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Chương 3
Tri thức và lập luận
Nội dung chính chương 3
I. Logic – ngôn ngữ của tư duy
Lecture 1
Lecture 2
Lecture 3,4
II. Logic mệnh đề (cú pháp, ngữ nghĩa,
sức mạnh biểu diễn, các thuật toán
suy diễn)
III. Prolog (cú pháp, ngữ nghĩa, lập trình
prolog, bài tập và thực hành)
IV. Logic cấp một (cú pháp, ngữ nghĩa,
sức mạnh biểu diễn, các thuật toán
suy diễn)
Lecture 4
Inference in First-Order Logic
Lecture 4: Inference in First-Order Logic - Outline
 Proofs (chứng minh)
 Unification (hợp nhất)
 Generalized modus ponens - GMP(Luật modus ponens tổng quát)
 Inference procedures based on GMP (các thủ tục dựa trên luật
modus ponens tổng quát)
• Forward chaining (suy diễn tiến)
• Backward chaining (suy diễn lùi)
 Incompleteness of GMP-based inferences (tính không đầy đủ của
suy diễn tiến và lùi).
 Inference procedures based on Resolution (thủ tục suy diễn hợp
giải, thủ tục có tính đầy đủ trong logic cấp một)
 Logic programming (lập trình logic)
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Entailment and proof - Hệ quả logic và chứng minh
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Remember:
propositional
logic
Nhớ lại:
Các định lý
trong logic
mệnh đề
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Proofs – chứng minh
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Proofs – Chứng minh (tiếp)
The three new inference rules for FOL (compared to propositional logic) are:
•
Universal Elimination (UE):
for any sentence , variable x and ground term ,
x 
{x/}
•
Existential Elimination (EE):
for any sentence , variable x and constant symbol k not in KB,
x 
{x/k}
•
Existential Introduction (EI):
for any sentence , variable x not in  and ground term g in ,

x {g/x}
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Proofs – Chứng minh (tiếp)
The three new inference rules for FOL (compared to propositional logic) are:
•
Universal Elimination (UE):
for any sentence , variable x and ground term ,
x 
e.g., from x Likes(x, Candy) and {x/Joe}
{x/}
we can infer Likes(Joe, Candy)
•
Existential Elimination (EE):
for any sentence , variable x and constant symbol k not in KB,
x 
e.g., from x Kill(x, Victim) we can infer
{x/k}
Kill(Murderer, Victim), if Murderer new symbol
•
Existential Introduction (EI):
for any sentence , variable x not in  and ground term g in ,

e.g., from Likes(Joe, Candy) we can infer
x {g/x}
x Likes(x, Candy)
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Example Proof
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Example Proof
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Example Proof
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Example Proof
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Search with primitive example rules
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Unification – Hợp nhất
• Unification is a "pattern matching" procedure
that takes two atomic sentences, called
literals, as input, and returns "failure" if they
do not match and a substitution list, Theta, if
they do match.
• That is,
unify(p,q) = Theta means subst(Theta, p) =
subst(Theta, q) for two atomic sentences p and q.
• Theta
is called the most general unifier (mgu)
• All variables in the given two literals are
implicitly universally quantified
• To make literals match, replace (universallyquantified) variables by terms
Unification – Hợp nhất
Goal of unification: finding σ
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Unification – Hợp nhất
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Unification Algorithm – Giải thuật hợp nhất
procedure unify(p, q, theta)
Scan p and q left-to-right and find the first
corresponding terms where p and q "disagree" ; where p
and q not equal
If there is no disagreement, return theta ; success
Let r and s be the terms in p and q, respectively, where
disagreement first occurs
If variable(r) then theta = union(theta, {r/s})
unify(subst(theta, p), subst(theta, q), theta)
else if variable(s) then theta = union(theta, {s/r})
unify(subst(theta, p), subst(theta, q), theta)
else return "failure" end
Unification…
• Examples
Literal 1
Literal 2
theta
parents(x, father(x),
mother(Bill))
parents(Bill,
father(Bill), y)
{x/Bill,
y/mother(Bill)}
parents(x, father(x),
mother(Bill))
parents(Bill,
father(y), z)
{x/Bill, y/Bill,
z/mother(Bill)}
parents(x, father(x),
mother(Jane))
parents(Bill,
father(y), mother(y))
Failure
More Unification Examples
VARIABLE
1 – unify(P(a,X), P(a,b))
2 – unify(P(a,X), P(Y,b))
3 – unify(P(a,X), P(Y,f(a))
4 – unify(P(a,X), P(X,b))
term
σ = {X/b}
σ = {Y/a, X/b}
σ = {Y/a, X/f(a)}
σ = failure
Note: If P(a,X) and P(X,b) are independent, then we can
replace X with Y and get the unification to work.
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Generalized Modus Ponens (GMP) - Luật Modus
Ponens tổng quát
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Soundness of GMP – Tính đúng đắn của GMP
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Horn form – Dạng câu Horn
• We convert sentences to Horn form as they are entered into the KB
• Using Existential Elimination and And Elimination
• e.g., x Owns(Nono, x)  Missile(x)
becomes
Owns(Nono, M)
Missile(M)
(with M a new symbol that was not already in the KB)
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Forward chaining procedure – Thủ tục suy diễn tiến
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Backward chaining procedure – Thủ tục suy diễn lùi
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An Example (from Konelsky)
• Nintendo example.
• Nintendo says it is Criminal for a programmer to provide
emulators to people. My friends don’t have a Nintendo 64, but
they use software that runs N64 games on their PC, which is
written by Reality Man, who is a programmer.
• Query: Has Reality Man done anything criminal?
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Forward Chaining
Programmer(x)  Emulator(y)  People(z)  Provide(x,z,y) Criminal(x)
Use(friends, x)  Runs(x, N64 games)  Provide(Reality Man, friends, x)
Software(x)  Runs(x, N64 games)  Emulator(x)
Programmer(Reality Man)
People(friends)
Software(U64)
Use(friends, U64)
Runs(U64, N64 games)
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
• Premises (6), (7) and (8) satisfy the implications fully.
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Forward Chaining
Programmer(x)  Emulator(y)  People(z)  Provide(x,z,y) Criminal(x)
Use(friends, x)  Runs(x, N64 games)  Provide(Reality Man, friends, x)
Software(x)  Runs(x, N64 games)  Emulator(x)
(1)
(2)
(3)
Programmer(Reality Man)
People(friends)
Software(U64)
Use(friends, U64)
Runs(U64, N64 games)
(4)
(5)
(6)
(7)
(8)
• So we can infer the consequents, which are now added to the
knowledge base (this is done in two separate steps).
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Forward Chaining
•
Programmer(x)  Emulator(y)  People(z)  Provide(x,z,y) Criminal(x)
(1)
Use(friends, x)  Runs(x, N64 games)  Provide(Reality Man, friends, x)
(2)
Software(x)  Runs(x, N64 games)  Emulator(x)
(3)
Programmer(Reality Man)
(4)
People(friends)
(5)
Software(U64)
(6)
Use(friends, U64)
(7)
Runs(U64, N64 games)
(8)
Provide(Reality Man, friends, U64)
(9)
Emulator(U64)
(10)
Addition of these new facts triggers further forward chaining.
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Forward Chaining
Programmer(x)  Emulator(y)  People(z)  Provide(x,z,y) Criminal(x)
(1)
Use(friends, x)  Runs(x, N64 games)  Provide(Reality Man, friends, x)
(2)
Software(x)  Runs(x, N64 games)  Emulator(x)
(3)
Programmer(Reality Man)
(4)
People(friends)
(5)
Software(U64)
(6)
Use(friends, U64)
(7)
Runs(U64, N64 games)
(8)
Provide(Reality Man, friends, U64)
(9)
Emulator(U64)
(10)
Criminal(Reality Man)
(11)
• Which results in the final conclusion: Criminal(Reality Man)
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Backward Chaining
• Question: Has Reality Man done anything criminal?
• We will use the same knowledge as in our forward-chaining version
of this example:
Programmer(x)  Emulator(y)  People(z)  Provide(x,z,y) Criminal(x)
Use(friends, x)  Runs(x, N64 games)  Provide(Reality Man, friends, x)
Software(x)  Runs(x, N64 games)  Emulator(x)
Programmer(Reality Man)
People(friends)
Software(U64)
Use(friends, U64)
Runs(U64, N64 games)
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Backward Chaining
• Question: Has Reality Man done anything criminal?
Criminal(x)
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Backward Chaining
• Question: Has Reality Man done anything criminal?
Criminal(x)
Programmer(x)
Yes, {x/Reality Man}
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Backward Chaining
• Question: Has Reality Man done anything criminal?
Criminal(x)
Programmer(x)
People(Z)
Yes, {x/Reality Man}
Yes, {z/friends}
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Backward Chaining
• Question: Has Reality Man done anything criminal?
Criminal(x)
Programmer(x)
Yes, {x/Reality Man}
Emulator(y)
People(Z)
Yes, {z/friends}
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Backward Chaining
•
Question: Has Reality Man done anything criminal?
Criminal(x)
Programmer(x)
Yes, {x/Reality Man}
Emulator(y)
People(z)
Yes, {z/friends}
Software(y)
Yes, {y/U64}
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Backward Chaining
•
Question: Has Reality Man done anything criminal?
Criminal(x)
Programmer(x)
Yes, {x/Reality Man}
Emulator(y)
People(z)
Yes, {z/friends}
Software(y)
Yes, {y/U64}
Runs(U64, N64 games)
yes, {}
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Backward Chaining
•
Question: Has Reality Man done anything criminal?
Criminal(x)
Programmer(x)
Yes, {x/Reality Man}
Emulator(y)
People(z)
Yes, {z/friends}
Provide
(reality man,
U64,
friends)
Software(y)
Runs(U64, N64 games)
Yes, {y/U64}
yes, {}
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Backward Chaining
•
Question: Has Reality Man done anything criminal?
Criminal(x)
Programmer(x)
Emulator(y)
Yes, {x/Reality Man}
People(z)
Yes, {z/friends}
Provide
(reality man,
U64,
friends)
Software(y)
Yes, {y/U64}
Use(friends, U64)
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Runs(U64, N64 games)
yes, {}
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Backward Chaining
• Backward Chaining benefits from the fact that it is
directed toward proving one statement or answering
one question.
• In a focused, specific knowledge base, this greatly
decreases the amount of superfluous work that needs to
be done in searches.
• However, in broad knowledge bases with extensive
information and numerous implications, many search
paths may be irrelevant to the desired conclusion.
• Unlike forward chaining, where all possible inferences
are made, a strictly backward chaining system makes
inferences only when called upon to answer a query.
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Completeness
Backward chaining
Generalized Modus Ponens
Forward chaining
• As explained earlier, Generalized Modus Ponens requires sentences to
be in Horn form:
• atomic, or
• an implication with a conjunction of atomic sentences as the antecedent
and an atom as the consequent.
• However, some sentences cannot be expressed in Horn form.
• e.g.:
x  bored_of_this_lecture (x)
• E.g.: x P(x) => Q(x)
• Cannot be expressed in Horn form due to presence of negation.
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Completeness
• A significant problem since Modus Ponens
cannot operate on such a sentence, and thus
cannot use it in inference.
• Knowledge exists but cannot be used.
• Thus inference using Modus Ponens is
incomplete.
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Completeness
• However, Kurt Gödel in 1930-31 developed the
completeness theorem, which shows that it is
possible to find complete inference rules.
• The theorem states:
• any sentence entailed by a set of sentences can be proven from
that set.
=> Resolution Algorithm which is a complete
inference method.
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Completeness in FOL
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Resolution Procedure… Thủ tục hợp giải
•
Proof by contradiction method
•
Given a consistent set of axioms KB and goal sentence Q, we want to show
that KB |= Q. This means that every interpretation I that satisfies KB, satisfies
Q. But we know that any interpretation I satisfies either Q or ~Q, but not both.
Therefore if in fact KB |= Q, an interpretation that satisfies KB, satisfies Q and
does not satisfy ~Q. Hence KB union {~Q} is unsatisfiable, i.e., that it's false
under all interpretations.
•
In other words,
•
What's the gain? If
unsatisfiable
•
Resolution procedure can be used to establish that a given sentence Q is
entailed by KB; however, it cannot, in general, be used to generate all logical
consequences of a set sentences. Also, the resolution procedure cannot be
used to prove that Q is not entailed by KB.
(KB |- Q) <=> (KB ^ ~Q |- False)
KB
union
~Q
is unsatisfiable, then some finite subset is
Resolution inference rule
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Resolution Algorithm – Giải thuật hợp giải
procedure resolution-refutation(KB, Q)
;; KB is a set of consistent, true FOL sentences
;; Q is a goal sentence that we want to derive
;; return success if KB |- Q, and failure otherwise
KB = union(KB, ~Q)
while false not in KB do
pick 2 sentences, S1 and S2, in KB that contain
literals that unify (if none, return "failure“)
resolvent = resolution-rule(S1, S2)
KB = union(KB, resolvent)
return "success"
Kinship Example
KB:
(1) father (art, jon)
(2) father (bob, kim)
(3) father (X, Y)  parent (X, Y)
Goal:parent (art, jon)?
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Refutation Proof/Graph
¬parent(art,jon)
¬ father(X, Y) \/ parent(X, Y)
\
/
¬ father (art, jon)
father (art, jon)
\
/
[]
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Problems yet to be addressed – Các vấn đề cần giải
quyết khi áp dụng hợp giải.
• Resolution rule of inference is only applicable with
sentences that are in the form P1 v P2 v ... v Pn, where
each Pi is a negated or nonnegated predicate and
contains functions, constants, and universally quantified
variables, so can we convert every FOL sentence into
this form?
• Resolution strategy
• How to pick which pair of sentences to resolve?
• How to pick which pair of literals, one from each sentence, to
unify?
Remember: normal forms
“product of sums of
simple variables or
negated simple variables”
“sum of products of
simple variables or
negated simple variables”
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Conjunctive normal form
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Skolemization
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Examples: Converting FOL sentences to clause form…
Convert the sentence
1. (x){P(x) => [(y)(P(y) => P(f(x,y))) ^ ¬(y)(Q(x,y) => P(y)))]
(like A => B ^ C)
(chú ý: để dễ nhìn, dùng “[“, “]”,”{“ và “}” thay cho”(“ và “)” trong một vài vị trí)
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)))]
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Examples: Converting FOL sentences to clause form…
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))))]
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))]
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Examples: Converting FOL sentences to clause form…
8. Create
¬P(x)
¬P(x)
¬P(x)
separate clauses
 ¬P(y)  P(f(x,y))
 Q(x,g(x))
 ¬P(g(x))
9. Standardize variables
¬P(x)  ¬P(y)  P(f(x,y))
¬P(z)  Q(z,g(z))
¬P(w)  ¬P(g(w))
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Resolution proof
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Resolution proof
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Problems yet to be addressed – Các vấn đề cần giải
quyết khi áp dụng hợp giải.
• Resolution rule of inference is only applicable with
sentences that are in the form P1 v P2 v ... v Pn, where
each Pi is a negated or nonnegated predicate and
contains functions, constants, and universally quantified
variables, so can we convert every FOL sentence into
this form?
• Resolution strategy
• How to pick which pair of sentences to resolve?
• How to pick which pair of literals, one from each sentence, to
unify?
Example of Refutation Proof
(in conjunctive normal form)
(1)
(2)
(3)
(4)
Cats like fish
Cats eat everything they like
Josephine is a cat.
Prove: Josephine eats fish.
cat (x)  likes (x,fish)
cat (y)  likes (y,z)  eats (y,z)
cat (jo)
eats (jo,fish)
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Backward Chaining – Hợp giải lùi
Negation of goal wff:  eats(jo, fish)
 eats(jo, fish)
 cat(y)  likes(y, z)  eats(y, z)
 = {y/jo, z/fish}
 cat(jo)  likes(jo, fish)
cat(jo)
=
 cat(x)  likes(x, fish)
 likes(jo, fish)
 = {x/jo}
 cat(jo)
cat(jo)
 (contradiction)
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Forward chaining – Hợp giải tiến
cat (X)  likes (X,fish)
cat (jo)
\
/
cat (Y)  likes (Y,Z)  eats (Y,Z)
likes (jo,fish)
\
/
cat (jo)  eats (jo,fish)
cat (jo)
\
/
eats (jo,fish)
\
 eats (jo,fish)
/
[]
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Logic programming, Prolog
 Language: Horn forms (at most 1 positive literal)
• KB: definite clauses: (Horn forms with exact 1 positive literal)
• Query (q): not(q) is a negative clause (Horn forms with 0
positive literal)
 Inference procedure: Backward chaining resolution
(actually, SLD Resolution – Selected literals Linear from
Definite clauses Resolution)
• Idea: Bắt đầu với Q=not(q), hợp giải với câu trong KB, được
câu hợp giải mới, câu mới này lại hợp giải với câu trong KB đến
khi nào câu [] tìm thấy
• Đlý: Vơi KB gồm positive clauses và Q là negative clause:
KB  Q |= [] khi và chỉ khi KB  Q |--
SDL
[]