Logics for Data and Knowledge Representation

Logics for Data and KnowledgeRepresentation Exercises: Propositional Logic Fausto Giunchiglia, Rui Zhang and Vincenzo Maltese

Outline • Syntax • Symbols and formation rules • Modeling a problem • Semantics • Assignments, models • Logical implication • Reasoning • Satisfiability and validity • Some nice problems

Symbols in PL SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL • Which of the following symbols are used in PL? ⊓  ⊤ ∨ ≡ ⊔ ⊑ → ↔ ⊥ ∧ ⊨ • Which of the following symbols are in well formed formulas? ⊓  ⊤ ∨ ≡ ⊔ ⊑ → ↔ ⊥ ∧ ⊨

Symbols in PL (solution) SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL • Which of the following symbols are used in PL? ⊓  ⊤∨ ≡ ⊔ ⊑ → ↔ ⊥ ∧ ⊨ • Which of the following symbols are in well formed formulas? ⊓  ⊤ ∨ ≡ ⊔ ⊑ → ↔ ⊥ ∧ ⊨ Remember the BNF grammar: <Atomic Formula> ::= A | B | ... | P | Q | ... | ⊥ | ⊤ <wff> ::= <Atomic Formula> | ¬<wff> |<wff>∧ <wff> | <wff>∨ <wff> | <wff> → <wff> | <wff> ↔ <wff> 4

Formation rules SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL • Which of the following is not a wff? •  MonkeyLow∨ BananaHigh •   MonkeyLow  BananaHigh • MonkeyLow   BananaHigh • MonkeyLow→ GetBanana NUM. 3 IS WRONG! 5

Modeling: Bananas SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL Bananas may differ in many ways. However, there are red and yellow bananas. I like bananas, but I eat only yellow bananas. If I do not eat at least a banana I get crazy. • Highlight relevant words, define a PL language and a theory for the problem below. 6

Modeling: Bananas (possible solution) SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL Bananas may differ in many ways. However, there are red and yellow bananas. I like bananas, but I eat only yellow bananas. If I do not eat at least a banana I get crazy. L = {RedBanana, YellowBanana, EatBanana, GetCrazy} T = {EatBanana → YellowBanana, EatBanana → GetCrazy} 7

Truth valuations and Truth Tables SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL • A truth valuation on a PL language L is a mapping ν that assigns to each formula P of L a truth value ν(P). • A truth table is composed of one column for each input variable and one (or more) final column for all of the possible results of the logical operation that the table is meant to represent. Each row of the truth table therefore contains one possible assignment of the input variables, and the result of the operation for those values. LOGICAL OPERATION VARIABLES POSSIBLE ASSIGNEMENTS

Example SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL • Calculate the Truth Table of the following formulas: • (1) A ∧ B; (2) P ∨ Q; (3) X ↔ Y. VARIABLES (1) (2) (3) POSSIBLE ASSIGNEMENTS

Exercise: Truth Valuations SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL • Compute truth valuations for the formulas: (A∨B)∧¬C (A∨B)→C (A∨B→C∨D∨E)∧(¬F↔A)∧(¬F∨G∧¬H∨F)∧(¬I→¬(D∧J))∧(¬J∨¬D∨E)∧F 10

Provide the models for the propositions SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL • RECALL: a truth valuation ν is a model for a proposition P iff ν(P) = true • List the models for the following formulas: • A ∧ ¬B • (A ∧ B) ∨ (B ∧ C) • A ∨ B → C • ¬A ↔ B ↔ C MODEL 11

Entailment SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL • RECALL:  ⊨ ψ iff all models satisfying the formulas in  also satisfy ψ • Let A, B, C be propositional sentences. (a) If A ⊨ B ∧ C, then A ⊨ B or A ⊨ C or both? (b) What if A ⊨ B ∨ C? Proof: The only model satisfying ‘B ∧ C’ is ν = {B=T, C=T}. If ‘A ⊨ B ∧ C’ then ν should be also a model of ‘A’. However, since ν assigns true to B, ν is a model of ‘B’. Similarly, ν is also a model of ‘C’. So, (a) above is true for both. Assume now that ‘A ⊨ B ∨ C’. A model of ‘B ∨ C’ is ν = {B=T, C=F}. ν is not a model of ‘C’, therefore A ⊭C. The other cases are similar.

Exercise: prove entailment SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL • Given that: P = (A ∨ B ) ∧ (C ∨ D ∨ E) Q1 = A ∨ B Q2 = (A ∨ B ∨ C) ∧ ((B ∧ C ∧ D) → E) Q3 = (A ∨ B) ∧ (D ∨ E) Does P ⊨ Qi ? Proof: Let X = A ∨ B, Y = D ∨ E, then we can rewrite: P = X ∧ (¬C ∨ Y); Q1 = X; Q2 = (X ∨ C) ∧ (¬B ∨ ¬C ∨ Y); Q3 = X ∧ Y P ⊨ Q1 is obvious. Since ‘X ⊨ X ∨ C’ and ‘(¬ C ∨ Y) ⊨ (¬B ∨ ¬C ∨ Y)’, then P ⊨ Q2. Since ‘Y ⊨ (¬C ∨ Y)’, then Q3 ⊨ P (and not vice versa).

Exercise: prove entailment using truth tables SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL • Given that: P = (A ∨ B ) ∧ (C ∨ D ∨ E) Q1 = A ∨ B Q2 = (A ∨ B ∨ C) ∧ (B ∧ C ∧ D → E) Q3 = (A ∨ B) ∧ (D ∨ E) (1) List all truth assignments such that P ⊨ Qi (2) Is there any assignment such that P ⊨ Qi for all i? • Solution to (1): First compute the truth tables for all the propositions above. Then, list all rows for which both P and Qi are true. • Solution to (2): Check whether there is any assignment for which all the sentences above are true. 14

Logical implication and deduction SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL • Deduction (⊢) is the process of inferring new knowledge from known facts. • Logical implication (→) is a way to include deduction directly in the language. It is an alternative way to implement deduction. For instance: ¬(A ∨ B) → (¬A ∧ ¬B) iff ¬(A ∨ B) ⊢ (¬A ∧ ¬B) • We provided some well known tautologies in the theoretical part (e.g. the De Morgan Law above) 15

Prove using truth tables, the following deductions SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL • Double negative elimination P ⊢ P • Conjunction introduction / elimination (a) {P, Q} ⊢ P∧Q; (b) P ∧ Q ⊢ P; (c) P ∧ Q ⊢ Q. • Disjunction introduction / elimination (a) P ⊢P ∨ Q; (a) Q ⊢ P ∨ Q; (c) {P ∨ Q, P → R, Q → R} ⊢ R • Bi-conditional introduction / elimination (P → Q) ∧ (Q → P) ⊢ (P ↔ Q) • De Morgan (a) (P ∧ Q) ⊢ P ∨ Q; (b) (P ∨ Q) ⊢ P ∧ Q 16

Proofs of the Deduction Rules SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL 17

Soundness and Completeness SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL • A deductive system is sound if any sentence P that is derivable from a set Г of sentences is also a logical consequence of that set Г. • A deductive system is complete if every sentence P that is a semantic consequence of a set of sentences Γ can be derived in the deduction system from that set. • A soundness property provides the initial reason for counting a logical system as desirable. The completeness property means that every validity (truth) is provable. 18

Prove validity (I) SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL • Suppose p, q, r, s are four propositional sentences, is the following sentence valid? A = (p → r) ∧ (q → s) ∧ (r ∨ s) →(p ∨ q) • A way to prove validity is to show that the proposition entails ⊤. This can be done by applying well known tautologies (e.g. De Morgan). In alternative we can show using truth tables that all the assignments are true. (p → r) ∧ (q → s) ∧ (¬r ∨ ¬s) → (¬p ∨ ¬q) = ¬((p → r) ∧ (q → s) ∧ (¬r ∨ ¬s)) ∨¬p ∨ ¬q = (p ∧ ¬r) ∨ (q ∧ ¬s) ∨ (r ∧ s) ∨¬p ∨ ¬q = = ⊤

Prove validity (II) SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL 20

Calculus with Tableaux SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL Γ = {A B, B, A} Γ = {B A, B} Γ = {(A  B), A} A B A B B  A A  B A B A B A  B closed closed A B closed closed 21

Calculus with Tableaux SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL Γ = {(A B)  (B  A), B} B (A  B)  ( B  A) (A  B)  B  A A  B B A closed A B closed 22

Are you ‘Sherlock Holmes’? SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL There was a robbery in which a lot of goods were stolen. The robber(s) left in a truck. It is known that : (1) Nobody else could have been involved other than A, B and C. (2) C never commits a crime without A's participation. (3) B does not know how to drive. Is A innocent or guilty? Proof: The 3 points above can be translated in PL as follows: (1) θ1 = A ∨ B ∨ C; (2) θ2 = C → A; (3) θ3 = B → (B ∧ A) ∨ (B ∧ C) Does {θ1 ,θ2 ,θ3}⊨ A ? Yes! We can prove it by showing that θ1 ∧θ2 ∧θ3 → A is a tautology.

Knights and Knaves SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL • A very special island is inhabited only by knights and knaves. Knights always tell the truth, and knaves always lie. You meet two inhabitants: Zoey and Mel. Zoey tells you that Mel is a knave. Mel says ‘Neither Zoey nor I are knaves’. Can you determine what are they? (who is a knight and who is a knave?) Proof: The two sentences above can be translated in PL as follows: (1) Z →M; (2) M → Z ∧ M. We can use truth tables to prove that there are two possible situations: - Both lie (they are both knaves) - Zoey tells the true (is a Knight) and Mel lies (is a knave)

Recall the DPLL algorithm SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL REVIEW :: REASONING IN PL :: THE DPLL PROCEDURE :: OBSERVTIONS :: CONCLUSIONS • Input: a proposition P in CNF • Output: true if "P satisfiable" or false if "P unsatisfiable" boolean function DPLL(P) { if consistent(P) then returntrue; if hasEmptyClause(P) then return false; foreach unit clause C in P do P = unit-propagate(C, P); foreach pure-literal L in P do P = pure-literal-assign(L, P); L = choose-literal(P); return DPLL(P  L) OR DPLL(P L); } Consistency Check Empty Close Test Unit Propagation Pure Literal Elimination Splitting rule 25

Use DPLL to prove satisfiability B  ¬A  (¬C  A)  (B  C) 1. Is it consistent? YES, there are no contradictions 2. Are there any empty clauses? No, go ahead. 3. Assign the right truth value to all literals and simplify the formula: Assign B = T, A = F and then the formula simplifies to ¬C 4. Assign the right value to pure literals: Assign C = F. Done. 5. No need for the splitting rule The formula is satisfiable at least for the assignment (a model for it): {A = F, B = T, C = F} SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL

Logics for Data and Knowledge Representation