		equiv Z \/ (exists i : R.i : T.i)
	Thm (conjunction distributes over existentials):
		      (exists i : R.i : Z /\ T.i)
		equiv Z /\ (exists i : R.i : T.i)
	Thm (empty range rule):
		(exists i : false : T.i) equiv false
	Thm (constant term rule):
		If the range is non-empty, then
		(exists i: R.i : Z) equiv Z
	Thm (dummy transformation rule):
		If f is invertible (i.e., i neq j equiv f.i neq f.j),
		(exists i : R.i : T.i) equiv (exists j : R.(f.j) : T.(f.j))
	Thm (nesting rule):
		      (exists i : P.i : (exists j : Q.i.j : T.i.j))
		equiv (exists i,j : P.i /\ Q.i.j : T.i.j)
	Thm (1-point rule):
		if T.Z is defined, then
		 (exists i : i equiv Z : T.i) equiv T.Z

	Thm (dual of Generalized De Morgan):
	       ~(exists i : R.i : ~T.i) equiv (all i : R.i : T.i)
	Thm (rule of instantiation):
		f.x ==> (exists i :: f.i)

	Q: Consider the following "rules", which are the converses to
	   the rules of instantiation.
	 	p.x ==> (all x :: p.x)     {universal generalization}
		(exists x :: q.x) ==> q.x  {existential specialization}
	   Using them, can you "prove" a contradiction?
		The existential one is really bad,
			the universal one is okay if x is fresh,
			and if used in a meta-mathematical way
			(do the proof for x, conclude holds for all x)

	Thm (trading):
		(exists i : R.i : T.i) equiv (exists i :: R.i /\ T.i)

** Arithmetic quantifications
	Summation is generalized plus.
	Numerical quantification is defined by the following
		(N i : R.i : T.i) = (PLUS i : R.i /\ T.i : 1)
	so it's the number of things such that R.i and T.i are true.