Logic in Artificial Intelligence this lecture is an informal introduction only

1
Logic in Artificial Intelligence(this lecture
is an informal introduction only)
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Human Logic

• Humans are, among other things, information
  processors.
• We acquire information about the world and use
  this information to further our ends.
• One of the strengths of human information
  processing is our ability to represent and
  manipulate logical information.

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A simple puzzle

• We are given some facts about the arrangement of
  five blocks in a stack, and we are asked to
  determine their exact arrangement.
• The sentences shown on the left constitute the
  premises of the problem.

• Conclusions
• The red block is on the green block.
• The green block is on the yellow block.
• The yellow block is on the blue block.
• The blue block is on the black block.
• The black block is directly on the table.

• The red block is on the green block.
• The green block is somewhere above the blue
  block.
• The green block is not on the blue block.
• The yellow block is on the green block or the
  blue block.
• There is some block on the black block.

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Proof

• Unfortunately, it is not always apparent which
  conclusions may be safely drawn from a given set
  of premises.
• What's more, even when we are given the
  conclusions, as in this example, their
  correctness may not be immediately obvious.
• In order to persuade others of a conclusion that
  we have drawn, it is useful to write down a
  proof, i.e. a series of intermediate conclusions,
  such that each is immediately obvious.
• As an example, consider the following informal
  proof
• We are told that the yellow block is on the green
  block or the blue block.
• We are also told that the red block is on the
  green block.
• Given the assumption that there can be only one
  block on another and that a block cannot be two
  colors at once, we can conclude that the yellow
  block is not on the green block.
• But then, by elimination, the yellow block must
  be on the blue block.

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Aristotle (384-322 B.C.E.)

• The concept of proof, in order to be meaningful,
  requires that we be able to recognize certain
  reasoning steps as immediately obvious.
• In other words, we need to be familiar with the
  reasoning atoms out of which complex proof
  molecules are built.
• One of Aristotle's great contributions to
  philosophy was his recognition that what makes a
  step of a proof immediately obvious is its form
  rather than its content.
• It does not matter whether you are talking about
  blocks or stocks or automobiles. What matters is
  the structure of the facts with which you are
  working.

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Examples

• All Accords are Hondas.
• All Hondas are Japanese.
• Therefore, all Accords are Japanese.
• All borogoves are slithy toves.
• All slithy toves are mimsy.
• Therefore, all borogoves are mimsy.
• What's more, in order to reach these conclusion,
  we do not need to know anything about Hondas and
  Accords or borogoves and slithy toves or what it
  means to be mimsy.
• What is interesting about these examples is that
  they share the same reasoning structure, with
  respect to the pattern shown below.
• All x are y.
• All y are z.
• Therefore, all x are z.

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Questions

• Which patterns are correct?
• How many patterns are enough?

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Unsound Patterns
Pattern All x are y. Some y are z. Therefore,
some x are z. Good Instance All Toyotas are
Japanese cars. Some Japanese cars are made in
America. Therefore, some Toyotas are made in
America. Not-So-Good Instance All Toyotas are
cars. Some cars are Porsches. Therefore, some
Toyotas are Porsches.
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Induction - Unsound
I have seen 1000 black ravens. I have never
seen a raven that is not black. Therefore, every
raven is black.
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Abduction - Unsound
If there is no fuel, the car will not
start. If there is no spark, the car will not
start. There is spark. The car will not
start. Therefore, there is no fuel. What if the
car is in a vacuum chamber? Abduction, or
abductive reasoning, is the process of inference
to the best explanation.
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Soundness

• What distinguishes a correct pattern from one
  that is incorrect is that it must always lead to
  correct conclusions, i.e. conclusions that are
  true whenever the premises are true.
• As we will see, this is the defining criterion
  for what we call deduction.

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Deduction - Sound

• Logical Entailment/Deduction
• Does not say that conclusion is true in general
• Conclusion true whenever premises are true
• Leibnitz
• The intellect is freed of all conception of the
• objects involved, and yet the computation yields
  the correct
• result.
• Russell
• Math may be defined as the subject in which we
• never know what we are talking about nor whether
  what we
• are saying is true.

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Formal Logic

• Algebra
• 1. Formal language for encoding information
• 2. Legal transformations
• Logic
• 1. Formal language for encoding information
• 2. Legal transformations

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Algebra Problem

• Xavier is three times as old as Yolanda. Xavier's
  age and
• Yolanda's age add up to twelve. How old are
  Xavier and
• Yolanda?
• x - 3y 0
• x y 12
• -4y -12
• y 3
• x 9

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Logic Problem

• If Amy loves Pat, then Amy loves Quincy.
• If it is Monday and raining, then Amy loves Pat
  or Quincy.
• If it is Monday and raining, does Amy love Quincy?

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Formalization
Simple sentences Amy loves Pat. loves(amy,
pat) Amy loves Quincy. loves(amy, quincy) It is
Monday. ismonday Its raining. raining
Premises If Amy loves Pat, Amy loves
Quincy. loves(amy,pat) Þ loves(amy,quincy) If
it Monday and raining, Amy loves Pat or
Quincy. ismonday ? raining Þ loves(amy,pat) Ú
loves(amy,quincy) Question If it is Monday and
raining, does Amy love Quincy?
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Rule of Inference (Resolution)
p1 Ù ... Ù pk Þ q1 Ú... Ú ql r1 Ù... Ù rm Þ s1 Ú
... Ú sn p1 Ù ... Ù pk Ù r1 Ù... Ù rm Þ q1 Ú... Ú
ql Ú s1 Ú... Ú sn

• If pi on the left hand side of one sentence is
  the same as qj in the right hand side of the
  other sentence, it is okay to drop the two
  symbols.

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Two Tricks

• A sentence
• Þ p
• asserts that p is true (i.e. a fact)
• A sentence
• p Þ
• asserts that p is false

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Modus Ponens - special case of Resolution
p Þ q p q
monday Þ alex is teaching AI monday alex teaching
AI

• Using the tricks
• p Þ q
• Þ p
• p Þ p Ù q
• q,
• i.e. q

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Solving Amy's love life

• p1 Ù ... Ù pk Þ q1 Ú... Ú ql
• r1 Ù... Ù rm Þ s1 Ú ... Ú sn
• p1 Ù ... Ù pk Ù r1 Ù... Ù rm Þ q1 Ú... Ú ql Ú s1
  Ú... Ú sn
• If pi on the left hand side of one sentence is
  the same as qj in the right hand side of the
  other sentence, drop the two symbols.

loves(amy,pat) Þ loves(amy,quincy) ismonday ?
raining Þ loves(amy,pat) Ú loves(amy,quincy) loves
(amy,pat) ? ismonday ? raining Þ
loves(amy,quincy) Ú loves(amy,pat) Ú
loves(amy,quincy) ismonday ? raining Þ
loves(amy,quincy) Ú loves(amy,quincy) ismonday ?
raining Þ loves(amy,quincy)
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Prolog as a Theorem Prover

• Prolog is by far the most widely used logic
  programming language.
• However, Prolog programs are sets of implications
  whose premise is a conjunction of positive
  literals and whose conclusion is a single
  positive literal.
• An implication with no premises simply asserts a
  given proposition sometimes called a fact.
• Hence, the Amys love life cant be solved with
  Prolog. We need a more powerful Theorem Prover.

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Kinship

• Art is the parent of Bob and Bud.
• Bob is the parent of Cal and Coe.
• A grandparent is a parent of a parent.
• Þ p(art,bob)
• Þ p(art,bud)
• Þ p(bob,cal)
• Þ p(bob,coe)
• p(x, y) Ù p(y, z) Þ g(x, z)

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Is Art the Grandparent of Coe?

• Lets assume g(art, coe) is not true, i.e. g(art,
  coe) Þ.
• 1. Þ p(art, bob)
• 2. Þ p(art, bud)
• 3. Þ p(bob,cal)
• 4. Þ p(bob,coe)
• p(x, y)Ù p(y, z) Þ g(x,z)
• 6. g(art, coe) Þ By assumption (see above)
• 7. p(art, y) Ù p(y,coe) Þ 5,6 (bind vars by
  unification)
• 8. p(bob,coe) Þ 1,7 (bind vars by unification)
• 9. Þ 4,8
• Nothing means contradiction!

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Who Are the Grandchildren of Art?

• 1. Þ p(art, bob)
• 2. Þ p(art, bud)
• 3. Þ p(bob,cal)
• 4. Þ p(bob,coe)
• p(x, y)Ù p(y, z) Þ g(x,z)
• 6. g(art, z) Þ goal(z)
• 7. p(art, y) Ù p(y, z) Þ goal(z) 5,6
• 8. p(bob, z) Þ goal(z) 1,7
• 9. p(bud, z) Þ goal(z) 2,7
• 10. goal(cal) 3,8
• 11. goal(coe) 4,8

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Databases
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Who Are the Grandchildren of Art? In SQL

• 1. Þ p(art, bob)
• 2. Þ p(art, bud)
• 3. Þ p(bob,cal)
• 4. Þ p(bob,coe)
• p(x, y)Ù p(y, z) Þ g(x,z)
• 6. g(art, z) Þ goal(z)

p(x, y) Ù p(y, z)Þg(x, z) CREATE VIEW g(x,y)
AS SELECT p1.x, p2.y FROM p AS p1, p AS
p2 WHERE p1.y p2.x
g(art, z)Þgoal(z) SELECT g.y FROM g WHERE g.x
'art'
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Agatha
Someone who lives in Dreadsbury Mansion killed
Aunt Agatha. Agatha, the butler, and Charles
live in Dreadsbury Mansion and are the only
people who live therein. A killer always hates
his victim, and is never richer than his
victim. Charles hates no one that Agatha hates.
Agatha hates everyone except the butler. The
butler hates everyone not richer than Aunt
Agatha. The butler hates everyone Agatha hates.
No one hates everybody. Aunt Agatha is not
the butler. Prove that Aunt Agatha killer
herself. gt Killed(Agatha,Agatha)
Killed(butler,Agatha) Killed(Charles,Agatha) gt
Is(x,Agatha) Is(x,butler) Is(x,Charles) Killed
(x,y) gt Hates(x,y) Killed(x,y) gt Richer(x,y)
Hates(Agatha,x) gt Hates(Charles,x) gtHates(Agat
ha,Agatha) gtHates(Agatha,Charles) gtHates(Agatha
,butler) Richer(x,Agatha) gt Hates(butler,x) Hate
s(Agatha,x) gt Hates(butler,x) gtHates(x,Agatha)
Hates(x,butler) Hates(x,Charles) gtIs(Agath
a,butler) gtKilled(Agatha,Agatha)
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Agatha into clausal form
Killed(Agatha(),Agatha()) Killed(butler(),Agatha
()) Killed(Charles(),Agatha()) Is(x,Agatha())
Is(x,butler()) Is(x,Charles()) Killed(x,y)
Hates(x,y) Killed(x,y) Richer(x,y)
Hates(Agatha(),x) Hates(Charles(),x)
Hates(Agatha(),Agatha()) Hates(Agatha(),Charles()
) Hates(Agatha(),butler()) Richer(x,Agatha())
Hates(butler(),x) Hates(Agatha(),x)
Hates(butler(),x) Hates(x,Agatha())
Hates(x,butler()) Hates(x,Charles())
Is(Agatha(),butler()) negated_conclusion Kille
d(Agatha(),Agatha())
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Theo Theorem Prover
Axioms 1gtKilledAgathaAgatha KilledbutlerAgatha
KilledCharlesAgatha 2 IsxAgatha Isxbutler
IsxCharles 3 gtKilledxy Hatesxy 4 gtKilledxy
Richerxy 5 gtHatesAgathax HatesCharlesx 6
gtHatesAgathaAgatha 7 gtHatesAgathaCharles 8
HatesAgathabutler 9 gtRicherxAgatha
Hatesbutlerx 10 gtHatesAgathax Hatesbutlerx 11
gtHatesxAgatha Hatesxbutler HatesxCharles 12
IsAgathabutler Negated conclusion 13SgtKilledAg
athaAgatha Phase 0 clauses used in
proof 14gt(13a1a) KilledbutlerAgatha
KilledCharlesAgatha 15 gt(9a4b) Hatesbutlerx
KilledxAgatha 17 gt(5b3b) HatesAgathax
KilledCharlesx 18gt(17b14b) HatesAgathaAgatha
KilledbutlerAgatha 19Sgt(18a6a)
KilledbutlerAgatha 20Sgt(19a15b)
Hatesbutlerbutler 21gt(20a11b)
HatesbutlerAgatha HatesbutlerCharles 22Sgt(19a3a
) HatesbutlerAgatha 23Sgt(22a21a)
HatesbutlerCharles 24Sgt(23a10b)
HatesAgathaCharles Phases 1 and 2 clauses used
in proof 25Sgt(24a,7a)
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History

• Theorem Provers have come up with novel
  mathematical results.
• The most famous of these concerns Robbins
  Algebra.
• In 1933, E. V. Huntington presented the following
  basis for Boolean algebra
• x y y x. commutativity
• (x y) z x (y z). associativity
• n(n(x) y) n(n(x) n(y)) x. Huntington
  equation
• Shortly thereafter, Herbert Robbins conjectured
  that the Huntington equation can be replaced with
  a simpler one
• n(n(x y) n(x n(y))) x. Robbins equation

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• The problem was first posed in the 1930s by Dr.
  Herbert Robbins, a New Jersey Professor of
  Mathematics at Rutgers University.
• Robbins said that he worked on the problem for
  some time, and then passed it on one of the
  century's most famous logicians, Dr. Albert
  Tarski of Stanford University. Tarski, who is now
  dead, worked on the problem, included it in a
  book, and handed it out to graduate students and
  visitors.
• Burris (University of Waterloo), for example said
  that Tarski suggested the problem to him in the
  early 1970s, while he was visiting Stanford for a
  couple of months. Tarski, he said, "liked to
  throw out challenging problems to people passing
  through."

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• While mathematicians were batting around
  Robbins's problem, computer scientists were
  striving to see if they could get computers to
  reason.
• Among them was Wos, who started working on
  automated reasoning in the 1960s. It was a time
  when computers were primitive, clunky and slow,
  and researchers were divided on how to proceed.
• On October 10, 1996, after eight days of
  computation, EQP (a version of OTTER) found a
  proof.