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Mathematical Foundations: (3) Lattice Theory --- Part I

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Mathematical Foundations: (3) Lattice Theory --- Part I « Mathematical foundations: (3) Lattice theory — Part I » Patrick Cousot Jerome C. Hunsaker Visiting Professor Posets Massachusetts Institute of Technology Department of Aeronautics and Astronautics cousot mit edu www.mit.edu/~cousot Course 16.399: “Abstract interpretation” http://web.mit.edu/afs/athena.mit.edu/course/16/16.399/www/ Course 16.399: “Abstract interpretation”, Thursday March 17th, 2005 — 1 — ľ P. Cousot, 2005 Course 16.399: “Abstract interpretation”, Thursday March 17th, 2005 — 3 — ľ P. Cousot, 2005 Binary relation – Given sets X1; X2;:::;Xn, the cartesian product is n def X1 X2 : : : Xn = x1; :::; xn ^ Xi ˆ ˆ ˆ fh i j 2 g i=1 – An n-ary relation r on X ; X ;:::;Xn is r }(X 1 2 2 1 ˆ X2 : : : Xn) i.e. r X1 X2 : : : Xn Garrett Birkhoff George Grätzer ˆ ˆ „ ˆ ˆ ˆ – If n =2, r is binary – A binary relation r on a set X is r }(X X) 2 ˆ – We write xry for x; y r h i2 Course 16.399: “Abstract interpretation”, Thursday March 17th, 2005 — 2 — ľ P. Cousot, 2005 Course 16.399: “Abstract interpretation”, Thursday March 17th, 2005 — 4 — ľ P. Cousot, 2005 Graph of a binary relation Poset – A relation can be seen as a graph where X is the set A poset X; is a set equipped with a partial order h »i » of vertices and r is the set of arcs. For example on X. a b Examples: – N; is a poset (where x; y N : x y z X = a; b; c; d h »i 8 2 » () 9 2 f g N : x + z = y) r = c; b ; b; c ; c d fh i h i – N; is a poset (where x; y N : x y x b; d ; d; c h –i 8 2 – () » h i h ig y) – Familiar relations on R are <, , =, = while on }(X), – 6 where X is a set, we have , , etc. „ ff Course 16.399: “Abstract interpretation”, Thursday March 17th, 2005 — 5 — ľ P. Cousot, 2005 Course 16.399: “Abstract interpretation”, Thursday March 17th, 2005 — 7 — ľ P. Cousot, 2005 Partial order Strict partial order – A partial order on a set X is a binary relation on – if is a partial order then @ is its strict version x @ » » v X which is y def= x y x = y, sometimes denoted . v ^ 6 v6 - reflexive i.e. x X : x x – @, is the negation of @ and 8 2 » 6 6v v - antisymetric i.e. x; y X : (x y y x) = – x @ y y @ x means that x and y are not comparable 8 2 » ^ » ) 6 ^ 6 x = y (sometimes written x y). k - transitive i.e. x; y; z X : (x y y z) = – A strict partial order < on a set X is a binary relation 8 2 » ^ » ) (x z) < on X which is » where x y formally means x; y . - irreflexive i.e. x X : (x < x) » h i 2 » 8 2 : - transitive i.e. x; y; z X : (x<y y < z) = 8 2 ^ ) (x < z) Course 16.399: “Abstract interpretation”, Thursday March 17th, 2005 — 6 — ľ P. Cousot, 2005 Course 16.399: “Abstract interpretation”, Thursday March 17th, 2005 — 8 — ľ P. Cousot, 2005 Correspondence between partial and Preorder strict partial orders – A preorder on a set X is a binary relation on X Theorem. If < is a strict partial order on X then — » » which is defined by x y x<y x = y is a partial order » () _ - reflexive i.e. x X : x x on X 8 2 — - transitive i.e. x; y; z X : (x y y z) = Proof. – x x x<x x = x = ¸ tt = tt » () _ _ 8 2 — ^ — ) – x y y x (x<y x = y) (y<x x = y) (x z) » ^ » () _ ^ _ — 1. if x = y antisymetry is proved (but not necessarily antisymetric) 2. if x = y we have x<y y<x whence x<x by transitivity, in 6 ^ Example: on ˚+~ defined by ff ff ff ff contradiction with irreflexivity, so this case is impossible. — — 0 () j j » j 0j – If x y y z then (x<y x = y) (y < z y = z) is a preorder but not a partial order (since e.g. ab bc » ^ » _ ^ _ 1. if x = y then x < z x = z so x z — _ » and bc ab but ab = bc). 2. if y = z then x < z x = z so x z — 6 _ » 3. Otherwise x<y y < z so by transitivity x < z ^ Course 16.399: “Abstract interpretation”, Thursday March 17th, 2005 — 9 — P. Cousot, 2005 Course 16.399: “Abstract interpretation”, Thursday March 17th, 2005 — 11 — P. Cousot, 2005 ľ ut ľ Theorem. If is a partial order on X then < defined def » Theorem. If is a preorder then x y = (x y) by x<y x y x = y is a strict partial order on — ” — ^ () » ^ 6 (y x) is a equivalence relation. X — Proof. def Proof. – x<x x x x = x = tt ¸ = ¸ – x x = (x x) (x x) = tt since is () » ^ 6 ^ ” — ^ — — – x<y y < z (x y x = y) (y z y = z) which implies reflexive ^ () » ^ 6 ^ » ^ 6 x z x = z by transitivity of since x = z would imply x = y = z, a def def » ^ 6 » – x y = (x y) (y x) (y x) (x y) = contradiction. ” — ^ — () — ^ — y x ut ” – x y y z def= (x y) (y x) (y z) (z y) ” ^ ” — ^ — ^ — ^ — (x y) (y z) (z y) (y x) = () — ^ def — ^ — ^ — ) (x z) (z x) = x z — ^ — ” ut Course 16.399: “Abstract interpretation”, Thursday March 17th, 2005 — 10 — ľ P. Cousot, 2005 Course 16.399: “Abstract interpretation”, Thursday March 17th, 2005 — 12 — ľ P. Cousot, 2005 Quotient 1 poset of a preorder Restriction of a poset to a subset If r is a binary relation on a set X and Y X then „ Theorem. Let be a preorder on a set X. Let be — ” def the equivalence relation defined by x y (x r Y = x; y r x; y Y ” () — j fh i2 j 2 g y) (y x). Let X= be the quotient of X by . ^ — ” ” Theorem. If X; is a poset and Y X then X; Define 2 on X= by h »i „ h »jY i —” ” is also a poset def [x] [y] = x y Proof. – If x Y then x x = x x = tt ” —” ” — 2 »jY » – If x;y Y then x y y x implies x y y x so x = y Then X= ; is the quotient poset of the preorder 2 »jY ^ »jY » ^ » h ” —”i – If x;y;z Y then x Y y y Y z implies x y z so x z on X hence X; . x z on2 Y since x;»j z Y^. »j » » » h —i »jY 2 def 1 Recall that if is an equivalence relation on a set X then the quotient X= = [x] x X is the set of ” def ” f ” j 2 g ut equivalence classes [x] = y X x y . ” f 2 j ” g 2 In general, is denoted for short. —” — Course 16.399: “Abstract interpretation”, Thursday March 17th, 2005 — 13 — ľ P. Cousot, 2005 Course 16.399: “Abstract interpretation”, Thursday March 17th, 2005 — 15 — ľ P. Cousot, 2005 Proof. – First remark that the definition of on X= is independent of ” Intervals the choice of the representants x and y of the—” classes [x] and [y] since ” ” x0 x and y0 y implies x0 x y y0 so x0 y0 by transitivity and ” ” — — — — It follows that if X; is a poset and a, b X, then reciprocally, if x0 y0 then x x0 y0 y so x y — — — — — def h »i 2 – We have x x so [x] [y] – [a; b] = x X a x b — ” —” ” – If [x] [y] and [y] [x] then x y y x so x y proving that def f 2 j » » g [x] =[” —y”] ” ” —” ” — ^ — ” – [a; b[ = x X a x < b ” ” f 2 j » g – If [x] [y] and [y] [z] then x y y z whence x z by def ” —” ” ” —” ” — ^ — — – ]a; b] = x X a < x b transitivity proving that [x] [z] ” ” def f 2 j » g —” – ]a; b[ = x X a<x<b ut f 2 j g are all posets for . » Course 16.399: “Abstract interpretation”, Thursday March 17th, 2005 — 14 — ľ P. Cousot, 2005 Course 16.399: “Abstract interpretation”, Thursday March 17th, 2005 — 16 — ľ P. Cousot, 2005 Equality Covering relation Theorem. Let X; be a poset. The covering relation is The only partial order which is also an equiv- h »i def alence relation is equality. x <y = (x<y) ( z X : x<z<y) ` ^: 9 2 Proof. Let be an equivalence relation which is a partial order ı We say that “y covers x” or “x is covered by y” and x y write x <y ı ` = x y y x by symmetry of equivalence ) ı ^ ı H I Examples: = x = y by antisymmetry of partial order def ) H I – The covering relation of N; or Z; is x < y = (y = x = y h »i h »i ` x + 1) = x y by reflexivity ) ı H I – The covering relation of R; is ¸ h »i – The covering relation of }(X); is X Y def= x Y X : h „i `„ 9 2 n ut Y = X x [ f g Course 16.399: “Abstract interpretation”, Thursday March 17th, 2005 — 17 — ľ P. Cousot, 2005 Course 16.399: “Abstract interpretation”, Thursday March 17th, 2005 — 19 — ľ P. Cousot, 2005 Inverse of a partial order If X; is a finite poset (i.e.
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