Inference in Propositional Logic

Transcription

1 Inference in Propositional Logic Deepak Kumar November 2017 Propositional Logic A language for symbolic reasoning Proposition a statement that is either True or False. E.g. Bryn Mawr College is located in Canada. Today is a sunny day. 23 is a prime number. This class is boring. 2 1

2 Propositional Logic - Syntax Sentences Well-formed formulas (wffs) Any atom is a wff [Atomic Sentences] e.g. P, Q, R, R3 Complex Sentences If ω 1 and ω 2 are wffs, then so are ω 1 ω 1 ω 2 disjunction ω 1 ω 2 conjunction ω 1 ω 2 implication ω 1 negation There are no other wffs. Literals are atomic wffs Examples P (P Q) P P Q P P P X etc. P, P 3 Propositional Logic - Semantics How symbols relate to the domain/world Atomic Sentences Interpretation of symbols & constants True is True False is False P I : Bryn Mawr is in Pennsylvania Meaning of a symbol is its truth value, given its interpretation. P is True, given P I Complex Sentences Use meaning of connectives. 4 2

3 Satisfiability More importantly, KnowledgeBase (KB) = a set of wffs An interpretation that satisfies each wff in a KB is called a model of the KB. In Propositional Logic, a model is just True/False assignments to all wffs. Model: A KB in which each wffs is True. 5 Unsatisfiable & Valid Wffs can be unsatisfiable i.e. no interpretation assigns it True E.g. False P P {P Q, P Q, P Q, P V Q } A wff is valid if it is True under all interpretations. E.g. True P P 6 3

4 Entailment ( ) Given We write Δ = set of wffs (e.g. a KB) Δ ω i.e. Δ logically entails ω If ω is True under all interpretations for which each wff in Δ is True. Also, we say ω is a logical consequence of Δ 7 Entailment ( ): Example 1 {P, P Q} Q P Q P Q False False True False True True True False False True True True 8 4

5 Entailment ( ): Example 2 {P Q} P P Q P Q False False False False True False True False False True True True 9 Towards Inference Entailment by model checking is too expensive for most KBs (2 n models!) How can we then produce entailed wffs quickly??? Answer: Use inference/theorem proving. Inference is a way of syntactically generating new wffs from a set of wffs. New wffs are generated using rules of inference. 10 5

6 Rules of Inference Name Rule Name Rule Modus Ponens ω 1 ω 1 ω 2 -Elimination ω 1 ω 2 ω 1 ω 2 ω 2 ω 1 ω 2 -Introduction ω 1 ω 2 ω 1 ω 2 -Introduction ω 1 ω 2 ω 1 ω 2 ω 1 ω 2 Commutativity of ω 1 ω 2 ω 2 ω 1 -Elimination ( ω 1 ) ω 1 11 Theorem Proving - Example Δ = {P, R, P Q} Is Q R a theorem of Δ? Proof: P P Q Q R Q R by Modus Ponens by -Introduction Proof Notation Δ R Q R 12 6

7 Soundness & Completeness Soundness For a set Δ and wff ω if Δ R ω imples Δ ω then R is sound. Completeness Whenever Δ ω, there exists a proof of ω from Δ using R i.e. Δ R ω then R is complete. Are the rules of inference introduced earlier sound? Complete? 13 Useful Metatheorems & Laws 1. Deduction Theorem If {ω 1, ω 2, ω n } ω then ω 1 ω 2 ω n ω is valid. That is, to see if α β we can check to see if α β is true in every model. Conversely, every valid implication describes a legitimate inference. 14 7

8 Useful Metatheorems & Laws 2. Reductio Ad Absurdum Proof by Contradiction Refutation. If Δ has a model but Δ { ω} does not then Δ ω 15 Useful Metatheorems & Laws 3. Law of Equivalence ( ) ω 1 ω 2 If their truth values are identical under all interpretations. E.g. (P Q) ( P Q) P Q P Q False False True False True True True False False True True True P Q True True False True 16 8

9 Useful Metatheorems & Laws 4. Associative Laws 5. Distributive Laws (ω 1 ω 2 ) ω 3 ω 1 (ω 2 ω 3 ) (ω 1 ω 2 ) ω 3 ω 1 (ω 2 ω 3 ) ω 1 (ω 2 ω 3 ) (ω 1 ω 2 ) (ω 1 ω 3 ) ω 1 (ω 2 ω 3 ) (ω 1 ω 2 ) (ω 1 ω 3 ) 6. DeMorgan s Laws (ω 1 ω 2 ) ω 1 ω 2 (ω 1 ω 2 ) ω 1 ω 2 17 De Morgan s Laws Prove: (ω 1 ω 2 ) ω 1 ω 2 ω 1 ω 2 (ω 1 ω 2 ) ω 1 ω 2 False False True True False True False False True False False False True True False False 18 9

10 Logic Syntax Semantics Rules of Inference Can be shown to be sound Completeness depends on the algorithm. (Needs a complete search algorithm!) Bigger question: Which and how many rules to use in a logic? 19 Rules of Inference Name Rule Name Rule Modus Ponens ω 1 ω 1 ω 2 -Elimination ω 1 ω 2 ω 1 ω 2 ω 2 ω 1 ω 2 -Introduction ω 1 ω 2 ω 1 ω 2 -Introduction ω 1 ω 2 ω 1 ω 2 ω 1 ω 2 Commutativity of ω 1 ω 2 ω 2 ω 1 -Elimination ( ω 1 ) ω

11 Resolution New rule of inference that combines Modus Ponens, as well as others, in one rule. Requires wffs to be in a special form: A wff is either a literal: P, P Or a clause a disjunction of literals: P Q R An empty set is equivalent to False: { } False Rule: Resolvent ω ω 1 ω k ω α 1 α m ω 1 ω k α 1 α m 21 Resolution Rule: ω ω 1 ω k ω α 1 α m ω 1 ω k α 1 α m Example: R P R P P Q P Q R Q R Q 22 11

12 Resolution Example: R R R P R P i.e. Modus Ponens! P P Example: P P Ø Unsatisfiable: P P { } False i.e. a contradiction! 23 Resolution Is a sound rule of inference But, it is not complete. Example: P P Q But, no way to produce: P P Q Use Reductio ad absurdum! 24 12

13 Resolution + Reduction ad Absurdum KB = {ω 1, ω 2, ω n } i.e. ω 1 ω 2 ω n Make sure all wffs are clauses Any wff can be converted into a conjunction of clauses (CNF) Use Reductio ad absurdum 25 How to use Resolution 1. Given, KB = {ω 1, ω 2, ω n } 2. Convert each wff can into CNF 3. Given a wff α, to prove KB α 4. Show, using Resolution on clauses from (1) that KB α leads to contradiction (i.e. derives empty clause). 5. Since KB α leads to contradiction, it must be that KB α is True. 6. Therefore, KB α 26 13

14 Resolution: Example KB = { P, Q, P R Q } KB? R Convert all wffs in KB to clauses: { P, Q, P R Q } 1. P 2. Q 3. P R Q 4. R Negation of conclusion 5. P Q From (3) & (4), resolving R and R 6. P 7. Ø From (1) & (6), resolving P and P Therefore, since KB R leads to a contradiction, it must be that KB R. 27 Example: Convert to CNF (P Q) (R P) Eliminate replace α β by α β Move inwards. Make literals. Use DeMorgan s Laws, if needed. Distribute over to get CNF Write each conjunct as a separate clause

15 Example: Convert to CNF (P Q) (R P) Eliminate ( P Q) ( R P) Move inwards. (P Q) ( R P) Distribute over to get CNF ( R P P) ( R P Q) ( R P) ( R P Q) Write each conjunct as a separate clause. { R P, R P Q} 29 15