1. L7. Logic for knowledge representation
and inference
知識表現 推理
Propositional logic versus first-order logic
(命題論理対一階述語論理 )
Apply first-order logic
e.g. A family relations
A proof

2. Sentence in propositional logic
Sentence  AtomicSentence | ComplexSentence
AtomicSentence  proposition symbols like P, Q, R
True | False
ComplexSentence  (Sentence)
| Sentence Connectives
like P  Q  (P  Q)  ( P   Q)
Connectives:  (not),  (and),  (or),  (implies), and  (equivalent)
原子文 複文
連結語
(必然的に）伴う

3. FOL: basic elements and sentences
Constant (定数) symbols: refer to the same object in the same interpretation
Predicate (述語) symbols: refer to a particular relation in the model.
Function (関数) symbols: refer to particular objects without using their names.
Logic connectives:  (not),  (and),  (or),  (implies), and  (equivalent)
Equality (対等): = e.g. Father(John) = Henry
Variables (変数): substitute (取り替える) for the name of an object.
e.g., x, y, a, b,… x, Cat(x)  Mammal(x) 　　　　　　　　　　//Mammal:哺乳動物.
　Quantifiers (数量詞):  (universal (一般) quantification symbol)
　　　　  (existential (存在に関する) quantification symbol)
　　　　　 x, for any x, … x, there is a x, …
Atomic sentences = predicate(term1, term2, …termn)
Term = function(term1, term2, …termn)
or constant or variable
Complex sentences: made from atomic sentences using connectives.

4. Seven inference rules for propositional Logic
R(1) Implication-Elimination
R(2) And-Elimination
R(3) And-Introduction
R(4) Or-Introduction
R(5) Double-Negation Elimination
R(6) Unit Resolution
R(7) Logic connectives: 
  ,  i
1  2 … n
1  2 … n
1, 2, …, n
1  2  …  n i 
   
  ,  
  ,    
  

5. The three new inference rules added to FOL
R(8) Universal-Elimination: For any sentence , variable v, and ground term g:
e. g.,  x Likes(x, IceCream), we can use the substitute {x/Rose} and
infer Like(Rose, IceCream).
R(9) Existential-Elimination: For any sentence , variable v, and constant symbol
k that does not appear elsewhere in the knowledge base:
e. g.,  x Kill(x, Victim), we can infer Kill(Murderer, Victim), as long as Murderer
does not appear elsewhere in the knowledge base.
R(10) Existential-Introduction: For any sentence , variable v that does not occur
in , and ground term g that does occur in :
e. g., from Likes(Rose, IceCream)
we can infer  x Likes(x, IceCream).
SUBST({v/g}, )
 v 
Ground term is a term that contains no variables. 
v/g
SUBST({v/k}, )
 v   k
 v SUBST({g/v}, ) 

6. Representing Simple Rules
Variables: x, P, C, ...
Generalised Facts: female(P), parent(P, C), ...
Conjunctions: parent(P, C)  female(P)
Rules:  P, C parent(P, C)  female(P)  mother(P, C)

7. Knowledge Bases: Rules
Family Relations
 GP, P, C
parent(P, C)  female(P)  mother(P, C)
parent(P, C)  male(P)  father(P, C)
parent(GP, P)  parent(P, C)  grandparent(GP, C)
parent(GP, P)  female(GP)  parent(P, C) grandmother(GP, C)
grandparent(GP, C)  male(GP)  grandfather(GP, C)

8. Knowledge Base Architecture
Updates
KNOWLEDGE
BASE (KB)
facts and rules
Query
Answer
INFERENCE MECHANISM

9. Example of proof in first-order logic
Bob is a buffalo | 1. Buffalo(Bob) f1
Pat is a pig | 2. Pig(Pat) f2
Buffaloes run faster than pigs | 3.  x, y Buffalo(x)  Pig(y)  Faster(x,y) --r1
To proof:
Bob runs faster than Pat
Apply R(3) to f1 And f | 4. Buffalo(Bob)  Pig(Pat) f3
(And-Introduction)
Apply R(8) to r1 {x/Bob, y/Pat} | 5. Buffalo(Bob)  Pig(Pat)  Faster(Bob,Pat) --f4
(Universal-Elimination)
Apply R(1) to f3 And f | 6. Faster(Bob,Pat) f5
(Inplication-Elimination )

10. Search with primitive inference rules
Operators are inference rules
States are sets of sentences
Goal test checks state to see if it contains query sentence
R(1) to R(10) are common inference pattern
1 2 3
Apply R(3) to 1 & 2
Apply R(8) to 3
Apply R(1) to 4 & 5

11. A Reasoning System Rule Base Working Memory Interaction with
Inference Engine
Matching
Fire a Rule
Select a Rule
Acting

12. Here is the answer: the gold is in [2,3].
Finding gold
The agent has been in [1,1]->[2,1]->[1,1]->[1,2] and
knows there is a wumpus in [1,3] and a pit in [3,1] then
goes to [2,2]
continues to [3,2] (forward since it perceives nothing special)
Now, percept sentences (add a new one):
there is a breeze in the square [3,2]  B3,2
Since knowledge is not enough to judge whether [4,2] and [3,3] are safe or not, the agent goes back to [2,2], turn left and goes to [2,3].
Now, percept sentences(add new ones):
　　there is a glitter in the square [2,3]  G2,3
The agent has been in [1,1]->[2,1]->[1,1]->[1,2] -> [2,2] -> [3,2] -> [2,2]-> [2,3]
Here is the answer: the gold is in [2,3].

13. Finding a way back to the start position
Go back along the path (5 steps)
[1,1]->[2,1]->[1,1]->[1,2] -> [2,2] -> [3,2] -> [2,2]-> [2,3]
(a doubly linked list would be a good data structure for recording
agent’s history path)
Using safe squares to build a tree, search a shortest path
from [2,3] to [1,1] (3 steps). s
s b b p w g p A p

14. Appendix (追加)
Apply first-order logic to making inference in the Wumpus world.
(If you have time, please take a look this appendix.)
A gold finding agent in the Wumpus world

15. Predicates and rules
At(p, l, S) : Agent p is at the location l in the situation S
Percept([Smell, b, g, u, v], l, S, t): Agent perceives smell at time t at the location l in the situation S
Perception rules
R1:  b, g, u, v, l, s, t Percept([Smell, b, g, u, v], s, t)  Smell(t)
R2: sm, g, u, v, l, s, t Percept([sm, Breeze, g, u, v], s, t)  Breeze(t)
R3: sm, b, u, v, l, s, t Percept([sm, b, Glitter, u, v], s, t)  AtGoldRoom(t)
R4: sm, b, u, v, l, s, t Percept([sm, b, g, Wall, v], s, t)  Wall(t)
R6: sm, b, g, u, v, l, s, t Percept([sm, b, g, u, v], s, t) 
 Smell(t)  Breeze(t)  AtGoldRoom(t)  Wall(t)
Action rules
R7:  i, j Result(Forward, li) = lj i  j  i, j Result(Turn, Sm) = Sn m  n
Orientation rules
R8:  s (Orientation(p, s ) =0  Face(p, east) )  (Orientation(p, s) = 90  Face(p, north )
 (Orientation(p, s) = 180  Face(p, west)  (Orientation(p, s) = 270  Face(p, south) )
R9:  a, d, p, s Orientation(p, Result(a, s)) =d  [(a=Turn(Right) d=Mod(Orientation(p,s) – 90, 360))
 (a=Turn(Left) d=Mod(Orientation(p,s) + 90, 360) )
 (Orientation(p,s)=d (a=Turn(Right)  a=Turn(Left)))]

16. R20: l1, l2, s At(Pit, l1)  Adjacent(,l1, l2)  Breeze(l2)
Location rules
R10:  l, l   x, y [x, y]
R11: (a)  x, y LocationToward([x, y], 0) = [x+1, y] (b)  x, y LocationToward([x, y], 90) = [x, y+1]
R12: (a)  x, y LocationToward([x, y], 180) = [x-1, y] (b)  x, y LocationToward([x, y], 270) = [x+1, y-1]
R13:  p, l, s At(p, l, s)  LocationAhead(p,S) = LocationToward(l, Orientation(p,s))
R14:  l1, l2 Adjacent(l1, l2)   d l1=LocationToward(l2, d)
R15:  x, y Wall([x, y])  (x=0  x=5  y=0  y=5)
R16:  a, d, p, s At(p, l, Result(a, s))  [(a=Forward  l = LocationAhead(p,s)  Wall(l))
 (At(p,l,s)  a  Forward)]
Hidden properties (rules)
R17: l, s At(Agent, l, s)  Smell(s)  Smell(l)
R18: l, s At(Agent, l, s)  Breeze(s)  Breeze(l)
R19: l1, s Breeze(l1)   l2 At(Pit, l2,s)  (l2, = l1  Adjacent(,l1, l2))
R20: l1, l2, s At(Pit, l1)  Adjacent(,l1, l2)  Breeze(l2)
R21:  l1, s Breeze(l1)   l2 At(Pit, l2, s )  (l2, = l1  Adjacent(,l1, l2))

17. R23: l1, l2, s At(Wumpus, l1)  Adjacent(,l1, l2)  Smell(l2)
Hidden properties (rules)
R22: l1, s Smell(l1)   l2 At(Wumpus, l2,s)  (l2, = l1  Adjacent(,l1, l2))
R23: l1, l2, s At(Wumpus, l1)  Adjacent(,l1, l2)  Smell(l2)
R24:  l1, s Smell(l1)   l2 At(Wumpus, l2, s )  (l2, = l1  Adjacent(,l1, l2))
R25: x, y, g, u, v, s, t Percept([None, None, g, u, v], t)  At(Agent, x, s)  Adjacent(,x, y)  OK(y)
R26: x, t ( At(Wumpus, x, t)  At(Pit, x, t))  OK(x)
Obtaining Gold
R27:  x, s AtGoldRoom( x, s)  Present(Gold, x, s)
R28:  x, s Present( x, s)  Portable(x) Holding(x, Result(Grab, s))
R29:  x, s Holding(x, Result(Release, s))
R30:  a, x, s Holding(x, s)  (aRelease)  Holding(x, Result(a, s))
R31:  a, x, s Holding(x, s)  (aGrab   (Present(x, s)  Portable(x))  Holding(x, Result(a, s))
P true afterwards  [an action made P true  P true already and no action made P false]
R32:  a, x, s Holding(x, Result(a, s)) 
[(a=Grab  (Present(x, s)  Portable(x))  (Holding(x, s)  aRelease) ]

18. Since the world is not changing, our inferences below drop time term.
Initial state
At(Agent, [1,1])
Apply And-Elimination rule to R6, we obtain
C1 Smell([1,1]) , Breeze([1,1]) , AtGoldRoom([1,1]) , Wall([1,1])
Apply And-Elimination rule to R21 and apply R15, we obtain
C At(Pit, [1,1]), At(Pit, [2,1]), At(Pit, [2,1]), Wall([1,0]), Wall([0,1])
Apply And-Elimination rule to R24, we obtain
C At(Wumpus, [1,1]), At(Wumpus, [2,1]), At(Wumpus, [1,2])
Apply R26 to C2 and C3, we obtain
C OK([1,1]), OK([2,1]), OK([1,2])
Apply R11(a) and R16, we obtain
At(Agent, [2,1])
Apply And-Elimination rule to R2, we obtain
C5 Smell([2,1]) , Breeze([2,1]) , AtGoldRoom([2,1]) , Wall([2,1])
Apply R19 to Breeze([2,1]), we obtain
C6 At(Pit, [1,1])  At(Pit, [2,1])  At(Pit, [2,2])  At(Pit, [3,1])

19. Apply unit resolution twice to C6, we obtain
C7 At(Pit, [2,2])  At(Pit, [3,1])
Use C4 and C7, (for a conservative agent, it would not go where is not sure), and apply R9, R12(a) and R16, we obtain
At(Agent, [1,1]), Orientation(Agent, [1,1])=180
Use C4 and apply R9, R11(b) and R16, we obtain
At(Agent, [1,2]), Orientation(Agent, [1,2])=90
Apply R22 to Smell([1,2]), we obtain
C8 At(Wumpus, [1,1])  At(Wumpus, [1,3])  At(Wumpus, [1,2])  At(Wumpus, [2,2])
Apply unit resolution twice to C8, we obtain
C9 At(Wumpus, [2,2])  At(Wumpus, [1,3])
Apply R24 and And-Elimination rule to Smell([2,1]), we obtain
C At(Wumpus, [1,1]), At(Wumpus, [1,2]), At(Wumpus, [2,2]), At(Wumpus, [2,1])
Apply unit resolution to C9, we obtain
C At(Wumpus, [1,3])
Apply R21 and And-Eleimination rule to Breeze([1,2]), we obtain
C12 At(Pit, [1,1]), At(Pit, [2,1]), At(Pit, [2,2]), At(Pit, [3,1])

20. Apply R26 to C10 and C12, we obtain
C OK([2,2]),
Use C13 and apply R9, R11(a) and R16, we obtain
At(Agent, [2,2]), Orientation(Agent, [2,2])=0
Use C4 and apply R9, R11(b) and R16, we obtain
At(Agent, [1,2]), Orientation(Agent, [1,2])=90
Apply R21 and And-Elimination rule to Breeze([2,2]), and
apply R24 and And-Elimination rule to Smell([2,2]), we obtain
C14 At(Pit, [2,1]), At(Pit, [1,2]), At(Pit, [2,2]), At(Pit, [2, 3], At(Pit, [2, 3])
C At(Wumpus, [2,1]), At(Wumpus, [1,2]), At(Wumpus, [2,2]), At(Wumpus, [2,3], At(Wumpus, [3,2])
Apply R26 to C14 and C15, we obtain
C OK([2,3]), OK([3,2])
Apply R11(a) and R16, we obtain
At(Agent, [3,2])
Apply R19 to Breeze([3,2]), we obtain
C17 At(Pit, [2,2])  At(Pit, [2,3])  At(Pit, [3,2])  At(Pit, [3,1])  At(Pit, [4,2])
Apply twice unit resolution to C9, we obtain
C At(Pit, [3,2])  At(Pit, [3,1])  At(Pit, [4,2])

21. The rest of question is how to go back safely?
For safe, agent goes back to [2, 2]
Use C13 and apply R9, R12(a) and R16, we obtain
At(Agent, [2,2]), Orientation(Agent, [2,2])=180
Use C16 and apply R9, R11(b) and R16, we obtain
At(Agent, [2,3]), Orientation(Agent, [2,3])=90
Apply R3, we obtain
C AtGoldRoom( [2,3])
Apply R32, we obtain
C20 Holding([2,3], Result(Grab))
The rest of question is how to go back safely?