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Intelligent Agents. Outline Agents and environments Rationality PEAS (Performance measure, Environment, Actuators, Sensors) Environment types Agent types.

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**Inference in first-order logic**

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**Outline Reducing first-order inference to propositional inference**

Unification Generalized Modus Ponens Forward chaining Backward chaining Resolution

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**Universal instantiation (UI)**

Every instantiation of a universally quantified sentence is entailed by it: v α Subst({v/g}, α) for any variable v and ground term g(without any variable) E.g., x King(x) Greedy(x) Evil(x) yields: King(John) Greedy(John) Evil(John) King(Richard) Greedy(Richard) Evil(Richard) King(Father(John)) Greedy(Father(John)) Evil(Father(John)) .

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**Existential instantiation (EI)**

For any sentence α, variable v, and constant symbol k that does not appear elsewhere in the knowledge base: v α Subst({v/k}, α) E.g., x Crown(x) OnHead(x,John) yields: Crown(C1) OnHead(C1,John) provided C1 is a new constant symbol, called a Skolem constant

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**Reduction to propositional inference**

Suppose the KB contains just the following: x King(x) Greedy(x) Evil(x) King(John) Greedy(John) Brother(Richard,John) Instantiating the universal sentence in all possible ways, we have: King(John) Greedy(John) Evil(John) King(Richard) Greedy(Richard) Evil(Richard) The new KB is propositionalized: proposition symbols are King(John), Greedy(John), Evil(John), King(Richard), etc.

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Reduction contd. Every FOL KB can be propositionalized so as to preserve entailment A ground sentence is entailed by new KB iff entailed by original KB Idea: propositionalize KB and query, apply resolution, return result Problem: with function symbols, there are infinitely many ground terms, e.g., Father(Father(Father(John)))

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Reduction contd. Theorem: Herbrand (1930). If a sentence α is entailed by an FOL KB, it is entailed by a finite subset of the propositionalized KB Idea: For n = 0 to ∞ do create a propositional KB by instantiating with depth-$n$ terms see if α is entailed by this KB Problem: works if α is entailed, loops if α is not entailed Theorem: Turing (1936), Church (1936) Entailment for FOL is semidecidable algorithms exist that say yes to every entailed sentence no algorithm exists that also says no to every nonentailed sentence.

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**Problems with propositionalization**

Propositionalization seems to generate lots of irrelevant sentences. Example from: x King(x) Greedy(x) Evil(x) King(John) y Greedy(y) Brother(Richard,John) it seems obvious that Evil(John), but propositionalization produces lots of facts such as Greedy(Richard) that are irrelevant With p k-ary predicates and n constants, there are p·nk instantiations.

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Unification We can get the inference immediately if we can find a substitution θ such that King(x) and Greedy(x) match King(John) and Greedy(y) θ = {x/John,y/John} works Unify(α,β) = θ if αθ = βθ p q θ Knows(John,x) Knows(John,Jane) Knows(John,x) Knows(y,OJ) Knows(John,x) Knows(y,Mother(y)) Knows(John,x) Knows(x,OJ) Standardizing apart eliminates overlap of variables, e.g., Knows(John,z27) Knows(z17,OJ) {x/Jane} {x/OJ, y/John} {x/Mother(John),y/John} No substitution possible yet.

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Unification We can get the inference immediately if we can find a substitution θ such that King(x) and Greedy(x) match King(John) and Greedy(y) θ = {x/John,y/John} works Unification finds substitutions that make different logical expressions look identical UNIFY takes two sentences and returns a unifier for them, if one exists UNIFY(p,q) = where SUBST(,p) = SUBST (,q) Basically, find a that makes the two clauses look alike

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**Unification Examples UNIFY(Knows(John,x), Knows(John,Jane)) = {x/Jane}**

UNIFY(Knows(John,x), Knows(y,Bill)) = {x/Bill, y/John} UNIFY(Knows(John,x), Knows(y,Mother(y))= {y/John, x/Mother(John) UNIFY(Knows(John,x), Knows(x,Elizabeth)) = fail Last example fails because x would have to be both John and Elizabeth We can avoid this problem by standardizing: The two statements now read UNIFY(Knows(John,x), Knows(z,Elizabeth)) This is solvable: UNIFY(Knows(John,x), Knows(z,Elizabeth)) = {x/Elizabeth,z/John}

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**Unification To unify Knows(John,x) and Knows(y,z)**

Can use θ = {y/John, x/z } Or θ = {y/John, x/John, z/John} The first unifier is more general than the second. There is a single most general unifier (MGU) that is unique up to renaming of variables. MGU = { y/John, x/z }

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**Unification Unification algorithm:**

Recursively explore the two expressions side by side Build up a unifier along the way Fail if two corresponding points do not match

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**The unification algorithm**

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**The unification algorithm**

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**Simple Example Brother(x,John)Father(Henry,y)Mother(z,John)**

Brother(Richard,x)Father(y,Richard)Mother(Eleanore,x)

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**Generalized Modus Ponens (GMP)**

p1', p2', … , pn', ( p1 p2 … pn q) qθ p1' is King(John) p1 is King(x) p2' is Greedy(y) p2 is Greedy(x) θ is {x/John,y/John} q is Evil(x) q θ is Evil(John) GMP used with KB of definite clauses (exactly one positive literal) All variables assumed universally quantified where pi'θ = pi θ for all i

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**Soundness of GMP Need to show that p1', …, pn', (p1 … pn q) ╞ qθ**

provided that pi'θ = piθ for all I Lemma: For any sentence p, we have p ╞ pθ by UI (p1 … pn q) ╞ (p1 … pn q)θ = (p1θ … pnθ qθ) p1', \; …, \;pn' ╞ p1' … pn' ╞ p1'θ … pn'θ From 1 and 2, qθ follows by ordinary Modus Ponens

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Storage and Retrieval Use TELL and ASK to interact with Inference Engine Implemented with STORE and FETCH STORE(s) stores sentence s FETCH(q) returns all unifiers that the query q unifies with Example: q = Knows(John,x) KB is: Knows(John,Jane), Knows(y,Bill), Knows(y,Mother(y)) Result is 1={x/Jane}, 2=, 3= {John/y,x/Mother(y)}

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**Storage and Retrieval First approach:**

Create a long list of all propositions in Knowledge Base Attempt unification with all propositions in KB Works, but is inefficient Need to restrict unification attempts to sentences that have some chance of unifying Index facts in KB Predicate Indexing Index predicates: All “Knows” sentences in one bucket All “Loves” sentences in another Use Subsumption Lattice (see below)

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**Storing and Retrieval Subsumption Lattice**

Child is obtained from parent through a single substitution Lattice contains all possible queries that can be unified with it. Works well for small lattices Predicate with n arguments has a 2n lattice Structure of lattice depends on whether the base contains repeated variables Knows(x,y) Knows(x,John) Knows(x,x) Knows(John,x) Knows(John,John)

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**Forward Chaining Forward Chaining Idea:**

Start with atomic sentences in the KB Apply Modus Ponens Add new atomic sentences until no further inferences can be made Works well for a KB consisting of Situation Response clauses when processing newly arrived data

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**Forward Chaining First Order Definite Clauses**

Disjunctions of literals of which exactly one is positive: Example: King(x) Greedy(x) Evil(x) King(John) Greedy(y) First Order Definite Clauses can include variables Variables are assumed to be universally quantified Greedy(y) means y Greedy(y) Not every KB can be converted into first definite clauses

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**Example knowledge base**

The law says that it is a crime for an American to sell weapons to hostile nations. The country Nono, an enemy of America, has some missiles, and all of its missiles were sold to it by Colonel West, who is American. Prove that Col. West is a criminal

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**Example knowledge base contd.**

... it is a crime for an American to sell weapons to hostile nations: American(x) Weapon(y) Sells(x,y,z) Hostile(z) Criminal(x) Nono … has some missiles, i.e., x Owns(Nono,x) Missile(x): Owns(Nono,M1) and Missile(M1) … all of its missiles were sold to it by Colonel West Missile(x) Owns(Nono,x) Sells(West,x,Nono) Missiles are weapons: Missile(x) Weapon(x) An enemy of America counts as "hostile“: Enemy(x,America) Hostile(x) West, who is American … American(West) The country Nono, an enemy of America … Enemy(Nono,America)

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**Forward chaining algorithm**

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**Forward chaining proof**

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**Forward chaining proof**

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**Forward chaining proof**

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**Properties of forward chaining**

Sound and complete for first-order definite clauses Datalog = first-order definite clauses + no functions FC terminates for Datalog in finite number of iterations May not terminate in general if α is not entailed This is unavoidable: entailment with definite clauses is semidecidable

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**Efficiency of forward chaining**

Incremental forward chaining: no need to match a rule on iteration k if a premise wasn't added on iteration k-1 match each rule whose premise contains a newly added positive literal Matching itself can be expensive: Database indexing allows O(1) retrieval of known facts e.g., query Missile(x) retrieves Missile(M1) Forward chaining is widely used in deductive databases

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Hard matching example Diff(wa,nt) Diff(wa,sa) Diff(nt,q) Diff(nt,sa) Diff(q,nsw) Diff(q,sa) Diff(nsw,v) Diff(nsw,sa) Diff(v,sa) Colorable() Diff(Red,Blue) Diff (Red,Green) Diff(Green,Red) Diff(Green,Blue) Diff(Blue,Red) Diff(Blue,Green) Colorable() is inferred iff the CSP has a solution CSPs include 3SAT as a special case, hence matching is NP-hard

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Backward Chaining Improves on Forward Chaining by not making irrelevant conclusions Alternatives to backward chaining: restrict forward chaining to a relevant set of forward rules Rewrite rules so that only relevant variable bindings are made: Use a magic set Example: Rewrite rule: Magic(x)American(x) Weapon(x) Sells(x,y,z) Hostile(z)Criminal(x) Add fact: Magic(West)

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**Backward Chaining Idea:**

Given a query, find all substitutions that satisfy the query. Algorithm: Work on lists of goals, starting with original query Algo finds every clause in the KB that unifies with the positive literal (head) and adds remainder (body) to list of goals

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**Backward chaining algorithm**

SUBST(COMPOSE(θ1, θ2), p) = SUBST(θ2, SUBST(θ1, p))

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**Backward chaining example**

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**Backward chaining example**

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**Backward chaining example**

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**Backward chaining example**

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**Backward chaining example**

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**Backward chaining example**

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**Backward chaining example**

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**Backward chaining example**

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**Properties of backward chaining**

Depth-first recursive proof search: space is linear in size of proof Incomplete due to infinite loops fix by checking current goal against every goal on stack Inefficient due to repeated subgoals (both success and failure) fix using caching of previous results (extra space) Widely used for logic programming

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**Logic programming: Prolog**

Algorithm = Logic + Control Basis: backward chaining with Horn clauses + bells & whistles Widely used in Europe, Japan (basis of 5th Generation project) Program = set of clauses: head :- literal1, … literaln. criminal(X) :- american(X), weapon(Y), sells(X,Y,Z), hostile(Z). Depth-first, left-to-right backward chaining Built-in predicates for arithmetic etc., e.g., X is Y*Z+3 Built-in predicates that have side effects (e.g., input and output predicates, assert/retract predicates) Closed-world assumption ("negation as failure") e.g., given alive(X) :- not dead(X). alive(joe) succeeds if dead(joe) fails

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**Prolog Appending two lists to produce a third: append([],Y,Y).**

append([X|L],Y,[X|Z]) :- append(L,Y,Z). query: append(A,B,[1,2]) ? answers: A=[] B=[1,2] A=[1] B=[2] A=[1,2] B=[]

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**Prolog Has problems with repeated states and infinite paths**

Example: Path finding in graphs path(X,Z) :- link(X,Z) path(X,Z) :- path(X,Y),link(Y,Z) path(a,c) A B C link(a,c) link(Y,c) fail path(a,Y) { Y/b} link(a,b) {Y/b }

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**Prolog Has problems with repeated states and infinite paths**

Example: Path finding in graphs path(X,Z) :- path(X,Y),link(Y,Z) path(X,Z) :- link(X,Z) path(a,c) A B C path(a,Y) link(Y,b) fail path(a,Y’) link(Y’,Y) path(a,Y) link(Y,b)

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Resolution Resolution for propositional logic is a complete inference procedure Existence of complete proof procedures in Mathematics would entail: All conjectures can be established mechanically All mathematics can be established as the logical consequence of a set of fundamental axioms Gődel 1930: Completeness Theorem for first order logic Any entailed sentence has a finite proof No algorithm given until J.A. Robinson’s resolution algorithm in 1965 Gődel 1931: Incompleteness Theorem: Any logical system with induction is necessarily incomplete There are sentences that are entailed, but not proof can be given

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**Resolution First order logic requires sentences in CNF**

Conjunctive Normal Form: Each clause is a disjunction of literals, but literals can contain variables, which are assumed to be universally quantified Example: Convert x American(x) Weapon(y) Sells(x,y,z) Hostile(z) Criminal(x) American(x) Weapon(y) Sells(x,y,z) Hostile(z) Criminal(x)

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**Conversion to CNF Everyone who loves all animals is loved by someone:**

x [y Animal(y) Loves(x,y)] [y Loves(y,x)] 1. Eliminate biconditionals and implications x [y Animal(y) Loves(x,y)] [y Loves(y,x)] 2. Move inwards: x p ≡ x p, x p ≡ x p x [y (Animal(y) Loves(x,y))] [y Loves(y,x)] x [y Animal(y) Loves(x,y)] [y Loves(y,x)] x [y Animal(y) Loves(x,y)] [y Loves(y,x)]

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Conversion to CNF Standardize variables: each quantifier should use a different one x [y Animal(y) Loves(x,y)] [z Loves(z,x)] Skolemize: a more general form of existential instantiation. Each existential variable is replaced by a Skolem function of the enclosing universally quantified variables: x [Animal(F(x)) Loves(x,F(x))] Loves(G(x),x) Drop universal quantifiers: [Animal(F(x)) Loves(x,F(x))] Loves(G(x),x) Distribute over : [Animal(F(x)) Loves(G(x),x)] [Loves(x,F(x)) Loves(G(x),x)]

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**Resolution Inference Rule**

Full first-order version: l1 ··· lk, m1 ··· mn Subst(θ ,l1 ··· li-1 li+1 ··· lk m1 ··· mj-1 mj+1 ··· mn)θ where Unify(li, mj) = θ. The two clauses are assumed to be standardized apart so that they share no variables. For example, Rich(x) Unhappy(x) Rich(Ken) Unhappy(Ken) with θ = {x/Ken} Apply resolution steps to CNF(KB α); complete for FOL

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Resolution Show KB ⊢ α by showing that KB α is unsatisfyable

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**Resolution Example Everyone who loves all animals is loved by someone**

Anyone who kills an animal is loved by no one. Jack loves all animals. Either Jack or Curiosity killed the cat, who is named Tuna All dogs kill a cats Rintintin is a dog Question: Did Curiosity kill the cat?

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**Resolution Example Everyone who loves all animals is loved by someone.**

Formulate in FOL x [y [Animal(y) Loves(x,y)]] [z Loves(z,x)] Remove Implications x [[y Animal(y) Loves(x,y)]] [z Loves(z,x)] x [[y Animal(y) Loves(x,y)]] [z Loves(z,x)] Move negation inward x [y Animal(y) Loves(x,y)] [z Loves(z,x)] x [y Animal(y) Loves(x,y)] [z Loves(z,x)] Skolemize x [ Animal(F(x)) Loves(x,F(x))] [Loves(G(x),x)] N.B.: Argument of Skolemization function are all universally qualified variables Drop universal quantifier [ Animal(F(x)) Loves(x,F(x))] [Loves(G(x),x)] Use distributive law (and get two clauses) Animal(F(x)) Loves(G(x),x); Loves(x,F(x)) Loves(G(x),x)

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**Resolution Example Anyone who kills an animal is loved by no one.**

Transfer to FOL x [y (Animal(y) Kills(x,y)] (z Loves(z,x) Remove Implications x [y (Animal(y) Kills(x,y)] (z Loves(z,x) Move negations inwards x [ y Animal(y) Kills(x,y)] (z Loves(z,x)) Remove quantifiers Animal(y) Kills(x,y) Loves(z,x)

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**Resolution Example Jack loves all animals. FOL form**

x [Animal(x) Loves(Jack, x)] Remove implications x [Animal(x) Loves(Jack, x)] Remove quantifier Animal(x) Loves(Jack, x)

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Resolution Example Either Jack or Curiosity killed the cat, who is named Tuna. FOL form Kills(Jack,Tuna) Kills(Curiosity,Tuna); Cat(Tuna)

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**Resolution Example All cats are animals FOL form Remove implications**

x [Cat(x) Animal(x) Remove implications x [Cat(x) Animal(x)] Remove quantifier Cat(x) Animal(x)

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**Resolution Example All dogs kill a cats FOL form Remove implications**

x [Dog(x) y[Cat(y) Kills(x,y)]] Remove implications x [Dog(x) y[Cat(y) Kills(x,y)]] Skolemize x [Dog(x) [Cat(H(x)) Kills(x,H(x))]] Drop universal quantifiers Dog(x) [Cat(H(x)) Kills(x,H(x))] Distribute (and obtain two clauses) Dog(x) Cat(H(x); Dog(x) Kills(x,H(x))]

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Resolution Example Rintintin is a dog FOL form Dog(Rintintin)

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**Resolution Example Animal(F(x)) Loves(G(x),x)**

Loves(x,F(x)) Loves(G(x),x) Animal(y) Kills(x,y) Loves(z,x) Animal(x) Loves(Jack, x) Kills(Jack,Tuna) Kills(Curiosity,Tuna) Cat(Tuna) Cat(x) Animal(x) Dog(x) Cat(H(x) Dog(x) Kills(x,H(x))] Dog(Rintintin)

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**Resolution Example Animal(F(x)) Loves(G(x),x)**

Loves(x,F(x)) Loves(G(x),x) Animal(y) Kills(x,y) Loves(z,x) Animal(x) Loves(Jack, x) Kills(Jack,Tuna) Kills(Curiosity,Tuna) Cat(Tuna) Cat(x) Animal(x) Dog(x) Cat(H(x) Dog(x) Kills(x,H(x))] Dog(Rintintin) Question: Did Curiosity kill the cat? Kills(Curiosity,Tuna)]

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**Resolution Example Animal(F(x)) Loves(G(x),x)**

Loves(x,F(x)) Loves(G(x),x) Animal(y) Kills(x,y) Loves(z,x) Animal(x) Loves(Jack, x) Kills(Jack,Tuna) Kills(Curiosity,Tuna) Cat(Tuna) Cat(x) Animal(x) Dog(x) Cat(H(x) Dog(x) Kills(x,H(x))] Dog(Rintintin) Kills(Curiosity,Tuna)] Cat(Tuna) , Cat(x) Animal(x) Unify(Cat(Tuna), Cat(x)) = { x/Tuna } First line thus resolves to: Animal(Tuna)

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**Resolution Example Animal(F(x)) Loves(G(x),x)**

Loves(x,F(x)) Loves(G(x),x) Animal(y) Kills(x,y) Loves(z,x) Animal(x) Loves(Jack, x) Kills(Jack,Tuna) Kills(Curiosity,Tuna) Cat(Tuna) Cat(x) Animal(x) Dog(x) Cat(H(x) Dog(x) Kills(x,H(x))] Dog(Rintintin) Kills(Curiosity,Tuna) Animal(Tuna) Kills(Jack,Tuna) Kills(Curiosity,Tuna), Kills(Curiosity,Tuna) Resolves to: Kills(Jack,Tuna)

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Resolution Example Animal(F(x)) Loves(G(x),x) Loves(x,F(x)) Loves(G(x),x) Animal(y) Kills(x,y) Loves(z,x) Animal(x) Loves(Jack, x) Kills(Jack,Tuna) Kills(Curiosity,Tuna) Cat(Tuna) Cat(x) Animal(x) Dog(x) Cat(H(x) Dog(x) Kills(x,H(x))] Dog(Rintintin) Kills(Curiosity,Tuna) Animal(Tuna) Kills(Jack,Tuna) Animal(y) Kills(x,y) Loves(z,x), Animal(Tuna) Unify(Animal(Tuna), Animal(y)) = {y/Tuna} Resolves to: Kills(x,Tuna) Loves(z,x),

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Resolution Example Animal(F(x)) Loves(G(x),x) Loves(x,F(x)) Loves(G(x),x) Animal(y) Kills(x,y) Loves(z,x) Animal(x) Loves(Jack, x) Kills(Jack,Tuna) Kills(Curiosity,Tuna) Cat(Tuna) Cat(x) Animal(x) Dog(x) Cat(H(x) Dog(x) Kills(x,H(x))] Dog(Rintintin) Kills(Curiosity,Tuna) Animal(Tuna) Kills(Jack,Tuna) Kills(x,Tuna) Loves(z,x), Loves(x,F(x)) Loves(G(x),x), Animal(z) Loves(Jack, z) Unify( Loves(x,F(x)) , Loves(Jack, z)) = { x / Jack, z / F(x)} Resolvent clause is obtained by substituting the unification rule Loves(G(Jack),Jack) Animal(F(Jack))

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Resolution Example Animal(F(x)) Loves(G(x),x) Loves(x,F(x)) Loves(G(x),x) Animal(y) Kills(x,y) Loves(z,x) Animal(x) Loves(Jack, x) Kills(Jack,Tuna) Kills(Curiosity,Tuna) Cat(Tuna) Cat(x) Animal(x) Dog(x) Cat(H(x) Dog(x) Kills(x,H(x))] Dog(Rintintin) Kills(Curiosity,Tuna) Animal(Tuna) Kills(Jack,Tuna) Kills(x,Tuna) Loves(z,x) Loves(G(Jack),Jack) Animal(F(Jack)) Animal(F(x)) Loves(G(x),x), Loves(G(Jack),Jack) Animal(F(Jack)) Unify(Animal(F(x)) , Animal(F(Jack)))= { x / Jack} Resolvent clause is obtained by substituting the unification rule Loves(G(Jack),Jack)

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**Resolution Example Kills(x,Tuna) Loves(z,x), Loves(G(Jack),Jack)**

Animal(F(x)) Loves(G(x),x) Loves(x,F(x)) Loves(G(x),x) Animal(y) Kills(x,y) Loves(z,x) Animal(x) Loves(Jack, x) Kills(Jack,Tuna) Kills(Curiosity,Tuna) Cat(Tuna) Cat(x) Animal(x) Dog(x) Cat(H(x) Dog(x) Kills(x,H(x))] Dog(Rintintin) Kills(Curiosity,Tuna) Animal(Tuna) Kills(Jack,Tuna) Kills(x,Tuna) Loves(z,x) Loves(G(Jack),Jack) Animal(F(Jack)) Loves(G(Jack),Jack) Kills(x,Tuna) Loves(z,x), Loves(G(Jack),Jack) Unify( Loves(z,x), Loves(G(Jack),Jack) ) = { x / Jack, z / G(Jack)} Resolvent clause is obtained by substituting the unification rule Loves(G(Jack),Jack)

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**Resolution Example Loves(G(Jack),Jack), Loves(G(Jack),Jack)**

Animal(F(x)) Loves(G(x),x) Loves(x,F(x)) Loves(G(x),x) Animal(y) Kills(x,y) Loves(z,x) Animal(x) Loves(Jack, x) Kills(Jack,Tuna) Kills(Curiosity,Tuna) Cat(Tuna) Cat(x) Animal(x) Dog(x) Cat(H(x) Dog(x) Kills(x,H(x))] Dog(Rintintin) Kills(Curiosity,Tuna) Animal(Tuna) Kills(Jack,Tuna) Kills(x,Tuna) Loves(z,x) Loves(G(Jack),Jack) Animal(F(Jack)) Loves(G(Jack),Jack) Loves(G(Jack),Jack) Loves(G(Jack),Jack), Loves(G(Jack),Jack) Resolvent clause is empty. Proof succeeded

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**Resolution Resolution is refutation-complete Theorem provers**

If a set of sentences is unsatisfiable, then resolution will be able to produce a contradiction Theorem provers Use control in order to be more efficient Focus of most research effort Separate control from knowledge base Example: Otter (Organized Technique for Theorem proving and Effective Research) A set of clauses known as the SoS - Set of Support The important facts about a problem Search if focused on resolving SoS with another axiom A set of usable axioms Background knowledge about problem field Rewrites / demodulators Rules to transform expressions into a canonical form Set of parameters or clauses that defines the control strategy to allow user to control search and filtering functions to eliminate useless subgoals

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**Theorem Prover Successes**

First formal proof of Gődel’s Incompleteness Theorem (1986) Robbins algebra (a simple set of axioms) is Boolean algebra (1996) Software verification: Remote agent spacecraft control program (2000)

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**Resolution proof: definite clauses**

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Inference in FOL Copyright, 1996 © Dale Carnegie & Associates, Inc. Chapter 9 Spring 2005.

Inference in FOL Copyright, 1996 © Dale Carnegie & Associates, Inc. Chapter 9 Spring 2005.

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