Announcements:
● No class Mar 14
● Lab 2 due soon
● Lab 3 will be due Apr 7, content is in the next lecture
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● Lab4willbedueApr28
PvRvW !P v Q v X
2.a !Human(Soc) v Moral(Soc)
1. P(Sk0(x), x)
2. !P(x, Soc) v Q(x) -> {x->y} !P(y, Soc) v Q(y)
Unify {x->Soc, y -> Sk0(Soc) P(Sk0(Soc), Soc)
!P(Sk0(Soc), Soc) v Q(Sk0(Soc))
3. Q(Sk0(Soc))
PvQ !P v !Q
– null clause
Sk conversion and resolution proof example
A vegetarian is someone who eats no meat.
V(x) – x is a vegetarian. E(x,y) – x eats y. M(y) – y is meat
∀x [V(x) ⇔ ¬∃y [E(x,y) ⋀ M(y)]]
∀x [V(x) => ¬∃y [E(x,y) ⋀ M(y)]] ^ [[¬∃y [E(x,y) ⋀ M(y)]] => V(x)]
∀x [¬V(x) v ¬∃y [E(x,y) ⋀ M(y)]] ^ [¬[¬∃y[E(x,y) ⋀ M(y)]] v V(x)]
∀x [¬V(x) v ∀y [¬E(x,y) v ¬M(y)]] ^ [∃y[E(x,y) ⋀ M(y)] v V(x)]
∀x [¬V(x) v ∀y [¬E(x,y) v ¬M(y)]] ^ [[E(x,Sk0(x)) ⋀ M(Sk0(x))] v V(x)] [¬V(x) v [¬E(x,y) v ¬M(y)]] ^ [[E(x,Sk0(x)) ⋀ M(Sk0(x))] v V(x)]
¬V(x) v ¬E(x,y) v ¬M(y) E(x,Sk0(x)) v V(x) M(Sk0(x)) v V(x)
Next a proof:
● A vegetarian is someone who eats no meat.
● John eats only carrots and tomatoes
● Carrots are not meat.
● Tomatoes are not meat.
● John is a vegetarian.
∀y E(John, y) => [C(y) v T(y)] ∀y C(y) => ¬M(y)
∀y T(y) => ¬M(y)
1. ¬ E(John, y) v C(y) v T(y)
2. ¬ C(y) v ¬M(y)
3. ¬ T(y) v ¬M(y)
4. ¬V(x) v ¬E(x,y) v ¬M(y)
5. E(x,Sk0(x)) v V(x)
6. M(Sk0(x)) v V(x)
7. ¬V(John)
Unify 1+2 under {}
8. ¬ E(John, y) v T(y) v ¬M(y)
Unify 5+8 {x -> John, y -> Sk0(John)}
9. V(John) v T(Sk0(John)) v ¬M(Sk0(John)) Unify 9+7 {}
10. T(Sk0(John)) v ¬M(Sk0(John))
Unify 6+10 {x -> John}
11. V(John) v T(Sk0(John))
Unify 10+3 {y->Sk0(John)} then factor
12. ¬M(Sk0(John)) Unify 12+6 {x->John} 13. V(John)
Unify 13+7 {}:
Null clause -> SUCCESS ∃y E(K, y) ^ H(y)
∀y H(y) => M(y) Proof: ¬ V(K)
1. ¬ C(y) v ¬M(y)
2. ¬ T(y) v ¬M(y)
3. ¬V(x) v ¬E(x,y) v ¬M(y)
4. E(x,Sk0(x)) v V(x)
5. M(Sk0(x)) v V(x)
6. E(K, Sk1)
8. ¬ H(y) v M(y)
Unify 7+8 {y -> Sk1} 10. M(Sk1)
Unify 10+3 {x->x, y->Sk1} 11. ¬V(x) v ¬E(x,Sk1)
Unify 11+9 {x->K} 12. ¬E(K,Sk1)
Unify 12+6 {}
Null clause -> SUCCESS
Basic probability
P(E,F) = P(F) * P(E|F) P(E,F) = P(E) * P(F|E)
P(F) * P(E|F) = P(E) * P(F|E)
P(C1)=3/6, P(C2)=1/6, P(C3)=2/6
P(H|C1)=.1, P(H|C2)=.3, P(H|C3)=.7 P(T|C1)=.9, P(T|C2)=.7, P(T|C3)=.3
P(H)=P(C1)* P(H|C1) + P(C2)* P(H|C2) + P(C3)* P(H|C3) = .05 + .05 + 7/30 = .333
P(C1|H) = P(C1) * P(H|C1) / P(H) = 3/6 * .1 / .333 P(C2|H) = P(C2) * P(H|C2) / P(H) = 1/6 * .3 / .333 P(C3|H) = P(C3) * P(H|C3) / P(H) = 2/6 * .7 / .333
# count the events
P(H|H) = P(C1|H)* P(H|C1,H) + P(C2|H)* P(H|C2,H) + P(C3|H)* P(H|C3,H) # drop independent probs
P(H|H) = P(C1|H)* P(H|C1) + P(C2|H)* P(H|C2) + P(C3|H)* P(H|C3)
# after class discussion
¬∃y [E(x,y) ⋀ M(y)]
¬∃y [M(y) => E(x,y)] -> ¬∃y [¬M(y) v E(x,y)]
Prove : P(x) v R(x) => Q(x) v Y(x) KB+![ P(x)vR(x)=>Q(x)vY(x)]
![ !( P(x) v R(x)) v (Q(x) v Y(x)) ] ( P(x) v R(x)) ^ (!Q(x) ^ !Y(x))
P(x) v R(x) !Q(x)
A(x) !A(x)
# BNF Grammar
Start | Pair => Qv!Wwo
Start [id] |
Pair [id] =>
((x => y) => (z v w)) (x, y, z)
X,y,z X,y X
List := List , element | element List := element , List | element
# easy case
(( P & !Q) & W) | (A | [B & C])
( P & !Q) | (A | [B & C]) W | (A | [B & C])
W | [(A | B) & (A | C)] P| [(A|B)&(A|C)] !Q| [(A|B)&(A|C)]
W|A|B W|A|C P|A|B P|A|C !Q | A | B
!Q | A | C
# hard case
(( P & !Q) | W) | (A | [B & C])
[(P | W) & (Q! | W)] | (A | [B & C])
(A | [B & C]) | (P | W) (A | [B & C]) | (Q! | W)
[(A | B) & (A | C)] | (P | W) [(A | B) & (A | C)] | (Q! | W)
P|W|A|B P|W|A|C Q! | W | A | B Q! | W | A | C
(A&B)|(C&D)
[A|(C&D)] & [B|(C&D)]
[A|(C&D)] [B|(C&D)]
A|C A|D B|C B|D
(P& Q) | W
P <=> Q <=> R [X + (y * z)] – g
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