1. First-Order Logic ECE457 Applied Artificial Intelligence Fall 2007 Lecture #7

2. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 2 Outline First-order logic (FOL) Sentences Russell & Norvig, chapter 8 Inference Russell & Norvig, chapter 9

3. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 3 First-Order Logic Propositional logic is limited Cannot represent information concisely We want something more expressive First-order logic Allows the representation of objects, functions on objects and relations between objects Allows us to represent almost any English sentence

4. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 4 New Symbols in FOL Constant symbol An individual object in the world Bob, James, Hat Predicate symbol A relation between two objects that can be true or false Brother(Bob, James), OnHead(Hat, Bob) Function symbol Special type of relation that maps one object to another Head(Bob) All symbols begin with uppercase letters

5. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 5 New Variables in FOL Begin with lowercase letters Stand-in for any symbol Brother(x,y) x is the brother of y

6. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 6 New Quantifiers in FOL Universal quantifier   x means “For all x…” Always true Usually used with   x means “Not all x…” Existential quantifier   x means “There exists an x…” True for at least one interpretation Usually used with   x means “There exists no x…”

7. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 7 Properties of New Quantifiers Nesting  x  y P(x,y) same as  y  x P(x,y)  x  y P(x,y) same as  y  x P(x,y)  x  y P(x,y) not same as  y  x P(x,y) Duality  x P(x) same as  x  P(x)  x P(x) same as  x  P(x)

8. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 8 Sentences in FOL Term Constant symbol, function symbol, or variable Atom (atomic sentence) Predicate symbol with value true or false Represents a relation between terms Sentence (complex sentence) Atom(s) joined together using logical connectives and/or quantifiers

9. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 9 Example Sentences Purple mushrooms are poisonous Purple(x)  Mushroom(x)  Poisonous(x) Some purple mushrooms are poisonous  x Purple(x)  Mushroom(x)  Poisonous(x) (bad!)  x Purple(x)  Mushroom(x)  Poisonous(x) No purple mushrooms are poisonous  x Purple(x)  Mushroom(x)  Poisonous(x)  x Purple(x)  Mushroom(x)   Poisonous(x) There are exactly two purple mushrooms  x  y  z Purple(x)  Mushroom(x)  Purple(y)  Mushroom(y)   (x=y)  Purple(z)  Mushroom(z)  (x=z)  (y=z)

10. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 10 Implication Truth Table If α is true, then β must be true for the implication to be true If α is false, then we have no information about β β might be true or false without affecting the implication β is not necessarily true If implication is true and α is true, then β must be true (Modus Ponens) αβ α  β TTT TFF FTT FFT

11. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 11 Universal Implication x = {,, } All purple mushrooms are poisonous  x Purple(x)  Mushroom(x)  Poisonous(x) Universal: all three implications are true is purple and a mushroom  is poisonous But the first one is the only one where the premise is true, therefore the only one where we know the consequent is true (Modus Ponens) Otherwise, we don’t know whether x is poisonous

12. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 12 Existential Implication x = {,, } Some purple mushrooms are poisonous  x Purple(x)  Mushroom(x)  Poisonous(x) Existential: at least one implication is true is purple and a mushroom  is poisonous But all those where the premise is false are true The existential implication doesn’t give us any information

13. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 13 Existential Conjunction x = {,, } Some purple mushrooms are poisonous  x Purple(x)  Mushroom(x)  Poisonous(x) Existential: at least one statement is true is purple and a mushroom and is poisonous The first statement is true, the other two are false The existential quantifier is satisfied The situation is consistent with our original meaning

14. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 14 Universal Conjunction x = {,, } All purple mushrooms are poisonous  x Purple(x)  Mushroom(x)  Poisonous(x) Universal: all three statements are true is purple and a mushroom and is poisonous “Everything is a poisonous purple mushroom” This is false, and not the original meaning of our sentence

15. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 15 Summary Universal quantifier  Usually used with  to create universal rules Using it with  makes blanket statements about all objects in the world Existential quantifier  Usually used with  to list properties of an object Using it with  creates sentences that do not say much

16. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 16 Example Sentences You can fool some people all the time  x  y Person(x)  Time(y)  CanBeFooled(x,y) Everyone who loves all animals is loved by someone  x  y Animal(y)  Love(x,y)   z Love(z,x) No living man can kill the Witch-King  x Living(x)  Man(x)  CanKill(x, Witch-King)

17. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 17 Wumpus World in FOL Predicate to represent adjacent rooms  x,y,a,b Adjacent([x,y],[a,b])  [a,b]  {[x+1,y], [x-1,y], [x,y+1], [x,y-1]} Sentences to represent the relationship between breezy rooms and pits  r Breeze(r)   s Adjacent(r,s)  Pit(s)  s Pit(s)  (  r Adjacent(r,s)  Breeze(r))

18. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 18 Inference in FOL We know how to do inference in propositional logic Can we reduce FOL to propositional logic and infer from there? Propositionalization Eliminate universal quantifiers Eliminate existential quantifiers

19. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 19 Universal Instantiation A universal variable x in a true sentence P(x) can be substituted by any (and all) ground terms g without variables in the domain of x, and P(g) will be true SUBST({x/g},P(x)) = P(g) Example x  {Bob, James, Tom}  x Strong(x)  Fast(x)  Athlete(x) Strong(Bob)  Fast(Bob)  Athlete(Bob) Strong(James)  Fast(James)  Athlete(James) Strong(Tom)  Fast(Tom)  Athlete(Tom)

20. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 20 Existential Instantiation Any existential variable x in a true sentence P(x) can be replaced by a constant symbol not already in the KB Skolem constant Example  x Purple(x)  Mushroom(x)  Poisonous(x) Purple(C)  Mushroom(C)  Poisonous(C)

21. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 21 Inference Example We know the following All of Bob’s dogs are brown. Fido is a dog. Bob owns Fido. Can we infer that Fido is brown? Translate into FOL KB  x Dog(x)  Own(Bob,x)  Brown(x) Dog(Fido) Own(Bob,Fido) Apply inference rules Universal Instantiation with {x/Fido}: Dog(Fido)  Own(Bob,Fido)  Brown(Fido) And-Introduction: Dog(Fido)  Own(Bob,Fido) Modus Ponens: Brown(Fido)

22. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 22 The set of ground substitution can be infinite Function symbols can be nested infinitely Example: Father(x) = Father(Bob), Father(Father(Bob)), Father(Father(Father(Bob))), … Algorithms can prove sentence is entailed Search every depth successively until sentence is found Algorithms cannot prove sentence is not entailed Will search deeper and deeper, forever Entailment in FOL is semidecidable Problem with Propositionalization

23. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 23 Problem with Propositionalization Modify the previous example  x Dog(x)  Own(Bob,x)  Brown(x) Dog(Fido), Dog(Barky), Dog(Dogbert), Dog(Princess), Dog(Lassie), Dog(K-9) Own(Bob,Fido) Inference with Propositionalization Universal Instantiation: Dog(Fido)  Own(Bob,Fido)  Brown(Fido) Dog(Barky)  Own(Bob,Barky)  Brown(Barky) Dog(Dogbert)  Own(Bob,Dogbert)  Brown(Dogbert) Dog(Princess)  Own(Bob,Princess)  Brown(Princess) Dog(Lassie)  Own(Bob,Lassie)  Brown(Lassie) Dog(K-9)  Own(Bob,K-9)  Brown(K-9) And-Introduction: Dog(Fido)  Own(Bob,Fido) Modus Ponens: Brown(Fido)

24. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 24 Problem with Propositionalization Brown(Fido) is obvious to us Universal Instantiation generates a lot of useless substitutions which are obviously unnecessary (to us) Recall Modus Ponens: (α  β), α β

25. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 25 Generalized Modus Ponens Combine And-Introduction, Universal Elimination, and Modus Ponens p 1 ’, p 2 ’,…, p n ’, (p 1  p 2  …  p n  q) SUBST( , q) p 1 = Dog(x) p 2 = Own(Bob,x) q = Brown(x) p 1 ’ = Dog(Fido) p 2 ’ = Own(Bob,Fido)  = {x/Fido} SUBST( , q) = Brown(Fido)

26. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 26 Inference Example FOL KB  x Dog(x)  Own(Bob,x)  Brown(x) Dog(Fido), Dog(Barky), Dog(Dogbert), Dog(Princess), Dog(Lassie), Dog(K-9) Own(Bob,Fido) Inference with Generalized Modus Ponens Dog(Fido), Own(Bob,Fido), Dog(x)  Own(Bob,x)  Brown(x)  = {x/Fido} Brown(Fido) Requires some form of pattern matching to discover 

27. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 27 Unification Unification algorithm takes two sentences p and q Returns the substitutions to make both sentences match, or failure if none is possible UNIFY(p,q) = , where  is the list of substitutions in p and q Find the matching that places the least restrictions on the variables Most general unifier (MGU)

28. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 28 Unification Given sentences p and q,  is empty Scan p and q left to right If they disagree on terms r and s If r is a variable If s is a complex term and r occurs in s Fail Else Add r/s to  Else if s is a variable (but r is a complex term) If s occurs in r Fail Else Add s/r to  Else (both r and s are complex term) Apply unification to r and s

29. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 29 Unification Rules Function symbols and predicate symbols unify with function symbols and predicate symbols that have identical names and number of arguments Constant symbols can only unify with identical constant symbols Variables can unify with other variables, constant symbols, or function symbols Variables cannot unify with a term in which that variable occurs. x cannot unify with P(x), as that leads to P(P(P(P(…P(x)…)))) That is an occur check error

30. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 30 Unification Examples p = Parents(x, Father(x), Mother(Bob)) q = Parents(Bob, Father(Bob), y) Consider first term of each sentence r = Parents(x, Father(x), Mother(Bob)) s = Parents(Bob, Father(Bob), y) Disagreement r and s are predicate symbols with identical names and number of arguments, so unification is possible Unify arguments r = x, s = Bob, x/Bob r = Father(Bob), s = Father(Bob), no substitution r = Mother(Bob), s = y, y/Mother(Bob)  = {x/Bob, y/Mother(Bob)}

31. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 31 Unification Examples p = Parents(x, Father(x), Mother(Bob)) q = Parents(Bob, Father(y), z) Unify arguments r = x, s = Bob, x/Bob r = Father(Bob), s = Father(y), y/Bob r = Mother(Bob), s = z, z/Mother(Bob)  = {x/Bob, y/Bob, z/Mother(Bob)}

32. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 32 Unification Examples p = Parents(x, Father(x), Mother(Jane)) q = Parents(Bob, Father(y), Mother(y)) Unify arguments r = x, s = Bob, x/Bob r = Father(Bob), s = Father(y), y/Bob r = Mother(Jane), s = Mother(Bob), Fail Fail

33. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 33 Unification Exercises p = Parents(x, Father(x), Mother(x)) q = Parents(Bob, SpermDonor(y), Mother(Bob)) Fail p = Parents(x, Father(x), Mother(x)) q = Parents(Anakin, Mother(Anakin)) Fail p = Parents(Bob, x, Mother(y)) q = Parents(Bob, Father(y), Mother(Bob)) {x/Father(Bob), y/Bob} p = Parents(Bob, x, Mother(y)) q = Parents(Bob, Father(x), Mother(Bob)) Fail

34. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 34 Forward / Backward Chaining Unification algorithm finds  that unifies (p 1 ’, …, p n ’) and (p 1  …  p n ) Makes Generalized Modus Ponens possible Recall from Propositional Logic Inference using Modus Ponens Either forward chaining or backward chaining We have Generalized Modus Ponens for FOL Inference is possible! Forward chaining or backward chaining

35. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 35 Forward / Backward Chaining Forward chaining Data-driven reasoning Start with known symbols Infer new symbols and add to KB Use new symbols to infer more new symbols Repeat until query proven or no new symbols can be inferred Backward chaining Goal-driven reasoning Start with query, try to infer it If there are unknown symbols in the premise of the query, infer them first If there are unknown symbols in the premise of these symbols, infer those first Repeat until query proven or its premise cannot be inferred

36. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 36 Resolution Recall from Propositional Logic ) (α  β), (¬β  γ) (α  γ) Resolution rule is both sound and complete Proof by contradiction Convert KB to conjunctive normal form (CNF) Add negation of query Then perform inference until we reach a contradiction resulting from the negated query Reaching contradiction means query is true

37. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 37 Resolution Steps 1. Convert problem into FOL KB 2. Convert FOL statements in KB to CNF 3. Add negation of query (in CNF) in KB 4. Use FOL resolution rule to infer new clauses from KB 5. Produce a contradiction that proves our query

38. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 38 Conversion to CNF Three new steps (4, 5, 6) & rules in step 3 1. Eliminate biconditionals: (α  β)  ((α  β)  (β  α)) 2. Eliminate implications: (α  β)  (¬α  β) 3. Move/Eliminate negations: ¬(¬α)  α, ¬(α  β)  (¬α  ¬β), ¬(α  β)  (¬α  ¬β), ¬  x P(x)   x ¬P(x), ¬  x P(x)   x ¬P(x) 4. Standardize variables:  x P(x)   x Q(x)   x P(x)   y Q(y) 5. Skolemize:  x P(x)  P(C) 6. Drop universal quantifiers:  x P(x)  P(x) 7. Distribute  over  : (α  (β  γ))  ((α  β)  (α  γ))

39. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 39 Conversion to CNF Everyone who loves all animals is loved by someone  x [  y Animal(y)  Love(x,y)]   y Love(y,x)  x ¬[  y Animal(y)  Love(x,y)]   y Love(y,x)  x  y ¬[Animal(y)  Love(x,y)]   y Love(y,x)  x  y ¬Animal(y)  ¬Love(x,y)   y Love(y,x)  x  y ¬Animal(y)  ¬Love(x,y)   z Love(z,x)  x ¬Animal(C 1 )  ¬Love(x,C 1 )  Love(C 2,x) ¬Animal(C 1 )  ¬Love(x,C 1 )  Love(C 2,x)

40. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 40 Conversion to CNF There is a book which everyone buys only if there is a class that requires it  x Book(x)  [  y Buy(y,x)]  [  y Class(y)  Requires(y,x)]  x Book(x)  {[  y Buy(y,x)]  [  y Class(y)  Requires(y,x)]}  {[  y Class(y)  Requires(y,x)]  [  y Buy(y,x)]}  x Book(x)  {¬[  y Buy(y,x)]  [  y Class(y)  Requires(y,x)]}  {¬[  y Class(y)  Requires(y,x)]  [  y Buy(y,x)]}  x Book(x)  {[  y ¬Buy(y,x)]  [  y Class(y)  Requires(y,x)]}  {[  y ¬Class(y)  ¬Requires(y,x)]  [  y Buy(y,x)]}  x Book(x)  {  y ¬Buy(y,x)  [  z Class(z)  Requires(z,x)]}  {[  z ¬Class(z)  ¬Requires(z,x)]   y Buy(y,x)} Book(C 1 )  {¬Buy(C 2,C 1 )  [Class(C 3 )  Requires(C 3,C 1 )]}  {[  z ¬Class(z)  ¬Requires(z,C 1 )]   y Buy(y,C 1 )} Book(C 1 )  {¬Buy(C 2,C 1 )  [Class(C 3 )  Requires(C 3,C 1 )]}  {¬Class(z)  ¬Requires(z,C 1 )  Buy(y,C 1 )} Book(C 1 )  [¬Buy(C 2,C 1 )  Class(C 3 )]  [¬Buy(C 2,C 1 )  Requires(C 3,C 1 )]  [¬Class(z)  ¬Requires(z,C 1 )  Buy(y,C 1 )]

41. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 41 FOL Resolution ) (α  β’), (¬β  γ) SUBST( , α  γ)  = UNIFY(β’, ¬β) Example Animal(Dog(x))  Love(Bob,x) ¬Love(y,z)  ¬Hunt(y,z)  = {y/Bob, z/x} Animal(Dog(x))  ¬Hunt(Bob,x)

42. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 42 Example of Resolution Consider the story Anyone passing their ECE457 exam and winning the lottery is happy. Anyone who studies or is lucky can pass all their exams. Bob did not study but is lucky. Anyone who’s lucky can win the lottery. Is Bob happy?

43. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 43 Example of Resolution Convert to FOL Anyone passing their ECE457 exam and winning the lottery is happy.  x Pass(x, ECE457)  Win(x,Lottery)  Happy(x) Anyone who studies or is lucky can pass all their exams.  x  y Study(x)  Lucky(x)  Pass(x,y) Bob did not study but is lucky. ¬Study(Bob)  Lucky(Bob) Anyone who’s lucky can win the lottery.  x Lucky(x)  Win(x,Lottery) Is Bob happy? Happy(Bob)

44. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 44 Example of Resolution Convert to CNF  x Pass(x, ECE457)  Win(x,Lottery)  Happy(x)  x ¬(Pass(x, ECE457)  Win(x,Lottery))  Happy(x)  x ¬Pass(x, ECE457)  ¬Win(x,Lottery)  Happy(x) ¬Pass(x, ECE457)  ¬Win(x,Lottery)  Happy(x)  x  y Study(x)  Lucky(x)  Pass(x,y)  x  y ¬(Study(x)  Lucky(x))  Pass(x,y)  x  y (¬Study(x)  ¬Lucky(x))  Pass(x,y) (¬Study(x)  ¬Lucky(x))  Pass(x,y) (¬Study(x)  Pass(x,y))  (¬Lucky(x)  Pass(x,y))

45. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 45 Example of Resolution Convert to CNF ¬Study(Bob)  Lucky(Bob)  x Lucky(x)  Win(x,Lottery)  x ¬Lucky(x)  Win(x,Lottery) ¬Lucky(x)  Win(x,Lottery) Negation of query for resolution ¬Happy(Bob)

46. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 46 Example of Resolution Content of KB ¬Pass(x 1, ECE457)  ¬Win(x 1,Lottery)  Happy(x 1 ) ¬Study(x 2 )  Pass(x 2,y 1 ) ¬Lucky(x 3 )  Pass(x 3,y 2 ) ¬Study(Bob) Lucky(Bob) ¬Lucky(x 4 )  Win(x 4,Lottery) ¬Happy(Bob)

47. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 47 Example of Resolution (¬Pass(x 1, ECE457)  ¬Win(x 1,Lottery)  Happy(x 1 ))  (¬Study(x 2 )  Pass(x 2,y 1 ))  (¬Lucky(x 3 )  Pass(x 3,y 2 ))  ¬Study(Bob)  Lucky(Bob)  (¬Lucky(x 4 )  Win(x 4,Lottery))  ¬Happy(Bob) {x 1 /Bob} (¬Pass(Bob, ECE457)  ¬Win(Bob,Lottery))  (¬Study(x 2 )  Pass(x 2,y 1 ))  (¬Lucky(x 3 )  Pass(x 3,y 2 ))  ¬Study(Bob)  Lucky(Bob)  (¬Lucky(x 4 )  Win(x 4,Lottery))  ¬Happy(Bob) {x 4 /Bob}

48. ECE457 Applied Artificial Intelligence R. Khoury (2007)Page 48 Example of Resolution (¬Pass(Bob, ECE457)  ¬Lucky(Bob))  (¬Study(x 2 )  Pass(x 2,y 1 ))  (¬Lucky(x 3 )  Pass(x 3,y 2 ))  ¬Study(Bob)  Lucky(Bob)  ¬Happy(Bob) No substitution needed ¬Pass(Bob, ECE457)  (¬Study(x 2 )  Pass(x 2,y 1 ))  (¬Lucky(x 3 )  Pass(x 3,y 2 ))  ¬Study(Bob)  Lucky(Bob)  ¬Happy(Bob) {x 3 /Bob, y 2 /ECE457} ¬Pass(Bob, ECE457)  (¬Study(x 2 )  Pass(x 2,y 1 ))  ¬Lucky(Bob)  ¬Study(Bob)  Lucky(Bob)  ¬Happy(Bob) No substitution needed