Propositional Logic: Why? We want to build a mathematical model of logical reasoning Starts with George Boole around 1850 Actually, of part of it, in particular, of part of mathematical reasoning In consequence, mathematics itself becomes a subject of study of mathematics Mathematical logics starts ... Logical reasoning is captured in terms of symbolic processes, like algebra We also talk of symbolic logic 1 • Algebra: x+y=2 -y + 2 x = 3 —————— 3x = 5 • Logic: (p → (r ∨ q)) (p ∧ ¬r) —————— q Computer science emerges from mathematical logic in the 1930s Mathematical logic relevant to CS today Important parts of CS have to do with symbolic processing, like symbolic logical reasoning Logic can help understand and support those symbolic processes in computers 2 Computers can implement logical reasoning, because it can be captured in symbolic terms Computers can be made to do logical reasoning, in particular, mathematical reasoning! Propositional logic can be used to represent knowledge Given a domain, one chooses propositional variables, e.g. p, q, r, s, ... to denote propositions about the domain of discourse Example: Courses at a university p: “Student takes 1805” q: “Student takes Advanced Programming” r: “Student takes Analysis of Algorithms”, etc. These are indivisible, atomic propositions More complex propositions can be represented (constructed) using the propositional variables and the logical connectives 3 This is the syntax of the symbolic languages of propositional logic 1. “If student takes Analysis of Algorithms, then it takes 1805” can be represented by (r → p) 2. “Student takes Analysis of Algorithms or Advanced Programming”: (q ∨ r) 3. “Student takes Analysis of Algorithms or Advanced Programming, but not both”: (q ⊕ r) 4. “Taking 1805 is a necessary condition to take Analysis of Algorithms”: (r → p) 5. “Taking 1805 is a suﬃcient condition to take Analysis of Algorithms”: (p → r) 6. “Not possible to take Analysis of Algorithms without 1805”: (¬p → ¬r) The contrapositive of 5. 7. “If 1805 is taken when Advanced Programming is taken, and Advanced Programming is taken, then 1805 is taken”: ((q → p) ∧ q) → p) 4 We know already how to assign truth values to complex propositions when truth values for the atomic propositions (propositional variables) have been assigned ... Whenever we talk about meaning, interpretations, truth (values), we are in the realm of semantics Truth tables! p 1 0 1 0 q p → q (p → q) ∧ p ((p → q) ∧ p) −→ q 1 1 1 1 1 1 0 1 0 0 0 1 0 1 0 1 The formula on the RHS is always true, independently from the truth values of its atomic propositions It is called a tautology It represents a class of common valid arguments 5 If Socrates is a man, then Socrates is a mortal. Socrates is a man. In consequence, Socrates is a mortal. Replace in this argument, the propositions Socrates is a man and Socrates is a mortal by any two propositions, and you will always get a valid argument p ¬p p ↔ ¬p 1 0 0 0 1 0 This formula is a contradiction: it is always false p 1 0 1 0 ¬p 0 1 0 1 q p → q (p → q) ∨ ¬p 1 1 1 1 1 1 0 0 0 0 1 1 The formula is neither a tautology nor a contradiction It is satisﬁable or consistent, i.e. it can be made true 6 Logical equivalence: • Based on truth tables • As logical validity of double implication • Based on basic algebraic laws for logical equivalence and the transitivity of logical equivalence Basic laws can be veriﬁed in general using truth tables and other previously derived laws for equivalence Logical equivalence can be used as the basis for an algebraic methodology to derive new (true) formulas from other formulas (that are assumed to be true) Example: Derive q from ((p → q) ∧ p), i.e. obtain that every time ((p → q) ∧ p) is true, also q is true One method: check with truth table that 7 • When the column for ((p → q) ∧ p) is true, also the column for q is true; or • (((p → q) ∧ p) → q) is a tautology (in consequence, the antecedent cannot be true without having the consequent true) An alternative (algebraic): We can use that • (ϕ ↔ ψ) ≡ (ϕ → ψ) ∧ (ψ → ϕ)) • (ϕ → (ψ ∧ χ)) ≡ ((ϕ → ψ) ∧ (ϕ → χ)) Now: (p → q) ∧ p ≡ (¬p ∨ q) ∧ p ≡ (¬p ∧ p) ∨ (q ∧ p) ≡ F ∨ (q ∧ p) ≡ (q ∧ p) That is ((p → q) ∧ p) ↔ (q ∧ p) is always true Then ((p → q) ∧ p) → (q ∧ p) is always true Then ((p → q) ∧ p) → q is always true 8 Still in propositional logic ... (will also hold for predicate logic) General facts: Assume we want to prove that the formula χ: (ϕ1 ∧ · · · ∧ ϕn) → ψ is always true (tautology), i.e. that ψ is a logical consequence of (ϕ1 ∧ · · · ∧ ϕn), i.e. ψ is true whenever (ϕ1 ∧ · · · ∧ ϕn) is true Alternatives: • Check that χ has always value 1 • Check that whenever all the ϕis have value 1, also ψ gets value 1 • Check that ϕ1 ∧ · · · ∧ ϕn ∧ ¬ψ is always false (a contradiction), i.e. it is not possible to have all the ϕis all simultaneously true with the negation of ψ also true Example: (p → q) ∧ p → q always true is equivalent to (p → q) ∧ p ∧ ¬q being always false 9 Example: If Joe fails to submit the a project in course CS414, then he fails the course. If Joe fails CS414, then he cannot graduate. Hence, if Joe graduates, he must have submitted a project. Want to represent this argument, and check it is valid It is of the form: F0 ∧ F1 → C Atomic formulas: • s: Joe submits the project in CS414. • f : Joe fails CS414. • g: Joe graduates. Then, • F0: ¬s → f • F1: f → ¬g • C: g → s And the argument is represented by (¬s → f ) ∧ (f → ¬g) 10 → (g → s) Check using a truth that • it is valid ; or alternatively, that • (g → s) is true every time (¬s → f ) ∧ (f → ¬g) is true; or ... • (¬s → f ) ∧ (f → ¬g) ∧ ¬(g → s) is always false (unsatisﬁable) Another method: algebraically ...; by transitivity of implication (¬s → f ) ∧ (f → ¬g) → (¬s → ¬g) (≡ T ) But (¬s → ¬g) ≡ (g → s) (contrapositive); replacing inside ... (¬s → f ) ∧ (f → ¬g) → (g → s) (≡ T ) 11 Example: If X is greater that zero, then if Y is zero, then Z is zero. Variable Y is zero. Hence, either X is greater than zero or Z is zero. Atomic propositions: • x: X is greater that zero. • y: Y is zero. • z: Z is zero. Formalization: (x → (y → z)) ∧ y → x ∨ z (*) Valid argument? Does not seem so ... (intuitively, y does not contribute to the conclusion, and the info about x and z is conditional ...) It is not: a counter example is the valuation that assigns x → 0; z → 0 (the consequent of (*) must be 0), and y → 1 (to make the antecedent true) This valuation makes (*) take value 0 12 Example: If Superman were able ad willing to prevent evil, he would do so. If Superman were unable to prevent evil, he would be impotent; if he were unwilling to prevent evil, he would be malevolent. Superman does not prevent evil. If Superman exists, he is neither impotent nor malevolent. Therefore, Superman does not exist. Introduce the following propositional variables to denote the atomic sentences • a: Superman is able to prevent evil. • w: Superman is willing to prevent evil. • i: Superman is impotent. • m: Superman is malevolent. • p: Superman prevents evil. • e: Superman exists. The main propositions in the argument are: • F0 : (a ∧ w) → p • F1 : (¬a → i) ∧ (¬w → m) • F2 : ¬p • F3 : e → (¬i ∧ ¬m) 13 The whole argument can be represented by the formula: (F0 ∧ F1 ∧ F2 ∧ F2 ∧ F3) → ¬e. The argument is valid: try to deduce ¬e reasoning backwards using the hypothesis .. Contrapositive of F3: ¬(¬i ∧ ¬m) → ¬e Try to prove ¬(¬i ∧ ¬m), i.e. i ∨ m (*) Now from F1 we get: ¬a → i, then instead of proving (*) it is good enough to prove ¬a ∨ m (¬a replaced i) Again from F1 we get ¬w → m, i.e. good enough to prove ¬a ∨ ¬w (same idea) , i.e. ¬(a ∧ w) From F0 we get ¬p → ¬(a ∧ w) (contrapositive) We need to prove p; but this given directly by F2 We are done ... Much easier than truth tables ... 14 Predicate Logic The argument: If Socrates is a man, then Socrates is a mortal. Socrates is a man. In consequence, Socrates is a mortal. Could be expressed as a valid (tautologic) argument in propositional logic Consider now: All men are mortal. Socrates is a man. Then, Socrates is a mortal. Can it be expressed in propositional logic? p: all men are mortal. q: Socrates is a man. r: Socrates is a mortal. (p ∧ q) → r ?? Not a tautology, and the argument seems to be a valid argument ... Propositional logic is not expressive enough to capture that argument 15 The connections between the elements of the argument is lost in propositional logic Here we are talking about general properties (also called predicates) and individuals of a domain of discourse who may or may not have those properties Instead of introducing names for complete propositions -like in propositional logic- we introduce: • names for the properties or predicates, • names for the individuals (or constants) • variables that can be used in combination with the predicates • quantiﬁers to express things like “all individuals have property ...” or “there exists an individual with the property ...” • And we also use all the logical connectives from propositional logic; in this sense predicate logic properly extends propositional logic 16 Example: The new argument around Socrates • Man(·): for “being a man” • Mortal (·): for “being a mortal” • Socrates: a symbolic name for the individual Socrates • Variables: x, y, z, ... • ∀: universal quantiﬁer; ∃: existential quantiﬁer Expressing the argument: ∀x(Man(x) → Mortal (x)) ∧ Man(Socrates) → Mortal (Socrates) We can use other names, capturing the same argument 17 ∀x(P (x) → Q(x)) ∧ P (c) → Q(c) (*) Here: P (x), Q(x), P (c), Q(c) are atomic formulas (the simplest formulas), and (*) is a complex formula The syntax of predicate logic tells us which strings of symbols are real formulas Does (*) express the original argument as a valid argument? Is it always true? This has to do with the semantics of predicate logic: meaning, interpretations, truth, ... 18 Languages of predicate logic are interpreted in structures consisting of: • A universe or domain of discourse • Interpretations of the predicates (predicate names) as properties over the domain of discourse • Interpretations of the constant (names for individuals) as concrete individuals in the domain Example (contd): the (speciﬁc) building blocks of the language are the predicates P (·), Q(·) and the constant c Possible interpretations: • – Universe U = the set of all the ancient Greeks, including humans and divinities, etc. – P (·): interpreted as the ancient male Greeks, i.e. as a subset of U – Q(·): ... the mortals among all the ancient Greeks, also as a subset of U – Socrates: ... as the philosopher Socrates, an element of (individual in) U 19 In this structure (the one of ancient Greeks), the formula - ∀x(P (x) → Q(x)) becomes true, because, all men are mortal. - P (Socrates) becomes true, because Socrates is a man. - Q(Socrates) becomes true, because Socrates is a man. (*) is of the form p ∧ q → r, with p, q, r true, then (*) is true (as expected) 20 • – Universe: U = living beings on earth – P (·): the plants, a subset of U – Q(·): the animals ... – c: a name for the canary Tweety In this structure, the formula - ∀x(P (x) → Q(x)) becomes false - P (Socrates) becomes false - Q(Socrates) becomes true (*) is of the form p ∧ q → r, with p, q false, and r true; then (*) is true (as expected) Actually (*) becomes true in every possible structure (or under every possible interpretation) It captures a valid argument (always true) ... 21 Example: We can use predicates, etc. to symbolically represent knowledge about a numerical domain (like algebra) Predicates: B(·, ·), M (·, ·, ·), S(·, ·, ·), E(·, ·) Constants: 0, 1 Atomic formulas: B(x, y), B(1, 0), B(1, 1), S(1, z, x), ... More complex formulas: - ∀x∀y∀z(S(x, y, z) → S(y, x, z)) - ∀x∀y(M (x, x, y) ∧ ¬E(x, 0) → B(y, 0)) - ∀x∀y∀z(B(x, 0)∧B(y, 0)∧B(z, 0)∧M (x, y, z) → (B(z, x) ∧ B(z, y)) ∨ E(z, 1)) - ∃x∃z(M (x, x, z) ∧ S(z, 1, 0)) Are these formulas true? 22 In general, we cannot answer yes or not: it depends on the interpretation of the symbols in it ... Only in certain cases we can say the formula is true, independently from the interpretation, e.g. ∀x(P (x) → P (x)) is always true (or valid) In other cases, we can say it is (always) false, e.g. ∃x(P (x) ∧ ¬P (x)) is always false (a contradiction) The formulas in the previous slide are neither valid nor contradictions (sure?) Their truth value depends upon the interpretation domain and the interpretation of the symbols on that domain (i.e. on the interpretation structure) The interpretation structures for that language require a domain (universe), two binary relations, and two ternary relations on it, and two distinguished elements in it (for which the 0 and 1 above are names) 23 We had the predicates: B(·, ·), M (·, ·, ·), S(·, ·, ·), E(·, ·), and the names 0, 1 Example: Structure N, >, $(·, ·, ·), +(·, ·, ·), =, 0, 1 N = {0, 1, 2, 3, ...} E.g. 9 > 3, ∗ (2, 3, 6), + (8, 1, 9), 5 = 5 - ∀x∀y∀z(S(x, y, z) → S(y, x, z)) becomes interpreted as a proposition about natural numbers and their usual operations, namely for every l, m, n ∈ N, it holds that if +(l, m, n), then +(m, l, n) Or simpler: l + m = n ⇒ m + l = n The original formula is true in that structure 24 - ∀x∀y(M (x, x, y) ∧ ¬E(x, 0) → B(y, 0)) becomes for all m, n ∈ N, if m2 = n and m = 0, then n>0 also true - ∀x∀y∀z(B(x, 0)∧B(y, 0)∧B(z, 0)∧M (x, y, z) → (B(z, x) ∧ B(z, y)) ∨ E(z, 1)) becomes for all ... if l > 0 and m > 0 and n > 0 and l $ m = n, then both n > l and n > m or n = 1 true ... - ∃x∃z(M (x, x, z) ∧ S(z, 1, 0)) becomes there is m ∈ N and there is n ∈ N such that false m2 = n and n + 1 = 0 This same language (formulas) of predicate logic can be interpreted in a diﬀerent structure 25 Example: Structure R, <, $, =, +, 0, 1 We can symbolically say other true things about the real numbers, e.g. - 2 has a square root in R - every number has an additive inverse - there is a multiplicative neutral element Example: Structure C, ∅, $, =, +, 0, 1 26 Exercise: (a) Check the formulas in R, <, $, −, π, e (b) Check the formulas in a non numerical domain (why not?) Free and bound variables In all the examples before, the variables in the formula are bound, i.e. they are aﬀected or under the scope of a quantiﬁer Formulas of predicate logic that have al their variables bound are called sentences However, formulas are allowed to contain free variables, i.e. not aﬀected by a quantiﬁer - ∃y(R(x, y) ∧ S(y, z)) has a bound variable y, and two free variables x, z Is the formula true? Now it depends on the structure and the values that we assign to the free variables in the domain 27 Example: Structure D, Flies, UsedTo 28 By means of the formula ∃y(R(x, y) ∧ S(y, z)) with the existentially quantiﬁed variable y that appears in the two tables, we are connecting the two tables A usual operation in relational data bases Predicate logic is an important tool in data bases: • To give meaning, i.e. semantics, to data • We can pose (symbolic) queries, that can be processed and answered by the data base management systems E.g. “Give me all the pairs of pilots and possible destinations for them”: Q(x, z) : ∃y(R(x, y)∧S(y, z)) The system returns all pairs of possible values for x and z, i.e. x and z are used as real variables, who get values from the database 29 • To specify that something must always happen is the database, for any update (integrity constraints) E.g. “Every aircraft in the table Flies must appear in the table UsedTo” ∀x(∃yFlies(y, x) → ∃zUsedTo(x, z)) Not satisﬁed in the data base above 30 (*) Some useful facts about formulas of predicate logic Since predicate logic extends the propositional logic: - we inherit all the “predicate logic like” tautologies now as valid formulas, i.e. formulas that are true in every interpretation (compatible with the language of the ﬂa.) E.g. ∀xP (x) ∨ ¬∀xP (x) (an instance of a tautology from propositional logic) - But there are some “new” valid formulas, that do not come from propositional logic, e.g. ∀xP (x) → P (c) P (c) → ∃yP (y) - the notion of logical equivalence applies basically as before: Two formulas ϕ, ψ (written in the same language of predicate logic) are logically equivalent, denoted ϕ ≡ ψ, iﬀ for every interpretation (compatible with the language) they take the same truth value 31 E.g. ∀xP (x) → ∀yQ(y) ≡ ¬∀xP (x) ∨ ∀yQ(y) ∀x∀y(P (x) → Q(y)) ≡ ∀x∀y(¬P (x) ∨ Q(y)) ∀x∀y(P (x) ∨ ¬Q(y)) ≡ ∀x∀y¬(¬P (x) ∧ Q(y)) Any new equivalences, not inherited from propositional logic? • ¬∀xϕ(x) ≡ ∃x¬ϕ(x) • ¬∃xϕ(x) ≡ ∀x¬ϕ(x) • ∀xϕ(x) ≡ ∀yϕ(y) On the RHS, y replaces x in ϕ But when y does not appear free on the LHS Quantiﬁed (bound) variables can be replaced by fresh variables 32 • ∀x(ϕ(x) ∧ ψ(x)) ≡ ∀xϕ(x) ∧ ∀xψ(x)) • ∃x(ϕ(x) ∨ ψ(x)) ≡ ∃xϕ(x) ∨ ∃xψ(x)) • ∀x(ϕ(x) ∨ ψ) ≡ ∀xϕ(x) ∨ ψ If x does not appear (free) in ψ • ∃x(ϕ(x) ∧ ψ) ≡ ∃xϕ(x) ∧ ψ) If x does not appear (free) in ψ Example: Verify that (*) is equivalent to ∀x∀y∃z(Flies(y, x) → UsedTo(x, z)) 33 Since ∀x∀y∃z(Flies(y, x) → UsedTo(x, z)) is false in the given data base, its negation is true in it I.e. the data base satisﬁes ¬∀x∀y∃z(Flies(y, x) → UsedTo(x, z)) That is equivalent to: Exercise: Verify that • ∀x(ϕ(x) ∨ ψ(x)) ≡ ∀xϕ(x) ∨ ∀xψ(x) • ∃x(ϕ(x) ∧ ψ(x)) ≡ ∃xϕ(x) ∧ ∃xψ(x) Hint: In each case, a counterexample must be found, i.e. a pair of formulas ϕ, ψ and a structure where the two formulas do not take the same truth values 34 Example: Transform the formula ∀x∃y(P (x, y) → (¬∃z∃yR(z, y) ∧ ¬∀xS(x))) into a (set of) formula(s) in prenex conjunctive normal form 35 Proofs The most important notion in logic is the one of logical consequence Given a set of formulas (premises, hypothesis, axioms, ...) Σ, a formula ψ is a logical consequence of Σ if whenever all the formulas in Σ are true, then also ψ becomes true We saw some examples in the context of propositional logic, e.g. the non existence of Superman In propositional logic, at least in principle, if there are ﬁnitely many propositional variables involved, logical consequence can be checked by means of truth tables (cumbersome) We also saw alternatives, e.g. backward reasoning In predicate logic, truth tables cannot be used in general (domains can be inﬁnite) 36 It is important, both in propositional logic and in predicate logic (in this case, crucial) to ﬁnd symbolic methodologies to establish logical consequence Or, equivalently, inconsistency: remember ϕ1 ∧ · · · ∧ ϕ n → ψ is valid (making ψ a logical consequence of Σ = {ϕ1, . . . , ϕn}) iﬀ ϕ1 ∧ · · · ∧ ϕn ∧ ¬ψ is a contradiction There are such formal deductive systems for symbolic (computational) manipulation of formulas that do the job Those systems have deduction rules that closely mimic the way we obtain conclusions in, e.g. mathematical reasoning 37 Deduction Rules: •ϕ→ψ ϕ −−− ψ Modus Ponens From the hypothesis ϕ → ψ and ϕ one can jump to the conclusion ψ Used all the time in mathematical and everyday reasoning If Socrates is man, then Socrates is a mortal Socrates is a man − − − − − − − − − − − − − − − − −− Socrates is a mortal Can be justiﬁed observing that ((ϕ → ψ) ∧ ϕ) → ψ is valid (always true) A variant ... 38 •ϕ→ψ ¬ψ −−− ¬ϕ Modus Tollens Can be obtained from Modus Ponens using the equivalent contrapositive of the top formula: ¬ψ → ¬ϕ If Socrates is man, then Socrates is a mortal Socrates is not a mortal − − − − − − − − − − − − − − − − −− Socrates is not a man • See a list of other deduction rules in the book Deduction rules are used in combination with certain hypothesis that are assumed to be true, plus other previously obtained conclusions, plus valid formulas that are always true and can always be used safely in reasoning 39 • A very useful deduction rule, implemented in many computational reasoning systems (¬)p ∨ q (¬)r ∨ ¬q −−−−− (¬)p ∨ (¬)r Resolution Rule (p, q are propositional variables) Can be generalized to a larger number of literals (propositional variables or negations of them) s ∨ ¬l ∨ ¬t u∨s∨q∨t −−−−−−−−− s ∨ ¬l ∨ u ∨ s ∨ q ¬p ∨ q p ∨ ¬q −−−−− ¬p ∨ p ¬p ∨ q p ∨ ¬q −−−−−−− ¬q ∨ q 40 There are diﬀerent proof strategies that have a sound logical basis; two of them are • Proofs by contradiction: Usually, theorems in mathematics are of the form: HYPOTHESIS (H) ⇒ THESIS (T ) (α) The proof can be direct (start from H and reach T ) or by contradiction In the second case, we try to prove: H and not T ⇒ F (β) where F is a contradiction (an always false proposition) If we establish (β), we know that (α) is true Why? Because (β) ⇒ (α) is true That is, in terms of propositional logic, ((H ∧ ¬T ) → F ) → (H → T ) is always true ... 41 H 0 1 0 1 T 0 0 1 1 F ¬T H ∧¬T (β) (α) (β) ⇒ (α) 0 1 0 1 1 1 0 1 1 0 0 1 0 0 0 1 1 1 0 0 0 1 1 1 • Proofs by cases: again we want to establish a theorem of the form (α) We consider two (or more) cases C1, C2 for H that cover all the possibilities (H ∧ C1) → T (H ∧ C2) → T C1 ∨ C2 −−−−−−− H→T Can be justiﬁed: ((H ∧ C1) → T ) ∧ ((H ∧ C2) → T ) ∧ (C1 ∨ C2) −→ (H → T ) is a tautology (check!) 42 There are also deductive rules for predicate logic • ∀x ϕ(x) − − −− ϕ(t) where t is a constant or a variable ... • etc. 43