1. 1 Logic Programming Technical Foundation Second part of Cmput325 Fall 2004 R Greiner + B Price

2. 2 Declarative Programming Motivation Warm Fuzzies What is Logic?... Logic Programming? Foundations: Terms, Substitution, Unification, Horn Clauses Proof (Resolution) Examples: List Processing; Integer Issues Logic / Theorem Proving Search Strategies Declarative/Procedural,... Other parts of Prolog “Impure'' Operators” --- NOT, ! Utilities ? Constraint Programming ? Bayesian Belief Nets

3. 3 What is Logic? Logic is formal system for reasoning Reasoning is inferring new facts from old Given { All men are mortal. Socrates is a man. } infer (conclude, reason that,...) Socrates is mortal. What is role of Logic within CS? Foundation of discrete mathematics Automatic theorem proving Hardware design/debugging Artificial intelligence (Cmput366) Components: Syntax (What does it look like?) Reasoning/ProofTheory (New facts from old) Semantics (What should it do?)

4. 4 Logic Programming Program  Logic Formula Execution of program  Proving a theorem User: 1. Specifies WHAT is true 2. Asks if something else follows Prolog: Answers question. By comparison, in Procedural Programming (C, JAVA, Fortran,...) user must … decide on data-structure explicitly write procedures: search, match, substitute write diff programs for father( X, tom) vs father(tom, Y )

5. 5 What is Logic? Syntax: Set of Expressions ? Well-formed or not? SEQUENCE of Symbols  Accept: The boys are at home. at(X, home) :- boy(X). Reject: boys.home the angrily democracy X(at), xY=, Boy(1X,( ) :- Accept Reject Given { All men are mortal. Socrates is a man. } infer (conclude, reason that,...) Socrates is mortal. Proof Process: Given Believed statements, Determine other Believed statements. { s 1,..., s n } ⊢ P s Semantics: Which expressions are Believed ? John's mother is Mary.  T John's mother is Fred.  F Colorless green ideas sleep furiously.  F

6. 6 Concept of PROLOG PROgramming in LOGic Sound Reasoning Engine 1. User asserts true statements. User asserts { All men are mortal. Socrates is a man. } 2. User poses query. “Is Socrates mortal?” “Who/what is mortal?” 3. Prolog provides answer (Y/N, binding). “Yes” “Socrates ''

7. 7 Tying Prolog to Logic Syntax: Horn Clauses (aka Rules + Facts) Terms Proof Process: Resolution Substitution Unification Semantics (From Predicate Calculus … + not )

8. 8 Terms BNF:  term  ::=  constant  |  variable  |  functor   constant  ::=  atom starting w/lower case   variable  ::=  atom starting w/upper case   functor  ::=  constant  (  tlist  )  tlist  ::= “” |  term  {,  tlist  } Examples of  term  : a1 b fred  constant  X Yc3 Fred _4  variable  married(fred) g(a, f(Yc3), b)  functor  Ground Term  term with no variables f(q) g(f(w), w1(b,c)) are ground f(Q) g(f(w), w1(B,c)) are not

9. 9 Substitution A Substitution is a set { t 1 /v 1 t 2 /v 2... t n /v n } where v i are distinct variables t i are terms that do not use any v j Examples: + { a/X } { a/X b/Y f(a,W)/Z } { W/X f(W)/Y W/Z } - { a/f(X) } { a/X b/X } { f(X)/X } { f(Y)/X g(q)/Y } - { f(g(Y))/X g(q)/Y }

10. 10 Applying a Substitution Given “t  ” is term, result of applying  to t Small Examples: X {a/X}  a f(X) {a/X}  f(a) Example, using t = f( a, h(Y,b), X ) t -- a term  -- a substitution { b/X }  f( a, h(Y,b), b ) { b/X f(Z)/Y }  f( a, h(f(Z),b), b ) { Z/X f(Z,a)/Y }  f( a, h(f(Z,a),b), Z ) { Z/W }  f( a, h(Y,b), X )

11. 11 Applying a Substitution f( a, h(Y,b), X ) { b/X } = f( a, h(Y,b), b ) f( a, h(Y,b), X ) { b/X f(Z)/Y } = f( a, h(f(Z),b), b ) f( a, h(Y,b), X ) { Z/X f(Z,a)/Y } = f( a, h(f(Z,a),b), Z ) f( a, h(Y,b), X ) { Z/W } = f( a, h(Y,b), X )  need not include all variables in t  can include variables not in t.

12. 12 Composition of Substitutions  ○  is composition of substitutions ,  For any term t, t [  ○  ] = (t  ) . Example: f(X) [{ Z/X} ○ { a/Z}] = (f(X) { Z/X}) {a/Z} = f(Z) { a/Z} =f(a)  ○  is typically a substitution Eg: { a/X} ○ { b/Y} = { a/X, b/Y} { Z/X} ○ { a/Z} = { a/X, a/Z}

13. 13 Examples: t 1 t 2 unifier term f(b,c) f(b,c) {} f(b,c) f(X,b) f(a,Y){ a/X b/Y } f(a,b) f(a,b) f(c,d) * f(a,b) f(X,X) * f(X,a) f(Y,Y){ a/X a/Y } f( a, a ) f(g(U),d) f(X,U){ d/U g(d)/X } f( g(d),d ) f(X) g(X) * f(X,g(X)) f(Y,Y) * f(X) f(Y) { Y/X } f(Y) Unifiers t 1 and t 2 are unified by  iff t 1  = t 2 . Then  is called a unifier t 1 and t 2 are unifiable NB t 1 and t 2 are symmetric -- either can have variables

14. 14 Multiple Unifiers Unifier for t 1 = f(X) and t 2 = f(Y) … { X/Y } and { Y/X } make sense, but { a/Y a/X } has irrelevant constant { X/Y g/W } has irrelevant variable ( W ) Adding irrelevant constants/bindings   unifiers  best one??  t 1  = t 2  = { X/Y } f(X) { Y/X } f(Y) { a/Y a/X } f(a) { g(b,Z))/Y g(b,Z)/X) } f(g(b,Z) ) { X/Y W/g) } f(X)

15. 15 Quest for Best Unifier Wish list: No irrelevant constants So {X/Y} preferred over { a/Y, a/X } No irrelevant bindings So {X/Y} preferred over { X/Y, f(4,Z)/W } Spse 1 has constant where 2 has variable 1 = { a/X, a/Y }, 2 = {Y/X} Then  substitution  s.t. 2 ○  = 1  = {a/Y} : {Y/X} ○ {a/Y} = { a/X, a/Y } Spse 1 has extra binding over 2 1 = {a/X, b/Y}, 2 = {a/X} Then  substitution  s.t. 2 ○  = 1  = {b/Y} : {a/X} ○ {b/Y} = {a/X, b/Y} INFERIOR unifier = composition of Good Unifier + another substitution

16. 16 Most General Unifier  is MGU for t 1 and t 2 iff  unifies t 1 and t 2, and  : unifier of t 1 and t 2,  substitution  s.t.  ○  =  Ie, for all terms t, t  = (t  )  Eg,  = {X/Y} is mgu for f(X) and f(Y) Consider unifier  = {a/X a/Y} Use  = { Y/a }: f(X)  = f(X) {a/X a/Y} = f(a) f(X)[  ○  ] = [f(X)  ]  = [f(X) {Y/X}] = f(Y) {a/X a/Y} = f(a) Similarly, f(Y)  = f(a) = f(Y) [  ○  ] (  is not a mgu, as ¬  ’s.t.  ○  ’=  )

17. 17 MGU --- Example#2 A mgu for f( W, Z, g(Z) ) & f( X, h(X), Y ) { W/X } W/X h(X)/Z h(W)/Z g(Z)/Y h(W)/Z g(h(W))/Y

18. 18 MGU (con't) Notes: If t 1 and t 2 are unifiable, then  mgu. Can be more than 1 mgu but they differ only in variable names. Not every unifier is a mgu. A mgu uses constants only as necessary. Implementation:  fast algorithm that computes a mgu of t 1 and t 2, if one exists; or reports failure. Slow part is verifying legal substitution: none of v i appear in any t j. Avoid by resetting Prolog's occurscheck parameter.

19. 19 MGU Procedure Recursive Procedure MGU( x, y ) If x=y then Return() If Variable(x) then Return( MguVar(x,y) ) If Variable(y) then Return( MguVar(y,x) ) If Constant(x) or Constant(y) then Return( False ) If Not(Length(x) = Length(y)) then Return( False ) g  [] For i = 0.. Length(x) s  MGU( Part(x,i), Part(y,i) ) g  Compose( g, s ) x  Substitute( x, g ) y  Substitute( y, g ) Return( g ) End Procedure MguVar (v,e) If Includes(v,e) then Return( False ) Return( [v/e] ) End

20. 20 Prolog's Syntax BNF:  Horn  ::=  literal . |  literal  :-  llist .  llist  ::=  literal  {,  llist  }  literal  ::=  term  Examples: father(john, sue). father(odin, X). parent(X, Y) :- father(X, Y). gparent(X, Z) :- parent(X, Y), parent(Y, Z). How to read as predicate calculus? father(john, sue)  X. father(odin, X).  X,Y. father(X,Y)  parent(X,Y)  X,Y,Z.parent(X,Y)} & parent(Y,Z)  gparent(X,Z)

21. 21 Relation to Predicate Calculus In general: t.   x 1, …, x m. t [called “atomic formula”] t :- t 1, …, t n.   x 1, …, x m. t 1, …, t n  t. [called “(production) rule”] Can write any Predicate Calculus Expression in Conjunctive Normal Form  Horn  clauses are CNF, with ONE Positive Literal  Horn   CNF  Predicate Calculus expressions that canNOT be written as Horn Clauses. Axioms A v B

22. 22 What User Really Types > sicstus SICStus 3.11.2 (x86-linux-glibc2.3): Wed Jun 2 11:44:50 CEST 2004 Licensed to cs.ualberta.ca | ?- [user]. % For user to enter ``Assert-fact'' mode. % consulting user... | on(aa,bb). | on(bb,cc). | above(X,Y) :- on(X,Y). | % Typing ^D exits ``assert'' mode. % consulted user in module user, 0 msec 608 bytes yes % Prolog's answer to most operations | ?- on(X,Y). X = aa, Y = bb ? ; % User types “;” to ask for ANOTHER answer X = bb, Y = cc ? ; no % Prolog's no means there are no other answers | ?- on(aa,bb). % User asks a question. yes % Prolog's answer | ?- on(aa,Y). % User's second question Y = bb ? % Prolog's answer: a binding list yes % Prolog's statement that there was answer | ?-

23. 23 What User Really Types -- II > sicstus SICStus 3.11.2 (x86-linux-glibc2.3): Wed Jun 2 11:44:50 CEST 2004 Licensed to cs.ualberta.ca | ?- [file1]. % file ”file1” contains facts, rules % file1 consulted 120 byes 0.3333 sec. yes | ?- on(aa,b10). no % Prolog's answer: not derivable | ?- on(X,b10). no % Again: NO entailments | ?- above(bb,cc). yes % More than simple look-up! | ?- above(bb,W). W = cc ; % Prolog finds answer; user wants another no % but this is only one… | ?-

24. 24 Prolog's Proof Process User provides KB: Knowledge Base (List of Horn Clauses --- axioms)  : Query (aka Goal, Theorem) -- Literal 1. Who is mortal? mortal(X). 2. Is Socrates mortal? mortal(socrates). Prolog finds a Proof of , from KB, if one exists & substitution for γ's variables: σ KB ⊢ P  { σ } KB 1 ⊢ P mortal(X){ socrates/X } Failure (otherwise)

25. 25 Proof Process (con’t) Prolog finds ``Top-Down'' (refutation) Proof … returns bindings Actually returns LIST of σ i s [one for each proof] { socrates/X } { plato/X } { freddy/X }...

26. 26 Examples of Proofs: I (DB retrieval) Using KB 1 = on(aa, bb). (1) on(bb, cc). (2) above(X, Y) :- on(X,Y). (3) Query  1 : on(aa, bb) on(aa, bb ) success (1) Hence KB 1 ├ P on(aa, bb){} empty substitution ( Like Data Base retrieval )

27. 27 Examples of Proofs: IIa (Variables) Query  2 : on(aa, Y) on(aa, Y) success –{bb/Y} (1) Y=bb KB 1 ├ P on(aa, Y) { bb/Y } Using KB 1 = on(aa, bb). (1) on(bb, cc). (2) above(X, Y) :- on(X,Y). (3)

28. 28 Examples of Proofs: IIb (Variables) Query  3 : on( X, Y ) KB 1 ├ P on(X, Y){ aa/X, bb/Y }... on(aa,bb) KB 1 ├ P on(X, Y){ bb/X, cc/Y }... on(bb,cc) Using KB 1 = on(aa, bb). (1) on(bb, cc). (2) above(X, Y) :- on(X,Y). (3) (2) X =bb, Y=cc success {bb/X,cc/Y} success {aa/X,bb/Y} (1) X =aa, Y=bb Either is sufficient!

29. 29 Ie, KB 1 ├ P on(aa,bb) Examples of Proofs: III (Failure) Query  4 : on(aa, b10) Using KB 1 = on(aa, bb). (1) on(bb, cc). (2) above(X, Y) :- on(X,Y). (3) Query  5 : on(X, b10) Ie, KB 1 ├ P on(X,b10) for any value of X. Attempts

30. 30 Examples of Proofs: IVa (Rules) Query  6 : above(bb, cc) Ie, KB 1 ├ P above(bb,cc) on(bb,cc) (3) X=bb, Y=cc success (2) Using KB 1 = on(aa, bb). (1) on(bb, cc). (2) above(X, Y) :- on(X,Y). (3)

31. 31 Examples of Proofs: IVb (Rules) Query  7 : above(bb, W) Ie, KB 1 ├ P above(bb,W){cc/W} on(bb, W) (3) X=bb, Y=W success –{cc/W} (2) Using KB 1 = on(aa, bb). (1) on(bb, cc). (2) above(X, Y) :- on(X,Y). (3)

32. 32 Examples of Proofs: V (big) Query  7 : above(aa, cc) on(aa,cc) (3) X=bb, Y=W Using KB 2 = on(aa, bb). (1) on(bb, cc). (2) above(X, Y) :- on(X,Y). (3) above(X,Y) :- on(X,Z), above(Z,Y). (4) on(aa,Z)&above(Z,cc) (4) X=aa, Y=cc (1) Z=bb above(bb,cc)

33. 33 Examples of Proofs: V (big) above(aa,cc) on(aa, bb). (1) on(bb, cc). (2) above(X, Y) :- on(X,Y). (3) above(X,Y):- on(X,Z),above(Z,Y). (4) on(aa,Z),above(Z,cc) (4) X=aa, Y=cc (1) Z=bb above(bb, cc) on(bb,cc) (3) X=bb, Y=cc (2) Z=cc above(cc, cc) (2) successon(cc,cc) (3) on(aa,cc) (3) X=bb, Y=cc on(bb,Z), above(Z, cc) (4) X=bb, Y=cc on(cc,Z),above(Z,cc) (4) X=cc, Y=cc aa bb cc

34. 34 Examples of Proofs: VI (many answers) Query  9 : above( X, Y) Answers { X=aa, Y=bb } above( aa, bb) (3), (1) { X=bb, Y=cc } above( bb, cc) (3), (2) { X=aa, Y=cc } above( aa, cc) (4), (1), (3), (2) { X=c1, Y=c2 } above( c1, c2) (5) Using KB 3 = on(aa, bb). (1) on(bb, cc). (2) above(X, Y) :- on(X,Y). (3) above(X,Y) :- on(X,Z), above(Z,Y). (4) above(d1, d2) (5)

35. 35 Prolog's Proof Process A goal is either a sequence of literals (conjunction), the special goal `` success '' (eg, ) on(X,Y)p(X,5), q(X)success Sequence of (sub)goals  G 1, G 2,..., G n  is a top-down proof of G 1 (from KB) iff 1. G n = success, and 2. G i is a SUBGOAL (in KB) of G i-1, i=2,3,... n

36. 36 Subgoals Subgoals of G = { g 1,..., g r } in KB: Rule 1: If  atomic axiom ``t '' in KB where t and g i have mgu , then { g 1 ,..., g i-1 , g i+1 ,..., g r  } is a subgoal of G. (If r=1, then `` success '' is subgoal of G.) Rule 2: If  rule ``t :- t 1,..., t_k '' in KB where t and g i have mgu , then { t 1 ,..., t k , g 1 ,..., g i-1 , g i+1 ,..., g r  } is a subgoal of G.

37. 37 Example of Subgoals -- I Subgoals of are  = {X/A, Y/B} using Rule 2, (3)  = {X/A, Y/B} using Rule 2, (4)  = {c1/A, c2/B} using Rule 1, (5) KB 3 = on(aa, bb). (1) on(bb, cc). (2) above(X, Y) :- on(X,Y). (3) above(X,Y) :- on(X,Z), above(Z,Y). (4) above(c1, c2) (5) above(A,B) on(A,B) on(A,Z), above(Z, B) success

38. 38 Example of Subgoals -- II Subgoals of are  = {aa/A, bb/Z1} using Rule 3, (1) [first literal]  = {bb/A, bb/Z1} using Rule 3, (2) [first literal]  = {c1/Z1, c2/B} using Rule 3, (5) [second literal]  = {X/Z1, B/Y} using Rule 4, (3) [second literal]  = {X/Z1, Y/B} using Rule 4, (4) [second literal] on(aa, bb). (1) on(bb, cc). (2) above(X, Y) :- on(X,Y). (3) above(X,Y):-on(X,Z),above(Z,Y). (4) above(c1, c2) (5) above(A,Z 1 ), above(Z 1, B) above(bb,B) above(cc, B) on(A,c1) on (Z 1,B), on(A, Z 1 ) on (Z 1,Z), above(Z,B), on(A, Z 1 )

39. 39 Comments wrt Prolog's Proof Procedure found during proof Prolog returns these overall mgu's 1-by-1. Prolog uses specific strategy called “SLD Resolution” Variable bindings Unifier