Proof Methods for Propositional Logic

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

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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Inference using inference rules

n  Inference using inference rules is sound. n  Is it complete? Depends if we have a rich enough set of

rules. n  Inference is a search problem. n  Resolution: a sound inference rule that when coupled

with a complete search method yields a complete inference algorithm

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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 Rule

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

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Resolution Rule

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 ∨ … ∨ mnl1 ∨…∨ li-1 ∨ li+1 ∨…∨ lk ∨ m1 ∨…∨ mj-1 ∨ mj+1 ∨...∨ mn where li and mj are complementary literals.

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Resolution rule is sound

For simplicity let’s consider clauses of length two:

l1 ∨ l2, ¬l2 ∨ l3

l1 ∨ l3

To demonstrate 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 Rule

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 knowledgebase can be expressed as a conjunction of disjunctions (conjunctive normal form, CNF).

n  Example: (A ∨ ¬B) ∧ (B ∨ ¬C ∨ ¬D)

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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)

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

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).

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¬P2,1 B1,1 ¬B1,1 P1,2 P2,1 ¬P1,2 B1,1 ¬B1,1 P1,2

¬P2,1 ¬P1,2P1,2 P2,1 ¬P2,1 ¬B1,1 P2,1 B1,1 P1,2 P2,1 ¬P1,2¬B1,1 P1,2 B1,1

^ ^ ^

^^ ^ ^ ^ ^ ^ ^

^

Proof by resolution

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

to the knowledgebase 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

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 ∧ ¬αnew ← {} 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

KB = conjunction of Horn clauses Horn clause =

propositional symbol; or (conjunction of symbols) ⇒ symbol

Example of a Horn clause: (C ∧ D) ⇒ B Example of a Horn KB: C ∧ ( B ⇒ A) ∧ ((C ∧ D) ⇒ B) Horn form is a special case of CNF where a clause can have at most one positive literal

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Inference for Horn clauses Horn Form

KB = conjunction of Horn clauses Horn clause =

propositional symbol; or (conjunction of symbols) ⇒ symbol

Example: C ∧ ( B ⇒ A) ∧ ((C ∧ D) ⇒ B)

Modus Ponens is a natural way to make inference in Horn KBs 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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Q

P

M

L

BA

Forward chaining example

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Forward chaining algorithm 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

Forward chaining is sound and complete for Horn KB

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Backward chaining

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

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Backward chaining example

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Backward chaining

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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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?

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

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Efficient propositional inference by model checking Two families of efficient algorithms for satisfiability: n  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 a knowledge base q  inference: deriving sentences from other sentences q  soundness: derivations produce only entailed sentences q  completeness: derivations can produce all entailed sentences

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Summary

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

clauses General observation: n  Propositional logic lacks expressive power

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Proof Methods for Propositional Logic

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