The document discusses knowledge-based agents and how they use inference to derive new representations of the world from their knowledge base in order to determine what actions to take. It provides the example of an agent exploring a cave, or "Wumpus world", where the goal is to locate gold and exit without being killed by the Wumpus monster or falling into pits. The agent uses its percepts and knowledge base along with inference rules to deduce its next action at each step.

1. Module 4
KNOWLEDGE
AND
REASONING

2. Knowledge based agent
• We design agents that can form representations of the
world
• Agents that use a process of inference to derive new
representations about the world
• and agents that use these new representations to
deduce what to do
• A knowledge-based agent needs to know many
things:
– the current state of the world
– how to infer unseen properties of the world from percepts
– how the world evolves over time
– what it wants to achieve
– what its own actions do in various circumstances
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3. Knowledge Base
(set of representations of facts about the world)
• The knowledge level or epistemological level is the most abstract;
we can describe the agent by saying what it knows. For example, an
automated taxi might be said to know that the Golden Gate Bridge
links San Francisco and Marin County.
• The logical level is the level at which the knowledge is encoded into
sentences. For example, the taxi might be described as having the
logical sentence Links(GGBridge, SF, Marin) in its knowledge base.
• The implementation level is the level that runs on the agent
architecture. It is the level at which there are physical representations
of the sentences at the logical level. A sentence such as
Links(GGBridge, SF, Marin) could be represented in the KB by the
string "Links (GGBridge, SF, Marin)" contained in a list of strings; or
by a "1" entry in a three-dimensional table indexed by road links and
location pairs; or by a complex set of pointers connecting machine
addresses corresponding to the individual symbols. The choice of
implementation is very important to the efficient performance of the
agent, but it is irrelevant to the logical level and the knowledge level.
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4. 4
• The Inference engine derives new sentences from the input
and KB
• The inference mechanism depends on representation in KB
• The agent operates as follows:
1. It receives percepts from environment
2. It computes what action it should perform (by IE and KB)
3. It performs the chosen action (some actions are simply
inserting inferred new facts into KB).
Knowledge
Base
Inference
Engine
Input from
environment
Output
(actions)
Learning
(KB update)
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5. Example – Knowledge Based Agent
• A classical AI problem
• An early computer game
• An agent had to explore a cave made of series of
interconnected rooms
• In one room, there was a Wumpus who would kill agent if it
entered the room
• Some rooms have pits – agent would die if it entered it
• Agent had 1 arrow to kill Wumpus
• Goal – Locate gold and return to start without getting killed.
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6. Shiwani Gupta 6

7. The Wumpus World
environment
• The Wumpus computer game
• The agent explores a cave consisting of rooms connected by
passageways.
• Lurking somewhere in the cave is the Wumpus, a beast that
eats any agent that enters its room.
• Some rooms contain bottomless pits that trap any agent that
wanders into the room.
• It is safe to enter a square with dead Wumpus.
• Occasionally, there is a heap of gold in a room.
• The goal is:
– to collect the gold and
– exit the world (climb out of cave)
– without being eaten Shiwani Gupta 7

8. Wumpus World PEAS description
• Performance measure:
– gold +1000, death -10000
– -1 per step, -10 for using the arrow
• Environment:
– Squares adjacent to wumpus are smelly
– Squares adjacent to pit are breezy
– Glitter iff gold is in the same square
– Shooting kills wumpus if you are facing it,
it screams
– Shooting uses up the only arrow
– Grabbing picks up gold if in same square
– Releasing drops the gold in same square
– You bump if you walk into a wall
• Actuators: Left turn, Right turn, Forward, Grab, Climb, Shoot
• Sensors: Stench, Breeze, Glitter, Bump, Scream
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9. Wumpus world characterization
• Fully Observable No – only local perception
• Deterministic Yes – outcomes exactly specified
• Episodic No – since no need to think ahead
• Static Yes – Wumpus and Pits do not move
• Discrete Yes
• Single-agent Yes – Wumpus is essentially a
natural feature
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10. The first step taken by the agent in the wumpus world,
(a)The initial situation, after percept [None, None, None, None, None].
(b)After one move, with percept [None, Breeze, None, None, None].
[ stench,  breeze,  glitter,  bump,  scream] [1,1]
[stench,  breeze,  glitter,  bump,  scream] [1,2]
[ stench, breeze,  glitter,  bump,  scream] [2,1] → Safe cell
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11. Two later stages in the progress of the agent,
(a) After the third move, with percept [Stench, None, None, None, None].
(b) After the fifth move, with percept [Stench, Breeze, Glitter, None, None].
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12. Propositional logic: Syntax
(limited expressiveness)
• Propositional logic is the simplest logic – illustrates basic ideas
• The proposition symbols P1, P2 etc are sentences
– If S is a sentence, S is a sentence (negation)
– If S1 and S2 are sentences, S1  S2 is a sentence
(conjunction)
– If S1 and S2 are sentences, S1  S2 is a sentence
(disjunction)
– If S1 and S2 are sentences, S1  S2 is a sentence
(implication)
– If S1 and S2 are sentences, S1  S2 is a sentence
(biconditional)
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13. Symbols of propositional logic
• Logical constants TRUE and FALSE are sentences
• Proposition symbols P and Q are sentences
• A, V, , , are logical connectives
• parenthesis ( )
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14. Ambiguity
• P A Q V R
(P A Q) V R
P A (Q V R)
• Order of precedence (highest to lowest)
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15. Propositional logic: Semantics
Each model specifies true/false for each proposition symbol
E.g. P1,2 P2,2 P3,1
false true false
With these symbols, 8 possible models, can be enumerated automatically.
Rules for evaluating truth with respect to a model m:
S is true iff S is false
S1  S2 is true iff S1 is true and S2 is true
S1  S2 is true iff S1is true or S2 is true
S1  S2 is true iff S1 is false or S2 is true
i.e., is false iff S1 is true and S2 is false
S1  S2 is true iff S1S2 is true and S2S1 is true
Simple recursive process evaluates an arbitrary sentence, e.g.,
P1,2  (P2,2  P3,1) = true  (true  false) = true  true = true
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16. Truth tables for connectives
If P then Q
If same then true else false
If P then Q and If Q then P
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17. Shiwani Gupta 17

18. Logical equivalence
• Two sentences are logically equivalent iff true in same models:
α ≡ ß iff α╞ β and β╞ α
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19. Inference Rules
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20. DNF (SOP) to CNF (POS)
for resolution refutation
Disjunctive Normal Form
(P  Q  R) V (X   Y  Z)
Conjunctive Normal Form
( P V X)  (P V  Y)  (P V Z)  (…..)
Every sentence of propositional logic is logically
equivalent to a conjunction of disjuncts of literals
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21. Conversion to CNF
• Eliminate α ß to (α  ß )  (ß  α )
• Eliminate α  ß to  α V ß
• CNF requires  to appear only with literals
For this apply
( α) ≡ α (double-negative elimination)
 (α  ß) ≡  α V  ß (de Morgan)
 (α V ß) ≡  α   ß (de Morgan)
• Now we have a sentence with nested  and V
operators applied to literals. Apply distributivity law
distributing V over 
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22. Resolution Inference Rule for CNF
( )
( )
( )
A B C
A
B C
 

− − − − − − − − − − − −
 
“If A or B or C is true, but not A, then B or C
must be true.”
( )
( )
( )
A B C
A D E
B C D E
 
  
− − − − − − − − − − −
   
“If A is false then B or C must be true,
or if A is true then D or E must be true,
hence since A is either true or false, B or
C or D or E must be true.”
( )
( )
( )
A B
A B
B B B

 
− − − − − − − −
  
Simplification
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23. Q ~Q v P ~P v ~Q v R premises
P
~Q v R
R theorem
Proof by resolution
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24. Horn Clauses
• Resolution in general can be exponential in space and time.
• If we can reduce all clauses to “Horn clauses” resolution is linear in
space and time
A clause with at most 1 positive literal. e.g.
• Every Horn clause can be rewritten as an implication with a
conjunction of positive literals in the premises and a single positive
literal as a conclusion.
e.g.
A B C   
B C A 
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25. Wumpus world in propositional logic
• Facts are propositions:
– e.g., W44 “Wumpus is at square (4,4)
– 96 propositions (16 each for wumpus, stench,
pit, breeze, gold, glitter) to represent a
particular cave
• General knowledge in sentences:
– e.g., W44  (S44 S43 S34) “if the wumpus is
at (4,4), there is stench at (4,4), (4,3) and
(3,4)”
– “many” sentences
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26. • Some Atomic Propositions
S12 = There is a stench in cell (1,2)
B34 = There is a breeze in cell (3,4)
W22 = The Wumpus is in cell (2,2)
V11 = We have visited cell (1,1)
OK11 = Cell (1,1) is safe.
etc
• Some rules
(R1) ~S11 => ~W11 ^ ~W12 ^ ~W21
(R2) ~S21 => ~W11 ^ ~W21 ^ ~W22 ^ ~W31
(R3) ~S12 => ~W11 ^ ~W12 ^ ~W22 ^ ~W13
(R4) S12 => W13 v W12 v W22 v W11
etc
Wumpus world in propositional logic
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27. Shiwani Gupta 27

28. Problems with propositional agent
• There are too many propositions to handle
eg. The simple rule “don’t go forward if wumpus
is in front of you” can be stated by set of 64
rules in propositional logic
• What if the world changes over time
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29. Question bank
1. Represent each of these sentences in propositional logic.
a. If I take History, I cannot take Economics.
b. I must take either Japanese or Italian but not both.
c. I must take at least two of cs520, cs552, and cs564.
2. Use truth tables to show the following sentences are valid, and thus that the
equivalences hold:
P ^ (Q ^ R) <=> (P ^ Q) ^ R Associativity of conjunction
P / (Q / R) <=> (P / Q) / R Associativity of disjunction
P ^ Q <=> Q ^ P Commutativity of conjunction
P / Q <=> Q / P Commutativity of disjunction
P ^ (Q / R) <=> (P ^ Q) / ( P ^ R) Distributivity of ^ over /
P / (Q ^ R) <=> (P / Q) ^ ( P / R) Distributivity of / over ^
~(P ^ Q) <=> ~Q / ~P de Morgan’s Law
~(P / Q) <=> ~Q ^ ~P de Morgan’s Law
P => Q <=> ~Q => ~P Contraposition
~ ~P <=> P Double negation
P => Q <=> ~P / Q
P <=> Q <=> (P => Q) ^ (Q => P)
P <=> Q <=> (P ^ Q) / (~Q ^ ~P)
P ^ ~P <=> False
P / ~P <=> True
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30. Question bank
• Explain the steps involved in converting
the propositional logic statement into CNF
with a suitable example
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31. Pros and cons of propositional logic
☺ Propositional logic is declarative: pieces of syntax
correspond to facts
☺ Propositional logic allows partial/disjunctive/negated
information (unlike most data structures and
databases)
☺Propositional logic is compositional: meaning of B1,1 
P1,2 is derived from meaning of B1,1 and of P1,2
☺ Propositional logic is context-independent (unlike
natural language, where meaning depends on
context)
Propositional calculus requires explicit enumeration
All elephants are gray: elephant1-is-gray^elephant1-is-gray^….
logic has very limited expressive power (unlike natural
language) Shiwani Gupta 31

32. First-order predicate logic
• Propositional logic assumes the world contains
facts
• First-order logic (like natural language)
assumes the world contains
– Objects: people, houses, numbers, colors, baseball
games, wars, …
– Relations: brother of, bigger than, inside, part of,
has color, occurred after, owns…
– Properties: red, round, bogus, prime, multistoried...
– Functions: father of, best friend, one more than,
plus, …
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33. Examples
• "One plus two equals three"
Objects: one, two, three, one plus two
Relation: equals
Function: plus. (One plus two is a name for the object that is
obtained by applying the function plus to the objects one and
two. Three is another name for this object.)
• "Squares neighboring the wumpus are smelly."
Objects: wumpus, square
Property: smelly
Relation: neighboring.
• "Evil King John ruled England in 1200."
Objects: John, England, 1200
Relation: ruled
Properties: evil, king.
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34. Syntax of FOL: Basic elements
• Constants KingJohn, 2, NUS,...
• Predicates Brother, GreaterThan, Round...
• Functions Sqrt, LeftLegOf, FatherOf, Cosin...
• Variables x, y, a, b,...
• Connectives , , , , 
• Equality =
• Quantifiers , 
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35. Syntax of predicate logic
Atomic sentence = predicate (term1,...,termn) or term1 = term2
Term = function (term1,...,termn) or constant or variable
Term with no variables is called ground term
E.g. Brother(KingJohn,RichardTheLionheart)
E.g. > (Length(LeftLegOf(Richard)), Length(LeftLegOf(KingJohn)))
E.g. GreaterThan ( Length ( LeftLegOf ( John ) ), ( Length (
LeftLegOf ( Eric ) ) ) )
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36. Complex sentences
• Complex sentences are made from atomic
sentences using connectives
S, S1  S2, S1  S2, S1  S2, S1  S2,
E.g. Sibling(KingJohn,Richard) 
Sibling(Richard,KingJohn)
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37. Equality
• term1 = term2 is true under a given interpretation if and
only if term1 and term2 refer to the same object
• e.g., definition of Sibling in terms of Parent:
x,y Sibling(x,y)  [(x = y)  m,f  (m = f) 
Parent(m,x)  Parent(f,x)  Parent(m,y) 
Parent(f,y)]
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38. Universal quantification
• <variables> <sentence>
Everyone at NUS is smart:
x At(x,NUS)  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 P
Example:
At(KingJohn,NUS) Smart(KingJohn)  At(Richard,NUS)
Smart(Richard)  At(Jesus,NUS) Smart(Jesus) …
Example: “All elephants are gray”
(x )(elephant(x) color(x, GRAY))
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39. Existential quantification
• <variables> <sentence>
Someone at NUS is smart:
x At(x,NUS)  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 P
Example:
At(KingJohn,NUS)  Smart(KingJohn)  At(Richard,NUS) 
Smart(Richard)  At(Jesus,NUS)  Smart(Jesus)  ...
Example: “Someone wrote Computer Chess”
(x write(x, COMPUTER-CHESS))
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40. Shiwani Gupta 42

41. x loves y Loves(x,y)
Everybody loves Jerry x Loves (x, Jerry)
Everybody loves somebody x y Loves (x, y)
There is somebody whom somebody loves y x Loves (x, y)
Nobody loves everybody  x y Loves (x, y)
There is somebody whom Rita doesn’t love y Loves (Rita, y)
There is exactly one person whom everybody loves
y(x Loves(x,y)  z((w Loves (w ,z) z=y))
There are exactly two people whom Lynn Loves
x y ((xy)  Loves(Lynn,x)  Loves(Lynn,y) z( Loves (Lynn
,z) (z=x  z=y)))
Everybody loves himself or herself x Loves(x,x)
There is someone who loves no one besides herself or himself
x y Loves(x,y)  (x=y)
Examples on FOPL
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42. Example
Tony, Mike, John belong to mountaineering
club. Every member of the mountaineering
club is either a skier or mountain climber or
both. No mountain climber likes rain, and all
skiers like snow. Mike dislikes whatever Tony
likes and likes whatever Tony dislikes.
Represent above statements in FOPL
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43. Wumpus world in FOL
• the objects in the environment
– terms: constants, variables, functions
• constants:
– times: 0, 1, 2, ...
– headings: R, L, D, U
– coordinates: 1, 2, 3, 4
– locations: 16 squares
– percepts: Stench, Breeze, Glitter, Bump, Scream, None
– actions: Turn(Left), Turn(Right), Forward, Grab, Shoot
– Agent, Wumpus
• functions:
– Square(x,y) // [x,y]
– Home(Wumpus)
– Perception(s, b, g, h, y)
– Heading(t), Location(t)
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44. Wumpus world in FOL
• predicates: (true or false)
– properties (of one term/object)
• Breezy(t) // agent feeling breeze at time t
• Breeze(s) // breeze blowing on square s
• Pit(s), Gold(s), etc.
• Time(x) // object x is a time
• Coordinate(x), Action(x), Heading(x), etc.
– relations (of multiple terms/objects)
• At(s,t) // agent on square s at time t
• Adjacent (r,s) // squares r and s are adjacent
• Alive(x,t) // x is alive at time t
• Percept(p,t) // perception at time t
• BestAction(a,t) // action a to take at time t
• term = term: (true or false)
– Home(Wumpus) = [3,3]
– Heading(5) = U
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45. Wumpus world in FOL
• the knowledge base
– General knowledge
– Facts that are fixed
– Facts that may change
over time
Atomic
sentences
Compound sentences
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46. Exploring Wumpus world in FOL
Deciding the best action: Need to reason about the cave conditions
Diagnostic rules (generate hidden causes from observed effects)
∀s Breezy(s) ⇒ ∃r Adjacent(r,s) Pit(r)
Causal rules (some fact causes certain percepts to be generated)
∀r Pit(r) ⇒ [∀s Adjacent(r,s) ⇒ Breezy(s)]
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47. First-Order Logic in Wumpus
World
• Suppose an agent perceives a stench,
breeze, glitter at time t = 5:
– Percept([Stench,Breeze,Glitter,None,None],5)
– [Stench,Breeze, Glitter,None,None] is a list
• Then want to query for an appropriate
action. Find an a (ask the KB):
Query:
∀ a BestAction(a,5)
Sol:
a/Grab Shiwani Gupta 49

48. Simplifying the percept and
deciding actions
• Simple Reflex Agent
• Agent Keeping Track of the World
• Reflex
)()],,,,,([,,
)()],,,,,([,,
)()],,,,,([,,
tAtGoldtnonenoneGlitterbsPercepttbs
tBreezetnonenonegBreezesPercepttgs
tStenchtnonenonegbStenchPercepttgb



),()( tGrabActiontAtGoldt 
),(),()( tGrabActiontGoldHoldingtAtGoldt 
),()( tGrabBestActiontGlittert  Shiwani Gupta 50

49. Using logic to deduce properties
• Define properties of locations:
• Diagnostic rule: infer cause from effect
• Causal rule: infer effect from cause
• Neither is sufficient: causal rule doesn’t say if squares far
from pits can be breezy. Leads to definition:
)()(),(,
)()(),(,
lBreezytBreezetlAtAgenttl
lSmellytStenchtlAtAgenttl


),()()( yxAdjacentxPitxyBreezyy 
)(),()(, yBreezyyxAdjacentxPityx 
),()()( yxAdjacentxPitxyBreezyy  Shiwani Gupta 51

50. Inference Rules
• Modus ponens
• Modus Tollens
• And elimination
• And introduction
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51. Inference Rules involving Quantifiers
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52. An Example Proof of Resolution
• 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 “West is a criminal” Criminal(West)
Eq. 1
Eq. 2
Eq. 3
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53. Assumed FOL
Missiles are weapons
An enemy of America
counts as hostile
West is an American
Nono is a country
Nono an enemy of America
Eq. 4
Eq. 5
Eq. 6
Eq. 7
Eq. 9
Eq. 8
Equations 1 to 9 form knowledge base
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54. • Apply Existential elimination to Eq. 2
• From Eq. 10 and AND elimination
• From Eq. 4 and Universal Elimination
• From Eq. 12, 13 and Modus Ponens
• From Eq. 3 and universal elimination
• From Eq. 15, 10 and Modus Ponens
• From eq. 1 and universal elimination (three times)
Eq. 10
Eq. 11
Eq. 12
Eq. 13
Eq. 14
Eq. 15
Eq. 16
Eq. 17
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55. • From Eq. 5 and universal elimination
• From Eq. 8, 18 and Modus Ponens
• From Eq. 6, 7, 14, 16, 19 and And-introduction
• From Eq. 17, 20 and Modus Ponens
Eq. 18
Eq. 19
Eq. 20
Eq. 21
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56. Crime problem revisited using
canonical form
Put original knowledge base to horn form
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57. The Proof
(just in four steps)
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58. Resolution (refutation)
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59. Unification
Q(x)
P(y) → FAIL
P(x)
P(y) → x/y
P(Marcus)
P(y) → Marcus/y
P(Marcus)
P(Julius) → FAIL
P(x,x)
P(y,y) → (y/x)
P(y,z) → (z/y , y/x)
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60. Algorithm: Unify(L1,L2)
1. If L1 or L2 is a variable or
constant, then:
a) If L1 and L2 are identical,
then return NIL.
b) Else if L1 is a variable, then if
L1 occurs in L2 then return
FAIL, else return {(L2/L1)}.
c) Else if L2 is a variable, then if
L2 occurs in L1 then return
FAIL, else return {(L1/L2)}.
d) Else return FAIL.
2. If the initial predicate symbols
in L1 and L2 are not
identical, then return FAIL.
3. If L1 and L2 have a different
number of arguments, then
return FAIL
4. Set SUBST to NIL.
5. For i  1 to number of
arguments in L1:
a) Call Unify with the ith
argument of L1 and the ith
argument of L2, putting result
in S.
b) If S = FAIL then return FAIL.
c) If S is not equal to NIL then:
i. Apply S to the remainder of
both L1 and L2.
ii. SUBST := APPEND(S,
SUBST).
6. Return SUBST.
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61. Shiwani Gupta 68

62. 1. Marcus was a man. man(Marcus)
2. Marcus was a Pompeian. Pompeian(Marcus)
3. All Pompeians were Romans.
x: Pompeian(x)→Roman(x)
4. Caesar was a ruler. ruler(Caesar)
5. All Romans were either loyal to Caesar or hated him.
x: Roman(x) → loyalto(x, Caesar) V hate(x,Caesar)
6. Everyone is loyal to someone. x: y: loyalto(x,y)
7. People only try to assassinate rulers they aren't loyal to.
x:y:person(x) ruler(y)  tryassassinate(x,y)→ loyalto(x,y)
8. Marcus tried to assassinate Caesar.
tryassassinate(Marcus, Caesar)
9. All men are people. x: man(x) → person(x)
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63. A Resolution Proof
Axioms in clause form:
1. man(Marcus)
2. Pompeian(Marcus)
3.  Pompeian(x1) v Roman(x1)
4. Ruler(Caesar)
5.  Roman(x2) v loyalto(x2, Caesar) v hate(x2,
Caesar)
6. loyalto(x3, f1(x3))
7.  man(x4) v  ruler(y1) v  tryassassinate(x4, y1) v
 loyalto (x4, y1)
8. tryassassinate(Marcus, Caesar)
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64. Resolution Proof cont.
Prove: hate(Marcus, Caesar)
hate(Marcus, Caesar)
Roman(Marcus) V loyalto(Marcus,Caesar)
Marcus/x2
5
3
2
7
1
4
8
Marcus/x1
Pompeian(Marcus) V loyalto(Marcus,Caesar)
loyalto(Marcus,Caesar)
Marcus/x4, Caesar/y1
man(Marcus) V  ruler(Caesar) V  tryassassinate(Marcus, Caesar)
 ruler(Caesar) V  tryassassinate(Marcus, Caesar)
 tryassassinate(Marcus, Caesar)
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65. Assignment
1 Marcus was a man
2 Marcus was a Pompeian
3 Marcus was born in 40 A.D.
4 All men are mortal
5 All Pompeians died when volcano erupted in 79 A.D.
6 No mortal lives longer than 150 years
7 It is now 1991
8 Alive means not dead
9 If someone dies, then he is dead at all later times
Prove by resolution that Marcus is not alive now
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66. Using Resolution with Equality and Reduce
Axioms in clause form:
1. man(Marcus)
2. Pompeian(Marcus)
3. Born(Marcus, 40)
4.  man(x1) V mortal(x1)
5.  Pompeian(x2) V died(x2,79)
6. erupted(volcano, 79)
7.  mortal(x3) V  born(x3, t1) V gt(t2—t1, 150) V dead(x3,
t2)
8. Now=1991
9.  alive(x4, t3) V dead (x4, t3)
10.  dead(x5, t4) V alive (x5, t4)
11.  died (x6, t5) V  gt(x6, t5) V dead(x6, t6)
Prove: alive(Marcus, now)
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67. Question Bank• Represent each of these sentences in first-order predicate calculus (FOPL).
– All basketballs are made of rubber.
– Every flower located in Sue’s yard is red.
– At least one question on HW4 was hard.
– All birds can fly except for penguins and ostriches or unless they have a broken wing.
– John’s boss is Mary.
– Selling a book changes who owns it but not who wrote it.
– James father is married to king Johan’s mother
– For all x and all y, if x is the parent of y then y is the child of x
– There is someone who is loved by everyone
– There is noone who does not like icecream
– Spot has atleast two sisters
– Not all students take History and Biology
– Only one student failed in both History and Biology
– There is a barber who shaves all men in town who do not shave themselves
– Well, I like Sandy and donot like Sandy
– One more outburst like that and you will be in contempt of court
– Only one student failed History.
– The best score in History was better than the best score in Biology.
– Every person who dislikes all vegetarians is smart.
– No person likes a smart vegetarian.
– There is a woman who likes all men who are not vegetarians.
– No person likes a professor unless the professor is smart.
– Politicians can fool some of the people all of the time, and they can fool all of the people some of
the time, but they can’t fool all of the people all of the time.
– All Germans speak the same languages
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68. University Questions
Consider the following axioms:
All people who are graduating are happy
All happy people smile
Someone is graduating
i. Represent these statements in FOL
ii.Convert each formula to clause form
iii.Prove that “Is someone smiling?” using
resolution technique. Draw the resolution
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69. Forward Chaining
• Data driven reasoning
• Agent derives conclusion from incoming
percepts
• It starts from known facts (positive literals)
• If all the premises of an implication are
known then its conclusion is added to set
of known facts
• This process continues till query added or
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70. Forward chaining proof
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71. Forward chaining proof
Nation(Nono)
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72. Forward chaining proof
Nation(Nono)
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73. Forward chaining algorithm
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74. Backward Chaining
• Goal directed reasoning
• Works backward from query
• Find those implications in KB that
conclude Q
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75. Backward chaining example
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76. Backward chaining example
Nation(z)
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77. Backward chaining example
Nation(z)
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78. Backward chaining example
Nation(z)
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79. Backward chaining example
Nation(z)
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80. Backward chaining example
Nation(z)
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81. Backward chaining example
Nation(Nono)
{ }
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82. Backward chaining algorithm
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83. Knowledge engineering Process in FOL
1. Identify the task (what to talk about)
2. Assemble the relevant knowledge
3. Decide on a vocabulary of predicates,
functions, and constants
4. Encode general knowledge about the domain
5. Encode a description of the specific problem
instance
6. Pose queries to the inference procedure and
get answers
7. Debug the knowledge base
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84. The digital circuit domain
One-bit full adder
Input bits
to be added
Carry bit
XOR gates
AND gates OR gate
Sum
Carry bit for
next adder
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85. The electronic circuit domain
One-bit full adder
0
1
1
1 0
0
1
1
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86. The electronic circuits domain
1. Identify the task
– Does the circuit actually add properly? (circuit verification)
2. Assemble the relevant knowledge
– Composed of wires and gates; Types of gates (AND, OR,
XOR, NOT)
– Irrelevant: size, shape, color, cost of gates
3. Decide on a vocabulary
– Alternatives:
Type(X1) = XOR
Type(X1, XOR)
XOR(X1)
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87. Gates or circuits can have one or more input terminals and one
or more output terminals.
E.g. the function In(1,X1) denotes the first input terminal for
gate X1.
Connectivity between gates can be represented by a
predicate:
Connected(Out(1,X1),In(1,X2))
Signal has to be on or off –
Signal(t) this function takes terminal as an argument and returns
the signal value for that terminal.
Two constants – 1 and 0.
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88. The electronic circuits domain
4- Encode general knowledge of the domain
• If two terminals are connected they have the same
signal
– t1,t2 Connected(t1, t2)  Signal(t1) = Signal(t2)
• Connected is a commutative predicate
– t1,t2 Connected(t1, t2)  Connected(t2, t1)
• The signal at every terminal is either 1 or 0
– t (Signal(t) = 1)  (Signal(t) = 0)
– 1 ≠ 0
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89. • An OR gate’s output is 1 iff any of its inputs is 1
– g Type(g) = OR  Signal(Out(1,g)) = 1
 n Signal(In(n,g)) = 1
• An AND gate’s output is 0 iff any of its inputs is 0
– g Type(g) = AND  Signal(Out(1,g)) = 0  n Signal(In(n,g))
= 0
• An XOR gate’s output is 1 iff its inputs are different
– g Type(g) = XOR  Signal(Out(1,g)) = 1  Signal(In(1,g)) ≠
Signal(In(2,g))
• A NOT gate’s output is different from its input
– g Type(g) = NOT  Signal(Out(1,g)) ≠ Signal(In(1,g))
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90. The electronic circuits domain
5. Encode the specific problem instance
Type(X1) = XOR Type(X2) = XOR
Type(A1) = AND Type(A2) = AND
Type(O1) = OR
Connected(Out(1,X1),In(1,X2)) Connected(In(1,C1),In(1,X1))
Connected(Out(1,X1),In(2,A2)) Connected(In(1,C1),In(1,A1))
Connected(Out(1,A2),In(1,O1)) Connected(In(2,C1),In(2,X1))
Connected(Out(1,A1),In(2,O1)) Connected(In(2,C1),In(2,A1))
Connected(Out(1,X2),Out(1,C1)) Connected(In(3,C1),In(2,X2))
Connected(Out(1,O1),Out(2,C1)) Connected(In(3,C1),In(1,A2))
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91. The electronic circuits domain
6 - Pose queries to the inference procedure
• What are the possible sets of values of the input of the adder
circuit that lead to a signal of 0 for the sum output bit and a
value of 1 for the carry output terminal?
i1,i2,i3 (Signal(In(1,C1))= i1)  (Signal(In(2,C1)) = i2)
(Signal(In(3,C1)) = i3)  (Signal(Out(1,C1)) = 0 
Signal(Out(2,C1)) = 1)
Answer: (i1 = 1; i2 = 1; i3 = 0) or ((i1 = 1; i2 = 0; i3 = 1) or (i1 = 0;
i2 = 1; i3 = 0)
• What are the possible sets of values of all the terminals of the
adder circuit?
i1,i2,i3,o1,o2 (Signal(In(1,C1)) = i1)  (Signal(In(2,C1)) = i2) 
(Signal(In(3,C1)) = i3)  (Signal(Out(1,C1)) = o1 
Signal(Out(2,C1)) = o2)
7 - Debug the knowledge base
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