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First-Order Logic (FOL) aka. predicate calculus

First-Order Logic (FOL) aka. predicate calculus. First-Order Logic (FOL) Syntax. User defines these primitives: Constant symbols (i.e., the "individuals" in the world) E.g., Mary, 3

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First-Order Logic (FOL) aka. predicate calculus

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  1. First-Order Logic (FOL)aka. predicate calculus

  2. First-Order Logic (FOL) Syntax • User defines these primitives: • Constant symbols (i.e., the "individuals" in the world) E.g., Mary, 3 • Function symbols (mapping individuals to individuals) E.g., father-of(Mary) = John, color-of(Sky) = Blue • Predicate symbols (mapping from individuals to truth values) E.g., greater(5,3), green(Grass), color(Grass, Green)

  3. First-Order Logic (FOL) Syntax… • FOL supplies these primitives: • Variable symbols. E.g., x,y • Connectives. Same as in PL: not (~), and (^), or (v), implies (=>), if and only if (<=>) • Quantifiers: Universal (A) and Existential (E)

  4. Quantifiers • Universal quantification corresponds to conjunction ("and") in that (Ax)P(x) means that P holds for all values of x in the domain associated with that variable. • E.g., (Ax)dolphin(x) => mammal(x) • Existential quantification corresponds to disjunction ("or") in that (Ex)P(x) means that P holds for some value of x in the domain associated with that variable. • E.g., (Ex) mammal(x) ^ lays-eggs(x) • Universal quantifiers are usually used with "implies" to form "if-then rules." • E.g., (Ax) cs15-381-student(x) => smart(x) means "All cs15-381 students are smart." • You rarely use universal quantification to make blanket statements about every individual in the world: (Ax)cs15-381-student(x) ^ smart(x) meaning that everyone in the world is a cs15-381 student and is smart.

  5. Quantifiers … • Existential quantifiers are usually used with "and" to specify a list of properties or facts about an individual. • E.g., (Ex) cs15-381-student(x) ^ smart(x) means "there is a cs15-381 student who is smart." • A common mistake is to represent this English sentence as the FOL sentence: (Ex) cs15-381-student(x) => smart(x) • Switching the order of universal quantifiers does not change the meaning: (Ax)(Ay)P(x,y) is logically equivalent to (Ay)(Ax)P(x,y). Similarly, you can switch the order of existential quantifiers. • Switching the order of universals and existentials does change meaning: • Everyone likes someone: (Ax)(Ey)likes(x,y) • Someone is liked by everyone: (Ey)(Ax)likes(x,y)

  6. First-Order Logic (FOL) Syntax… • Sentences are built up of terms and atoms: • A term (denoting a real-world object) is a constant symbol, a variable symbol, or a function e.g. left-leg-of ( ). For example, x and f(x1, ..., xn) are terms, where each xi is a term. • An atom (which has value true or false) is either an n-place predicate of n terms, or, if P and Q are atoms, then ~P, P V Q, P ^ Q, P => Q, P <=> Q are atoms • A sentence is an atom, or, if P is a sentence and x is a variable, then (Ax)P and (Ex)P are sentences • A well-formed formula (wff) is a sentence containing no "free" variables. I.e., all variables are "bound" by universal or existential quantifiers. • E.g., (Ax)P(x,y) has x bound as a universally quantified variable, but y is free.

  7. Translating English to FOL • Every gardener likes the sun.(Ax) gardener(x) => likes(x,Sun) • You can fool some of the people all of the time.(Ex)(At) (person(x) ^ time(t)) => can-fool(x,t) • You can fool all of the people some of the time.(Ax)(Et) (person(x) ^ time(t) => can-fool(x,t) • All purple mushrooms are poisonous.(Ax) (mushroom(x) ^ purple(x)) => poisonous(x)

  8. Translating English to FOL… • No purple mushroom is poisonous.~(Ex) purple(x) ^ mushroom(x) ^ poisonous(x) or, equivalently,(Ax) (mushroom(x) ^ purple(x)) => ~poisonous(x) • Deb is not tall.~tall(Deb) • X is above Y if X is on directly on top of Y or else there is a pile of one or more other objects directly on top of one another starting with X and ending with Y.(Ax)(Ay) above(x,y) <=> (on(x,y) v (Ez) (on(x,z) ^ above(z,y)))

  9. Inference Rules for FOL • Inference rules for PL apply to FOL as well. For example, Modus Ponens, And-Introduction, And-Elimination, etc. • New sound inference rules for use with quantifiers: • Universal EliminationIf (Ax)P(x) is true, then P(c) is true, where c is a constant in the domain of x. For example, from (Ax)eats(Ziggy, x) we can infer eats(Ziggy, IceCream). • The variable symbol can be replaced by any ground term, i.e., any constant symbol or function symbol applied to ground terms only. • Existential IntroductionIf P(c) is true, then (Ex)P(x) is inferred. • For example, from eats(Ziggy, IceCream) we can infer (Ex)eats(Ziggy, x). • All instances of the given constant symbol are replaced by the new variable symbol. Note that the variable symbol cannot already exist anywhere in the expression. • Existential EliminationFrom (Ex)P(x) infer P(c). • For example, from (Ex)eats(Ziggy, x) infer eats(Ziggy, Cheese). • Note that the variable is replaced by a brand new constant that does not occur in this or any other sentence in the Knowledge Base. In other words, we don't want to accidentally draw other inferences about it by introducing the constant. All we know is there must be some constant that makes this true, so we can introduce a brand new one to stand in for that (unknown) constant.

  10. Inference Rules for FOL • Generalized Modus Ponens (GMP) • Combines And-Introduction, Universal-Elimination, and Modus Ponens • E.g.: from P(c), Q(c), and (Ax)(P(x) ^ Q(x)) => R(x), derive R(c) • A substitution list Theta = {v1/t1, v2/t2, ..., vn/tn} means to replace all occurrences of variable symbol vi by term ti. • Substitutions are made in left-to-right order in the list. • E.g.: subst({x/IceCream, y/Ziggy},eats(y,x)) = eats(Ziggy, IceCream)

  11. Incompleteness of the generalized modus ponens inference rule Backward chaining Forward chaining Generalized Modus Ponens x P(x) => Q(x) x P(x) => Q(x) x Q(x) => S(x) x R(x) => S(x)  Cannot be converted to Horn Want to conclude S(A) S(A) is true if Q(A) or R(A) is true, and one of those must be true because either P(A) or P(A)

  12. Generalized Modus Ponens in Horn FOL • Generalized Modus Ponens (GMP) is complete for KBs containing only Horn clauses • A Horn clause is a sentence of the form:(Ax) (P1(x) ^ P2(x) ^ ... ^ Pn(x)) => Q(x)where there are 0 or more Pi's, and the Pi's and Q are positive (i.e., un-negated) literals • Horn clauses represent a subset of the set of sentences representable in FOL. For example, P(a) v Q(a) is a sentence in FOL but is not a Horn clause. • Natural deduction using GMP is complete for KBs containing only Horn clauses. Proofs start with the given axioms/premises in KB, deriving new sentences using GMP until the goal/query sentence is derived. This defines a forward chaining inference procedure because it moves "forward" from the KB to the goal.

  13. Example of forward chaining • Example: KB = All cats like fish, cats eat everything they like, and Ziggy is a cat. In FOL, KB = • (Ax) cat(x) => likes(x, Fish) • (Ax)(Ay) (cat(x) ^ likes(x,y)) => eats(x,y) • cat(Ziggy) • Goal query: Does Ziggy eat fish? Proof: • Use GMP with (1) and (3) to derive: 4. likes(Ziggy, Fish) • Use GMP with (3), (4) and (2) to derive eats(Ziggy, Fish) • So, Yes, Ziggy eats fish. Data-driven

  14. Backward chaining • Backward-chaining deduction using GMP is complete for KBs containing only Horn clauses. Proofs start with the goal query, find implications that would allow you to prove it, and then prove each of the antecedents in the implication, continuing to work "backwards" until we get to the axioms, which we know are true. • Example: Does Ziggy eat fish? • To prove eats(Ziggy, Fish), first see if this is known from one of the axioms directly. Here it is not known, so see if there is a Horn clause that has the consequent (i.e., right-hand side) of the implication matching the goal. Proof: Goal Driven • Goal matches RHS of Horn clause (2), so try and prove new sub-goals cat(Ziggy) and likes(Ziggy, Fish) that correspond to the LHS of (2) • cat(Ziggy) matches axiom (3), so we've "solved" that sub-goal • likes(Ziggy, Fish) matches the RHS of (1), so try and prove cat(Ziggy) • cat(Ziggy) matches (as it did earlier) axiom (3), so we've solved this sub-goal • There are no unsolved sub-goals, so we're done. Yes, Ziggy eats fish.

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