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Cooperating Intelligent Systems

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Cooperating Intelligent Systems

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Reduce to propositional logic

- Reduce the first order logic sentences to propositional (boolean) logic.
- Use the inference systems in propositional logic.
We need a system for transfering sentences with quantifiers to sentences without quantifiers

FOL inference rules

All the propositional rules (Modus Ponens, And Elimination, And introduction, etc.) plus:

Universal Instantiation (UI)

Where the variable x is replaced by the ground term a everywhere in the sentence w.

Example:

∀x P(x,f(x),B) ⇒ P(A,f(A),B)

Existential Instantiation (EI)

Where the variable x is replaced by a ground term a (that makes the sentence true) in the sentence w.

Example:

∃x Q(x,g(x),B) ⇒ Q(A,g(A),B)

A must be a new symbol.

Ground term = a term without variables

Propositionalization

Apply Universal Instantiation (UI) and Existential Instantiation (EI) so that every FOL KB is made into a propositional KB.

⇒ We can use the tools from propositional logic to prove theorems.

Problem with function constants: Father(A), Father(Father(A)), Father(Father(Father(A))), etc. ad infinitum...infinite number of sentences...how can we prove this in finite time?

Theorem: We can find every entailed sentence [Gödel, Herbrand], but the search is not guaranteed to stop for nonentailed sentences.(”Solution”: negation-as-failure, stop after a certain time and assume the sentence is false)

Inefficient...generalized (lifted) inference rules better

Notation: Substitution

Subst(q,a) = Apply the substitution q to the sentence a.

Example:

- = {x/John} (replace x with John)
a = (King(x) ∧ Greedy(x)) ⇒ Evil(x)

(King(John) ∧ Greedy(John)) ⇒ Evil(John)

General form: q = {v/g} where v is a variable and g is a ground term.

p1 = King(John)

q1 = King(x)

∀x (King(x) ∧ Greedy(x) ⇒ Evil(x))

p2 = Greedy(John)

q2 = Greedy(x)

q = {x/John}

r = Evil(x)

Generalized (lifted) Modus PonensFor atomic sentences pi, qi, and r where there exists a substitution q such that Subst(q,pi) = Subst(q,qi) for all i

We have John who is King and is Greedy. If someone is King and Greedy then he/she/it is also Evil.

p1 = King(John)

q1 = King(x)

∀x (King(x) ∧ Greedy(x) ⇒ Evil(x))

p2 = Greedy(John)

q2 = Greedy(x)

q = {x/John}

r = Evil(x)

Generalized (lifted) Modus PonensFor atomic sentences pi, qi, and r where there exists a substitution q such that Subst(q,pi) = Subst(q,qi) for all i

Subst(q,p1) = Subst(q,q1)

p1 = King(John)

q1 = King(x)

∀x (King(x) ∧ Greedy(x) ⇒ Evil(x))

p2 = Greedy(John)

q2 = Greedy(x)

q = {x/John}

r = Evil(x)

Generalized (lifted) Modus PonensFor atomic sentences pi, qi, and r where there exists a substitution q such that Subst(q,pi) = Subst(q,qi) for all i

Subst(q,p2) = Subst(q,q2)

p1 = King(John)

q1 = King(x)

∀x (King(x) ∧ Greedy(x) ⇒ Evil(x))

p2 = Greedy(John)

q2 = Greedy(x)

q = {x/John}

r = Evil(x)

King(John), Greedy(John)

Subst(q,r) = Evil(John)

⇒Evil(John)

Generalized (lifted) Modus Ponensp1 = King(John)

q1 = King(x)

∀x (King(x) ∧ Greedy(x) ⇒ Evil(x))

p2 = Greedy(John)

q2 = Greedy(x)

q = {x/John}

r = Evil(x)

King(John), Greedy(John)

Subst(q,r) = Evil(John)

⇒Evil(John)

Generalized (lifted) Modus PonensLifted inference rules make only the necessary substitutions

Forward chaining example

KB:

- All cats like fish
- Cats eat everything they like
- Ziggy is a cat

Example from Tuomas Sandholm @ CMU

Forward chaining example

KB:

- All cats like fish
- Cats eat everything they like
- Ziggy is a cat

Example from Tuomas Sandholm @ CMU

Eats(Ziggy,Fish)

Cat(x) ∧ Likes(x,y) ⇒ Eats(x,y)q = {x/Ziggy, y/Fish}

Likes(Ziggy,Fish)

Cat(x) ⇒ Likes(x,Fish)q = {x/Ziggy}

Cat(Ziggy)

Example: Arms dealer

KB in Horn Form

Forward chaining generates all inferences (also irrelevant ones)

We have proved thatWest is a criminal

Criminal(West)

American(x) ∧ Weapon(y) ∧ Hostile(z)∧ Sells(x,y,z) ⇒ Criminal(x)q = {x/West, y/M, z/NoNo}

Weapon(M)

Sells(West,M,NoNo)

Hostile(NoNo)

Missile(x) ∧ Owns(NoNo,x) ⇒Sells(West,x,NoNo)q = {x/M}

Missile(x) ⇒ Weapon(x)q = {x/M}

Enemy(x,America) ⇒ Hostile(x)q = {x/NoNo}

American(West)

Missile(M)

Owns(NoNo,M)

Enemy(NoNo,America)

Example: Financial advisor

KB in Horn Form

- SavingsAccount(Inadequate) ⇒ Investments(Bank)
- SavingsAccount(Adequate) ∧ Income(Adequate) ⇒ Investments(Stocks)
- SavingsAccount(Adequate) ∧ Income(Inadequate) ⇒ Investments(Mixed)
- ∀x (AmountSaved(x) ∧∃y (Dependents(y) ∧ Greater(x,MinSavings(y))) ⇒ SavingsAccount(Adequate))
- ∀x (AmountSaved(x) ∧∃y (Dependents(y) ∧¬Greater(x,MinSavings(y))) ⇒ SavingsAccount(Inadequate))
- ∀x (Earnings(x,Steady) ∧∃y (Dependents(y) ∧ Greater(x,MinIncome(y))) ⇒ Income(Adequate))
- ∀x (Earnings(x,Steady) ∧∃y (Dependents(y) ∧¬Greater(x,MinIncome(y))) ⇒ Income(Inadequate))
- ∀x (Earnings(x,UnSteady) ⇒ Income(Inadequate))
- AmountSaved($22000)
- Earnings($25000,Steady)
- Dependents(3)

MinSavings(x) ≡ $5000•xMinIncome(x) ≡ $15000 + ($4000•x)

Example from G.F. Luger, ”Artificial Intelligence” 2002

Investments(Mixed)

SavingsAccount(Adequate)

Income(Inadequate)

¬Greater($25000,MinIncome)

Greater($22000,MinSavings)

MinIncome = $27000MinSavings = $15000

AmountSaved($22000)

Dependents(3)

Earnings($25000,Steady)

FOL CNF (Conjunctive Normal Form)

Literal = (possibly negated) atomic sentence, e.g., ¬Rich(Me)

Clause = disjunction of literals, e.g. ¬Rich(Me) ∨ Unhappy(Me)

The KB is a conjunction of clauses

Any FOL KB can be converted to CNF as follows:

- Replace (P ⇒ Q) by (¬P ∨ Q) (implication elimination)
- Move ¬ inwards, e.g., ¬∀x P(x) becomes ∃x ¬P(x)
- Standardize variables apart, e.g., (∀x P(x) ∨∃x Q(x)) becomes (∀x P(x) ∨∃y Q(y))
- Move quantifiers left, e.g., (∀x P(x) ∨∃y Q(y)) becomes ∀x ∃y (P(x) ∨ Q(y))
- Eliminate ∃ by Skolemization
- Drop universal quantifiers
- Distribute ∧ over ∨, e.g., (P ∧ Q) ∨ R becomes (P ∨ R) ∧ (Q ∨ R)

Slide from S. Russel

CNF example

”Everyone who loves all animals is loved by someone”

∀x [∀y Animal(y) ⇒ Loves(x,y)] ⇒ ∃y Loves(y,x)

Implication elimination∀x ¬[∀y ¬Animal(y) ∨ Loves(x,y)] ∨ ∃y Loves(y,x)

Move ¬ inwards (¬∀y P becomes ∃y ¬P)∀x [∃y ¬(¬Animal(y) ∨ Loves(x,y))] ∨ ∃y Loves(y,x)∀x [∃y (Animal(y) ∧¬Loves(x,y))] ∨ ∃y Loves(y,x)∀x [∃y Animal(y) ∧¬Loves(x,y)] ∨ ∃y Loves(y,x)

Standardize variables individually∀x [∃y Animal(y) ∧¬Loves(x,y)] ∨ ∃z Loves(z,x)

Skolemize (Replace ∃ with constants)∀x [Animal(F(x)) ∧¬Loves(x,F(x))] ∨ Loves(G(x),x)Why not ∀x [Animal(A) ∧¬Loves(x,A)] ∨ Loves(B,x) ??

Drop ∀[Animal(F(x)) ∧¬Loves(x,F(x))] ∨ Loves(G(x),x)

Distribute ∨ over ∧[Animal(F(x)) ∨ Loves(G(x),x)] ∧ [¬Loves(x,F(x)) ∨ Loves(G(x),x)]

FOL resolution inference rule

First-order literals are complementary if one unifies with the negation of the other

Where Unify(li,¬mj) = q.

Note that li and mj are removed

Resolution proves KB ⊨a by proving (KB ∧¬a) is unsatisfiable

Start fromthe top

¬

Arms dealer example

Resolution proves KB ⊨a by proving (KB ∧¬a) is unsatisfiable

∀x (American(x)∧Weapon(y)∧Hostile(z)∧Sells(x,y,z)⇒Criminal(x))

Translate to CNF:

∀x (¬(American(x)∧Weapon(y)∧Hostile(z)∧Sells(x,y,z))∨Criminal(x))

∀x ((¬American(x)∨¬Weapon(y)∨¬Hostile(z)∨¬Sells(x,y,z))∨Criminal(x))

∀x (¬American(x)∨¬Weapon(y)∨¬Hostile(z)∨¬Sells(x,y,z)∨Criminal(x))

¬American(x)∨¬Weapon(y)∨¬Hostile(z)∨¬Sells(x,y,z)∨Criminal(x)

- Any FOL KB can be converted to CNF as follows:
- Replace (P ⇒ Q) by (¬P ∨ Q) (implication elimination)
- Move ¬ inwards, e.g., ¬∀x P(x) becomes ∃x ¬P(x)
- Standardize variables apart, e.g., (∀x P(x) ∨ ∃x Q(x)) becomes (∀x P(x) ∨ ∃y Q(y))
- Move quantifiers left, e.g., (∀x P(x) ∨ ∃y Q(y)) becomes ∀x ∃y (P(x) ∨ Q(y))
- Eliminate ∃ by Skolemization
- Drop universal quantifiers
- Distribute ∧ over ∨, e.g., (P ∧ Q) ∨ R becomes (P ∨ R) ∧ (Q ∨ R)

¬

Arms dealer example

Resolution proves KB ⊨a by proving (KB ∧¬a) is unsatisfiable

∀x (American(x)∧Weapon(y)∧Hostile(z)∧Sells(x,y,z)⇒Criminal(x))

Translate to CNF:

∀x (¬(American(x)∧Weapon(y)∧Hostile(z)∧Sells(x,y,z))∨Criminal(x))

∀x ((¬American(x)∨¬Weapon(y)∨¬Hostile(z)∨¬Sells(x,y,z))∨Criminal(x))

∀x (¬American(x)∨¬Weapon(y)∨¬Hostile(z)∨¬Sells(x,y,z)∨Criminal(x))

¬American(x)∨¬Weapon(y)∨¬Hostile(z)∨¬Sells(x,y,z)∨Criminal(x)

- Any FOL KB can be converted to CNF as follows:
- Replace (P ⇒ Q) by (¬P ∨ Q) (implication elimination)
- Move ¬ inwards, e.g., ¬∀x P(x) becomes ∃x ¬P(x)
- Standardize variables apart, e.g., (∀x P(x) ∨ ∃x Q(x)) becomes (∀x P(x) ∨ ∃y Q(y))
- Move quantifiers left, e.g., (∀x P(x) ∨ ∃y Q(y)) becomes ∀x ∃y (P(x) ∨ Q(y))
- Eliminate ∃ by Skolemization
- Drop universal quantifiers
- Distribute ∧ over ∨, e.g., (P ∧ Q) ∨ R becomes (P ∨ R) ∧ (Q ∨ R)

¬

Arms dealer example

Resolution proves KB ⊨a by proving (KB ∧¬a) is unsatisfiable

Where Unify(li,¬mj) = q.

¬

Arms dealer example

Resolution proves KB ⊨a by proving (KB ∧¬a) is unsatisfiable

Where Unify(li,¬mj) = q.

l1 = ¬American(x)

l2 = ¬Weapon(y)

l3 = ¬Sells(x,y,z)

l4 = ¬Hostile(z)

l5 = Criminal(x)

m1 = Criminal(West)

Unify(l5,¬m1) = q= {x/West}

Subst(q,l1∨ l2∨ l3∨ l4) =...

¬

Arms dealer example

Resolution proves KB ⊨a by proving (KB ∧¬a) is unsatisfiable

l1 = ¬American(x)

l2 = ¬Weapon(y)

l3 = ¬Sells(x,y,z)

l4 = ¬Hostile(z)

l5 = Criminal(x)

m2 = American(West)

Unify(l1,¬m2) = q= {x/West}

Subst(q,l2∨ l3∨ l4) =...

¬

Arms dealer example

Resolution proves KB ⊨a by proving (KB ∧¬a) is unsatisfiable

l2 = ¬Weapon(y)

l3 = ¬Sells(x,y,z)

l4 = ¬Hostile(z)

m3 = Weapon(x)

m4 = Missile(x)

Unify(l2,¬m3) = q= {y/x}

Subst(q,l2∨ l3∨ l4 ∨ m4) =...

¬

Arms dealer example

Resolution example II

- Problem Statement: Tony, Shikuo and Ellen belong to the Hoofers Club. Every member of the Hoofers Club is either a skier or a mountain climber or both. No mountain climber likes rain, and all skiers like snow. Ellen dislikes whatever Tony likes and likes whatever Tony dislikes. Tony likes rain and snow.
- Query: Is there a member of the Hoofers Club who is a mountain climber but not a skier?

Example from Charles Dyer (referenced by Tuomas Sandhom @ CMU)

KB

The rules only apply to members of the Hoofers club (our domain).

Problem Statement: Tony, Shikuo and Ellen belong to the Hoofers Club. Every member of the Hoofers Club is either a skier or a mountain climber or both. No mountain climber likes rain, and all skiers like snow. Ellen dislikes whatever Tony likes and likes whatever Tony dislikes. Tony likes rain and snow.

Query

Query: Is there a member of the Hoofers Club who is a mountain climber but not a skier?

(KB ∧¬Q) to Clause form...(I)

(KB ∧¬Q) to Clause form...(II)

(KB ∧¬Q) to Clause form...(III)

(KB ∧¬Q) to Clause form...(IV)

(KB ∧¬Q) in Clause form

We drop the universal quantifiers...

But introduce different notation to keepbetter track...

2

3

4

5

6

7

8

9

10

11

12

13

14

We have proved that there is amember of the Hoofers club whois a mountain climber but not a skier.

Unify(p13, ¬p14) = q= {x/Ellen}