First-Order Logic
ECE457 Applied Artificial IntelligenceFall 2007
Lecture #7
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Outline First-order logic (FOL) Sentences
Russell & Norvig, chapter 8 Inference
Russell & Norvig, chapter 9
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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
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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
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New Variables in FOL Begin with lowercase letters Stand-in for any symbol Brother(x,y)
x is the brother of y
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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…”
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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)
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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
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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)
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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)
α β α β
T T T
T F F
F T T
F F T
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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 is purple and a mushroom is poisonous 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
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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 is purple and a mushroom is poisonous 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
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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 is purple and a mushroom and is poisonous 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
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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 is purple and a mushroom and is poisonous 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
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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
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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)
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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))
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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
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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)
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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)
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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)
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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
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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)
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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
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Generalized Modus Ponens Combine And-Introduction, Universal
Elimination, and Modus Ponens p1’, p2’,…, pn’, (p1 p2 … pn q)
SUBST(, q) p1 = Dog(x) p2 = Own(Bob,x) q =
Brown(x) p1’ = Dog(Fido) p2’ = Own(Bob,Fido) = {x/Fido} SUBST(, q) = Brown(Fido)
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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
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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)
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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
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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
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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)}
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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)}
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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
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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
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Forward / Backward Chaining
Unification algorithm finds that unifies (p1’, …, pn’) and (p1 … pn) 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
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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
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Resolution Recall from Propositional Logic
(αβ), (¬βγ) (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
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Resolution Steps
1. Convert problem into FOL KB2. Convert FOL statements in KB to
CNF3. Add negation of query (in CNF) in
KB4. Use FOL resolution rule to infer
new clauses from KB5. Produce a contradiction that
proves our query
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Conversion to CNF Three new steps (4, 5, 6) & rules in step 3
1. Eliminate biconditionals: (αβ) ((αβ)(βα))
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 (βγ)) ((αβ) (αγ))
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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(C1) ¬Love(x,C1) Love(C2,x) ¬Animal(C1) ¬Love(x,C1) Love(C2,x)
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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(C1) {¬Buy(C2,C1) [Class(C3) Requires(C3,C1)]}
{[z ¬Class(z) ¬Requires(z,C1)] y Buy(y,C1)} Book(C1) {¬Buy(C2,C1) [Class(C3) Requires(C3,C1)]}
{¬Class(z) ¬Requires(z,C1) Buy(y,C1)} Book(C1) [¬Buy(C2,C1) Class(C3)] [¬Buy(C2,C1)
Requires(C3,C1)] [¬Class(z) ¬Requires(z,C1) Buy(y,C1)]
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FOL Resolution (αβ’), (¬βγ)
SUBST(, αγ)) = 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)
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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?
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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)
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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))
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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)
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Example of Resolution Content of KB
¬Pass(x1, ECE457) ¬Win(x1,Lottery) Happy(x1)
¬Study(x2) Pass(x2,y1) ¬Lucky(x3) Pass(x3,y2) ¬Study(Bob) Lucky(Bob) ¬Lucky(x4) Win(x4,Lottery) ¬Happy(Bob)
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Example of Resolution (¬Pass(x1, ECE457) ¬Win(x1,Lottery) Happy(x1))
(¬Study(x2) Pass(x2,y1)) (¬Lucky(x3) Pass(x3,y2)) ¬Study(Bob) Lucky(Bob) (¬Lucky(x4) Win(x4,Lottery)) ¬Happy(Bob)
(¬Pass(x1, ECE457) ¬Win(x1,Lottery) Happy(x1)) (¬Study(x2) Pass(x2,y1)) (¬Lucky(x3) Pass(x3,y2)) ¬Study(Bob) Lucky(Bob) (¬Lucky(x4) Win(x4,Lottery)) ¬Happy(Bob)
{x1/Bob} (¬Pass(Bob, ECE457) ¬Win(Bob,Lottery))
(¬Study(x2) Pass(x2,y1)) (¬Lucky(x3) Pass(x3,y2)) ¬Study(Bob) Lucky(Bob) (¬Lucky(x4) Win(x4,Lottery)) ¬Happy(Bob)
{x4/Bob}
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Example of Resolution (¬Pass(Bob, ECE457) ¬Lucky(Bob))
(¬Study(x2) Pass(x2,y1)) (¬Lucky(x3) Pass(x3,y2)) ¬Study(Bob) Lucky(Bob) ¬Happy(Bob) No substitution needed
¬Pass(Bob, ECE457) (¬Study(x2) Pass(x2,y1)) (¬Lucky(x3) Pass(x3,y2)) ¬Study(Bob) Lucky(Bob) ¬Happy(Bob) {x3/Bob, y2/ECE457}
¬Pass(Bob, ECE457) (¬Study(x2) Pass(x2,y1)) ¬Lucky(Bob) ¬Study(Bob) Lucky(Bob) ¬Happy(Bob) No substitution needed
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