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Proof Methods for Propositional Logic

Russell and Norvig Chapter 7

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Logical equivalence

n  Two sentences are logically equivalent iff they are true in the same models: α ≡ ß iff α╞ β and β╞ α

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Validity and satisfiability A sentence is valid (a tautology) if it is true in all models

e.g., True, A ∨¬A, A ⇒ A, (A ∧ (A ⇒ B)) ⇒ B

Validity is connected to inference via the Deduction Theorem: KB ╞ α if and only if (KB ⇒ α) is valid

A sentence is satisfiable if it is true in some model e.g., A ∨ B

A sentence is unsatisfiable if it is false in all models e.g., A ∧ ¬A

Satisfiability is connected to inference via the following: KB ╞ α if and only if (KB ∧¬α) is unsatisfiable (known as proof by contradiction)

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Inference rules

n  Modus Ponens A ⇒ B, A B Example: “raining implies soggy courts”, “raining” Infer: “soggy courts”

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Example

n  KB: {A⇒B, B ⇒C, A} n  Is C entailed? n  Yes.

Given Rule Inferred

A⇒B, A Modus Ponens B

B ⇒C, B Modus Ponens C

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Inference rules (cont.)

n  Modus Tollens A ⇒ B, ¬B ¬A Example: “raining implies soggy courts”, “courts not soggy” Infer: “not raining”

n  And-elimination A ∧ B A

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Reminder: The Wumpus World

n  Performance measure q  gold: +1000, death: -1000 q  -1 per step, -10 for using the arrow

n  Environment q  Squares adjacent to wumpus: smelly q  Squares adjacent to pit: breezy q  Glitter iff gold is in the same square q  Shooting kills wumpus if you are facing it q  Shooting uses up the only arrow q  Grabbing picks up gold if in same square

n  Sensors: Stench, Breeze, Glitter, Bump n  Actuators: Left turn, Right turn, Forward, Grab, Release,

Shoot

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Inference in the wumpus world

Given: 1.  ¬B1,1 2.  B1,1 ⇔ (P1,2 ∨ P2,1) Let’s make some inferences: 1.  (B1,1 ⇒ (P1,2 ∨ P2,1)) ∧ ((P1,2 ∨ P2,1) ⇒ B1,1 ) (By

definition of the biconditional) 2.  (P1,2 ∨ P2,1) ⇒ B1,1 (And-elimination)

3.  ¬B1,1 ⇒ ¬(P1,2 ∨ P2,1) (equivalence with contrapositive) 4.  ¬(P1,2 ∨ P2,1) (modus ponens) 5.  ¬P1,2 ∧ ¬P2,1 (DeMorgan’s rule) 6.  ¬P1,2 (And-elimination) 7.  ¬P2,1 (And-elimination)

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More inference

n  Recall that when we were at (2,1) we could not decide on a safe move, so we backtracked, and explored (1,2), which yielded ¬B1,2. This yields ¬P2,2 ∧ ¬P1,3

n  Now we can consider the implications of B2,1

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ABG

PS

W

= Agent = Breeze = Glitter, Gold

= Pit = Stench

= Wumpus

OK = Safe square

V = Visited

A

OK 1,1 2,1 3,1 4,1

1,2 2,2 3,2 4,2

1,3 2,3 3,3 4,3

1,4 2,4 3,4 4,4

OKOKB

P?

P?A

OK OK

OK 1,1 2,1 3,1 4,1

1,2 2,2 3,2 4,2

1,3 2,3 3,3 4,3

1,4 2,4 3,4 4,4

V

(a) (b)

BB P!

A

OK OK

OK 1,1 2,1 3,1 4,1

1,2 2,2 3,2 4,2

1,3 2,3 3,3 4,3

1,4 2,4 3,4 4,4

V

OK

W!

VP!

A

OK OK

OK 1,1 2,1 3,1 4,1

1,2 2,2 3,2 4,2

1,3 2,3 3,3 4,3

1,4 2,4 3,4 4,4

V

S

OK

W!

V

V V

BS G

P?

P?

(b)(a)

S

ABG

PS

W

= Agent = Breeze = Glitter, Gold

= Pit = Stench

= Wumpus

OK = Safe square

V = Visited

Resolution

1.  ¬P2,2, ¬P1,1 2.  B2,1 ⇔ (P1,1 ∨ P2,2 ∨ P3,1) 3.  B2,1 ⇒ (P1,1 ∨ P2,2 ∨ P3,1) (biconditional elimination) 4.  P1,1 ∨ P2,2 ∨ P3,1 (modus ponens) 5.  P1,1 ∨ P3,1 (resolution rule) 6.  P3,1 (resolution rule)

The resolution rule: if there is a pit in (1,1) or (3,1), and it’s not in (1,1), then it’s in (3,1).

P1,1 ∨ P3,1, ¬P1,1 P3,1 CS440 Fall 2015 10

Resolution

Unit Resolution inference rule: l1 ∨… ∨ lk, m

l1 ∨ … ∨ li-1 ∨ li+1 ∨ … ∨ lk

where li and m are complementary literals. Full Resolution

l1 ∨… ∨ lk, m1 ∨ … ∨ mn l1 ∨…∨ li-1 ∨ li+1 ∨…∨ lk ∨ m1 ∨…∨ mj-1 ∨ mj+1 ∨...∨ mn where li and mj are complementary literals.

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Resolution is sound

For simplicity let’s consider clauses of length two:

l1 ∨ l2, ¬l2 ∨ l3 l1 ∨ l3

To derive the soundness of resolution consider the values l2 can take: q  If l2 is True, then since we know that ¬l2 ∨ l3 holds, it must

be the case that l3 is True. q  If l2 is False, then since we know that l1 ∨ l2 holds, it must

be the case that l1 is True.

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Resolution

n  Properties of the resolution rule: q  Sound q  Complete

n  Resolution can be applied only to disjunctions of literals. How can it be complete?

n  Turns out any knowledge base can be expressed as a conjunction of disjunctions (conjunctive normal form, CNF).

n  Example: (A ∨ ¬B) ∧ (B ∨ ¬C ∨ ¬D) CS440 Fall 2015 13

Conversion to CNF

B1,1 ⇔ (P1,2 ∨ P2,1) 1.  Eliminate ⇔, replacing α ⇔ β with (α ⇒ β)∧(β ⇒α).

(B1,1 ⇒ (P1,2 ∨ P2,1)) ∧ ((P1,2 ∨ P2,1) ⇒ B1,1) 2.  Eliminate ⇒, replacing α ⇒ β with ¬α ∨ β.

(¬B1,1 ∨ P1,2 ∨ P2,1) ∧ (¬(P1,2 ∨ P2,1) ∨ B1,1)

3.  Move ¬ inwards using de Morgan's rules and double-negation: (¬B1,1 ∨ P1,2 ∨ P2,1) ∧ ((¬P1,2 ∧ ¬P2,1) ∨ B1,1)

4.  Apply distributive law (∧ over ∨) and flatten: (¬B1,1 ∨ P1,2 ∨ P2,1) ∧ (¬P1,2 ∨ B1,1) ∧ (¬P2,1 ∨ B1,1)

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Converting to CNF

n  Every sentence can be converted to CNF 1.  Replace α ⇔ β with (α ⇒ β) ∧(β ⇒ α) 2.  Replace α ⇒ β with (¬ α v β) 3.  Move ¬ “inward”

1.  Replace ¬(¬ α) with α 2.  Replace ¬(α ∧ β) with (¬ α v ¬ β) 3.  Replace ¬(α v β) with (¬ α ∧ ¬ β)

4.  Replace (α v (β ∧ γ)) with (α v β) ∧(α v γ)

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Converting to CNF (II)

n  While converting expressions, note that q  ((α v β) v γ) is equivalent to (α v β v γ) q  ((α ∧ β) ∧ γ) is equivalent to (α ∧ β ∧ γ)

n  Why does this algorithm work? q  Because ⇒ and ⇔ are eliminated q  Because ¬ is always directly attached to literals q  Because what is left must be ∧’s and v’s, and they

can be distributed over to make CNF clauses

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Using resolution

n  Even if our KB entails a sentence α, resolution is not guaranteed to produce α.

n  To get around this we use proof by contradiction, i.e., show that KB∧¬α is unsatisfiable.

n  Resolution is complete with respect to proof by contradiction.

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Example of proof by resolution

KB = (B1,1 ⇔ (P1,2∨ P2,1)) ∧¬ B1,1 in CNF… (¬B1,1 ∨ P1,2 ∨ P2,1) ∧ (¬P1,2 ∨ B1,1) ∧ (¬P2,1 ∨ B1,1)

α = ¬P1,2

Resolution yielded the empty clause. The empty clause is False (a disjunction is True only if

at least one of its disjuncts is true). CS440 Fall 2015 18

Automated Theorem Proving

n  How do we automate the inference process? q  Step 1: assume the negation of the consequent and add it

to the knowledge base q  Step 2: convert KB to CNF

n  i.e. a collection of disjunctive clauses

q  Step 3: Repeatedly apply resolution until: n  It produces an empty clause (contradiction), in which case the

consequent is proven, or n  No more terms can be resolved, in which case the consequent

cannot be proven

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Resolution Algorithm

function PL-RESOLUTION(KB, α) returns true or false inputs: KB, the knowledge base, a sentence in propositional logic α, the query, a sentence in propositional logic clauses ← the set of clauses in the CNF representation of KB ∧ ¬α new ← {} loop do for each pair of clauses Ci,Cj in clauses do resolvents ← PL-RESOLVE(Ci,Cj) if resolvents contains the empty clause then return true new ← new U resolvents if new ⊆ clauses then return false clauses ← clauses U new

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Another example

n  If it rains, I get wet. n  If I’m wet, I get mad. n  Given that I’m not mad, prove that it’s not

raining.

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Inference for Horn clauses Horn Form (special form of CNF)

KB = conjunction of Horn clauses Horn clause = disjunction of literals of which at most one is

positive Example: C ∨¬ B ∨¬A

Modus Ponens is a natural way to make inference in

Horn KBs (recall a⇒b is equivalent to ¬a ∨ b) α1, … ,αn, α1 ∧ … ∧ αn ⇒ β

β

n  Successive application of modus ponens leads to algorithms that are sound and complete, and run in linear time

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Forward chaining

n  Idea: fire any rule whose premises are satisfied in the KB q  add its conclusion to the KB, until query is found

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And-or graph

Forward chaining example

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Forward chaining example

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Forward chaining example

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Forward chaining example

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Forward chaining example

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Forward chaining example

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Forward chaining example

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Forward chaining example

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Forward chaining algorithm Forward chaining is sound and complete for Horn KB function PL-FC-ENTAILS?(KB, q) returns true or false

inputs: KB, knowledge base, a set of propositional definite clauses q, the query, a propositional symbol count ← a table, where count[c] is the number of symbols in c’s premise inferred ← a table, where inferred[s] is initially false for all symbols agenda ← a queue of symbols, initially symbols known to be true in KB while agenda is not empty do p ← POP(agenda) if p=q then return true if inferred[p]=false then inferred[p] ← true for each clause c in KB where p is in c.PREMISE do decrement count[c] if count[c]=0 then add c.CONCLUSION to agenda return false

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Backward chaining

aka Goal Directed reasoning Idea: work backwards from the query q:

check if q is known already, or prove by backward chaining all premises of some rule

concluding q Avoid loops:

check if new subgoal is already on the goal stack

Avoid repeated work: check if new subgoal has already been proved true, or has already failed

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Backward chaining example

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Backward chaining example

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Backward chaining example

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Backward chaining example

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Backward chaining example

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Backward chaining example

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Backward chaining example

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Backward chaining example

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Backward chaining example

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Backward chaining example

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Forward vs. backward chaining

n  FC is data-driven n  May do lots of work that is irrelevant to the goal

n  BC is goal-driven, appropriate for problem-solving, q  e.g., What courses do I need to take to graduate? How do I

get into a PhD program?

n  Complexity of BC can be much less than linear in size of KB

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Efficient propositional inference

Two families of efficient algorithms for satisfiability: n  Complete backtracking search algorithms:

q  DPLL algorithm (Davis, Putnam, Logemann, Loveland) n  Local search algorithms

q  WalkSAT algorithm: n  Start with a random assignment n  At each iteration pick an unsatisfied clause and pick a symbol in the

clause to flip; alternate between: q  Pick the symbol that minimizes the number of unsatisfied clauses q  Pick a random symbol

Is WalkSAT sound? Complete?

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In the wumpus world

A wumpus-world agent using propositional logic: ¬P1,1 ¬W1,1 Bx,y ⇔ (Px,y+1 ∨ Px,y-1 ∨ Px+1,y ∨ Px-1,y) Sx,y ⇔ (Wx,y+1 ∨ Wx,y-1 ∨ Wx+1,y ∨ Wx-1,y) W1,1 ∨ W1,2 ∨ … ∨ W4,4 ¬W1,1 ∨ ¬W1,2 ¬W1,1 ∨ ¬W1,3 …

⇒ 64 distinct proposition symbols, 155 sentences

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Summary

n  Logical agents apply inference to a knowledge base to derive new information and make decisions

n  Basic concepts of logic: q  syntax: formal structure of sentences q  semantics: truth of sentences wrt models q  entailment: truth of one sentence given another q  inference: deriving sentences from other sentences q  soundness: derivations produce only entailed sentences q  completeness: derivations can produce all entailed sentences

n  Resolution is complete for propositional logic n  Forward, backward chaining are linear-time, complete for Horn

clauses n  Propositional logic lacks expressive power

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