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## Herbrand Interpretations

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**Let us think of interpretations for FOPC that are more**like interpretations for prop logic Herbrand Interpretations • Herbrand Universe • All constants • Rao,Pat • All “ground” functional terms • Son-of(Rao);Son-of(Pat); • Son-of(Son-of(…(Rao)))…. • Herbrand Base • All ground atomic sentences made with terms in Herbrand universe • Friend(Rao,Pat);Friend(Pat,Rao);Friend(Pat,Pat);Friend(Rao,Rao) • Friend(Rao,Son-of(Rao)); • Friend(son-of(son-of(Rao),son-of(son-of(son-of(Pat)) • We can think of elements of HB as propositions; interpretations give T/F values to these. Given the interpretation, we can compute the value of the FOPC database sentences If there are n constants; and p k-ary predicates, then --Size of HU = n --Size of HB = p*nk But if there is even one function, then |HU| is infinity and so is |HB|. --So, when there are no function symbols, FOPC is really just syntactic sugaring for a (possibly much larger) propositional database**But what about Godel?**• In First Order Logic • We have finite set of constants • Quantification allowed only over variables… • Godel’s incompleteness theorem holds only in a system that includes “mathematical induction”—which is an axiom schema that requires infinitely many FOPC statements • If a property P is true for 0, and whenever it is true for number n, it is also true for number n+1, then the property P is true for all natural numbers • You can’t write this in first order logic without writing it once for each P (so, you will have to write infinite number of FOPC statements) • So, a finite FOPC database is still semi-decidable in that we can prove all provably true theorems**Proof-theoretic Inference in first order logic**• For “ground” sentences (i.e., sentences without any quantification), all the old rules work directly (think of ground atomic sentences as propositions) • P(a,b)=> Q(a); P(a,b) |= Q(a) • ~P(a,b) V Q(a) resolved with P(a,b) gives Q(a) • What about quantified sentences? • May be infer ground sentences from them…. • Universal Instantiation (a universally quantified statement entails every instantiation of it) • Existential instantiation (an existentially quantified statement holds for some term (not currently appearing in the KB). • Can we combine these (so we can avoid unnecessary instantiations?) Yes. Generalized modus ponens • Needs UNIFICATION**UI can be applied several**times to add new sentences --The resulting KB is equivalent to the old one EI can only applied once --The resulting DB is not equivalent to the old one BUT will be satisfiable only when the old one is**How about knows(x,f(x)) knows(u,u)?**x/u; u/f(u)leads to infinite regress (“occurs check”)**GMP can be used in the “forward” (aka “bottom-up”)**fashion where we start from antecedents, and assert the consequent or in the “backward” (aka “top-down”) fashion where we start from consequent, and subgoal on proving the antecedents.**Apt-pet**• An apartment pet is a pet that is small • Dog is a pet • Cat is a pet • Elephant is a pet • Dogs, cats and skunks are small. • Fido is a dog • Louie is a skunk • Garfield is a cat • Clyde is an elephant • Is there an apartment pet?**Efficiency can be improved by re-ordering subgoals**adaptively e.g., try to prove Pet before Small in Lilliput Island; and Small before Pet in pet-store.**Similar to “Integer Programming” or “Constraint**Programming”**Generate compilable**matchers for each pattern, and use them**y/z;x/Rao**~loves(z,Rao) z/SK(rao);x’/rao Example of FOPC Resolution.. Everyone is loved by someone If x loves y, x will give a valentine card to y Will anyone give Rao a valentine card?**Finding where you left your key..**Atkey(Home) V Atkey(Office) 1 Where is the key? Ex Atkey(x) Negate Forall x ~Atkey(x) CNF ~Atkey(x) 2 Resolve 2 and 1 with x/home You get Atkey(office) 3 Resolve 3 and 2 with x/office You get empty clause So resolution refutation “found” that there does exist a place where the key is… Where is it? what is x bound to? x is bound to office once and home once. so x is either home or office**Existential proofs..**• Are there irrational numbers p and q such that pq is rational? This and the previous examples show that resolution refutation is powerful enough to model existential proofs. In contrast, generalized modus ponens is only able to model constructive proofs.. Rational Irrational**Existential proofs..**• The previous example shows that resolution refutation is powerful enough to model existential proofs. In contrast, generalized modus ponens is only able to model constructive proofs.. • (We also discussed a cute example of existential proof—is it possible for an irrational number power another irrational number to be a rational number—we proved it is possible, without actually giving an example).**GMP vs. Resolution Refutation**• While resolution refutation is a complete inference for FOPC, it is computationally semi-decidable, which is a far cry from polynomial property of GMP inferences. • So, most common uses of FOPC involve doing GMP-style reasoning rather than the full theorem-proving.. • There is a controversy in the community as to whether the right way to handle the computational complexity is to • a. Develop “tractable subclasses” of languages and require the expert to write all their knowlede in the procrustean beds of those sub-classes (so we can claim “complete and tractable inference” for that class) OR • Let users write their knowledge in the fully expressive FOPC, but just do incomplete (but sound) inference. • See Doyle & Patil’s “Two Theses of Knowledge Representation”