first order logic (syntax, semantics and inference)
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First Order Logic(Syntax, Semantics and Inference)
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Knowledge based agents can represent the world it is in and can deduce the actions to take
In most programming languages the data and operations are tied closely
Each update to a data structure is done via a domain-specific procedure whose details are derived by the programmer from his own knowledge of the domain
Is there a way to say “pit is in [1,2] or [2,1]” ? Propositional logic could represent the partial
information, with negation and disjunction
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Propositional Logic is “compositional” Propositional logic is less expressive when the world
consists many objects (why?) But english/.../.. are very much expressive
Why not use engilsh for our representation? Ambiguous & not compositional
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First-order logic
Propositional logic assumes the world contains facts, (ontological commitment)
First-order logic (like natural language) assumes the world contains– Objects: people, houses, numbers, colors, baseball
games, wars, … – Relations: red, round, prime, brother of, bigger than,
part of, comes between, …– Functions: father of, best friend, one more than, plus,
…
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Ontological Commitments Propositional Logic : facts which hold ot does not. FOL : Objects and relations
Epistemological Commitments (possible states of knowledge w.r.t each facts)
Propositional Logic : true/false FOL : true/false
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Syntax of FOL: Basic elements
Constants KingJohn, 2, ... Predicates Brother, >,... Functions Sqrt, LeftLegOf,... Variables x, y, a, b,... Connectives , , , , Equality = Quantifiers ,
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Atomic sentences
Sentence = AtomicSentence | (Sentence Connective Sentence) | Quantifier Variables,…. Sentence | Sentence
Atomic sentence = predicate (term1,...,termn) | term1 = term2
Term = function (term1,...,termn) | constant | variable
E.g., Brother(KingJohn,RichardTheLionheart) > (Length(LeftLegOf(Richard)), Length(LeftLegOf(KingJohn)))
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Truth in first-order logic
Sentences are true with respect to a model (possible world) and an interpretation
Model contains objects (domain elements) and relations among them
Interpretation specifies referents forconstant symbols → objectspredicate symbols → relationsfunction symbols → functional relations
An atomic sentence predicate(term1,...,termn) is true iff the objects referred to by term1,...,termn are in the relation referred to by predicate
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Universal quantification
<variables> <sentence>
Everyone at England is smart:x At(x, England) Smart(x)
x P is true in a model m iff P is true with x being each possible object in the model
Roughly speaking, equivalent to the conjunction of instantiations of PAt(KingJohn,England) Smart(KingJohn)
At(Richard, England) Smart(Richard) At(England, England) Smart(England) ...
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A common mistake to avoid
Typically, is the main connective with Common mistake: using as the main connective
with :x At(x, England) Smart(x)
means “Everyone is at England and everyone is smart”
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Existential quantification
<variables> <sentence>
Someone at England is smart:x At(x, England) Smart(x)
x P is true in a model m iff P is true with x being some possible object in the model
Roughly speaking, equivalent to the disjunction of instantiations of PAt(KingJohn, England) Smart(KingJohn)
At(Richard, England) Smart(Richard) At(England, England) Smart(England) ...
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Another common mistake to avoid
Typically, is the main connective with
Common mistake: using as the main connective with :
x At(x,England) Smart(x)
is true if there is anyone who is not at England!
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Properties of quantifiers
x y is the same as y x x y is the same as y x
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Properties of quantifiers
x y is the same as y xx y is the same as y x
x y is not the same as y xx y Loves(x,y)
“There is a person who loves everyone in the world”y x Loves(x,y)
“Everyone in the world is loved by at least one person”
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Properties of quantifiers
x y is the same as y xx y is the same as y x
x y is not the same as y xx y Loves(x,y)
“There is a person who loves everyone in the world”y x Loves(x,y)
“Everyone in the world is loved by at least one person”
Quantifier duality: each can be expressed using the otherx Likes(x,IceCream) x Likes(x,IceCream)x Likes(x,Broccoli) x Likes(x,Broccoli)
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Interacting with FOL KBs
Suppose a wumpus-world agent is using an FOL KB and perceives a smell and a breeze (but no glitter) at t=5:Tell(KB,Percept([Smell,Breeze,None],5))Ask(KB,a BestAction(a,5))
i.e., does the KB entail some best action at t=5? Answer: Yes, {a/Shoot} ← substitution (binding list) Given a sentence S and a substitution σ, Sσ denotes the result of plugging σ into S; e.g.,
S = Smarter(x,y)σ = {x/Hillary,y/Bill}Sσ = Smarter(Hillary,Bill)
• Ask(KB,S) returns some/all σ such that KB╞ Sσ
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Universal instantiation (UI)
Every instantiation of a universally quantified sentence is entailed by it:
v αSubst({v/g}, α)
for any variable v and ground term g 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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. … substitutions are {x/Richard} , {x/John}, {x/ Father(John)} We can infer any sentence by replacing the variable with the ground term
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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 αSubst({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)King(John)Greedy(John)Brother(Richard,John)
The new KB is propositionalized: proposition symbols are King(John), Greedy(John), Evil(John), King(Richard), etc.
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Reduction contd.
• 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)))
• 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.
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Reduction contd.
• 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, but 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.
E.g., from:x King(x) Greedy(x) Evil(x)King(John)y Greedy(y)Brother(Richard,John)
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Forward Chaining
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} worksUnify(α,β) = θ if αθ = βθ
p q θ Knows(John,x) Knows(John,Jane) {x/Jane}Knows(John,x) Knows(y,OJ) {x/OJ, y/John}Knows(John,x) Knows(y,Mother(y)) {y/Jane, x/Mother(y)}Knows(John,x) Knows(x,OJ) {Fail}
Standardizing apart eliminates overlap of variables, e.g., Knows(z17,OJ)
Unification
To unify Knows(John,x) and Knows(y,z),θ = {y/John, x/z } or θ = {y/John, x/John, z/John}
The first unifier is more general than the second, places fewer restriction on the values of the variables
There is a single most general unifier (MGU) that is unique up to renaming of variables.
MGU = { y/John, x/z }
First Order Definite Clauses
Exactly one positive literalAtomic literalImplication whose antecedent is conjunction of positive literal and a single positive literal as consequent
ExamplesKing(x) Λ Greedy(x) Evil(x)King(John)Greedy(y)
Generalized Modus Ponens
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)
Lifted version of modus ponens GMP used with KB of definite clausesAll variables assumed universally quantified
An Example KB
Consider the passageThe 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.
The KB should answer the queryIs Col. West a criminal?
An Example KB (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 WestMissile(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)
No Function Symbols in the KB>> Datalog KB
Forward Chaining Algorithm
Forward chaining - Example
Forward chaining - Example
Forward chaining - Example
Forward Chaining Algorithm - discussion
No new sentences can be added to the KB after it has generated criminal(west)Fixed point of inference process
SoundComplete (?)Datalog KB
• k- maximum arity of predicates, n – constant symbols, p- predicates pnk distinct ground facts
KB with function symbols• Use Herbrand’s theorem, if the query has answer.
Remember, entailment with FOL is semidecidable
Forward Chaining Algorithm - discussion
Efficiency concernsNotice the inner-loop generate all possible θ
• Expensive pattern matching
Algorithm rechecks every rule in every iteration to see if the premises are satisfiedGenerates many facts which are irrelevant to the goal
Addressing Efficiency ConcernsMatching against known rules only
To apply the rule, Missile(x) Weapon (x)Look for the rules that unify only with Missile(x)Use Indexed KB
Missile(x) Λ Owns(Nono,x) Sells(West, x, Nono)Owns(Nono,x)Nono may own thousands of objectsFind all missiles first, then see if these missiles are owned by Nono : conjunct orderingRemember heuristic : Most constrained variable
Addressing Efficiency ConcernsMatching against known rules only
Colorable() is inferred iff the CSP has a solutionCSPs include 3SAT as a special case, hence matching is NP-hardForward chaining has an NP Hard Matching Problem in its inner loop
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)
Addressing Efficiency ConcernsMatching against known rules only
> Most Rules in real world are small and simple, upper bounds on the rule size and arity.> Data complexity> Consider only subclasses of databases for which matching is efficient datalog KBUse better algorithm that avoids redundant matchings
Addressing Efficiency ConcernsIncremental Forward Chaining
Every fact inferred at iteration t will use at least one fact derived at iteration t-1At iteration t, consider only rules whose premise include conjuncts which unifies with a fact derived at iteration t-1With indexing, we may find out all rules which may trigger at iteration t
Rete Algorithm
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Backward Chaining
Backward chaining algorithm
SUBST(COMPOSE(θ1, θ2), p) = SUBST(θ2, SUBST(θ1, p))
Backward chaining example
Backward chaining example
Backward chaining example
Backward chaining example
Backward chaining example
Backward chaining example
Backward chaining example
Backward chaining example
Properties of backward chaining
Depth-first recursive proof search: space is linear in size of proofIncomplete 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
Resolution
Converting sentences to CNF
1. Eliminate all ↔ connectives (P ↔ Q) ((P Q) ^ (Q P))
2. Eliminate all connectives (P Q) (P Q)
3. Reduce the scope of each negation symbol to a single predicate P P
(P Q) P Q
(P Q) P Q
(x)P (x)P
(x)P (x)P
4. Standardize variables: rename all variables so that each quantifier has its own unique variable name
Converting sentences to clausal form
5. Eliminate existential quantification by introducing Skolem constants/functions(x)P(x) P(c)
c is a Skolem constant (a brand-new constant symbol that is not used in any other sentence)(x)(y)P(x,y) (x)P(x, f(x))since is within the scope of a universally quantified variable, use a Skolem function f to construct a new value that depends on the universally quantified variable
f must be a brand-new function name not occurring in any other sentence in the KB.
E.g., (x)(y)loves(x,y) (x)loves(x,f(x)) In this case, f(x) specifies the person that x loves
Converting sentences to clausal form
6. Remove universal quantifiers by (1) moving them all to the left end; (2) making the scope of each the entire sentence; and (3) dropping the “prefix” partEx: (x)P(x) P(x)
7. Put into conjunctive normal form (conjunction of disjunctions) using distributive and associative laws(P Q) R (P R) (Q R)(P Q) R (P Q R)
8. Split conjuncts into separate clauses9. Standardize variables so each clause contains only variable names
that do not occur in any other clause
An example
(x)(P(x) ((y)(P(y) P(f(x,y))) (y)(Q(x,y) P(y)))) 2. Eliminate
(x)(P(x) ((y)(P(y) P(f(x,y))) (y)(Q(x,y) P(y))))
3. Reduce scope of negation(x)(P(x) ((y)(P(y) P(f(x,y))) (y)(Q(x,y) P(y))))
4. Standardize variables(x)(P(x) ((y)(P(y) P(f(x,y))) (z)(Q(x,z) P(z))))
5. Eliminate existential quantification(x)(P(x) ((y)(P(y) P(f(x,y))) (Q(x,g(x)) P(g(x)))))
6. Drop universal quantification symbols(P(x) ((P(y) P(f(x,y))) (Q(x,g(x)) P(g(x)))))
Example
7. Convert to conjunction of disjunctions(P(x) P(y) P(f(x,y))) (P(x) Q(x,g(x))) (P(x) P(g(x)))
8. Create separate clausesP(x) P(y) P(f(x,y))
P(x) Q(x,g(x))
P(x) P(g(x))
9. Standardize variablesP(x) P(y) P(f(x,y))
P(z) Q(z,g(z))
P(w) P(g(w))
Resolution: brief summary
Full first-order version:l1 ··· lk, m1 ··· mn
(l1 ··· li-1 li+1 ··· lk m1 ··· mj-1 mj+1 ··· mn)θwhere 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
Resolution proof: definite clauses
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