Logic. Part I: Propositional Logic. Max Schäfer. Formosan Summer School on Logic, Language, and Computation 2010
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1 Logic Part I: Propositional Logic Max Schäfer Formosan Summer School on Logic, Language, and Computation / 39
2 Organisation Lecturer: Lectures: Homework: Exam Max Schäfer (Schaefer) Monday 28th June, 2:00pm 5:00pm Wednesday 30th June, 9:30am 12:30pm Friday 2nd July, 9:30am 12:30pm after every lecture Friday 9th July, 9:30am 12:30pm material of first two lectures only 2 / 39
3 What Is Logic? this course is about formal logic investigate principles of reasoning, independently of particular language, mindset, or philosophy different logical systems for different kinds of reasoning three basic components of a logical system: 1. formal language 2. semantics 3. deductive system 3 / 39
4 Principles of Classical Logic classical logic aims to model reasoning about truth logical formulas represent statements that are either true or false proving a formula means showing that it is true sometimes this is easy 2 Q sometimes it is hard n.n > 2 ( a, b, c.a n + b n = c n ) proving a formula does not make it true, it just demonstrates its truth 4 / 39
5 Propositional Logic 命題邏輯 5 / 39
6 Principles of Propositional Logic propositional logic talks about propositions a proposition is a sentence that is either true or false some propositions are atomic; represented by capital letters P, Q, R,... other propositions are composed from simpler ones using connectives such as,,... 6 / 39
7 The Formal Language of Propositional Logic assume we have an alphabet R of propositional letters, assumed to contain at least the capital letters P, Q, R,... the language of formulas of propositional logic is given by the following grammar: ϕ ::= R ϕ ϕ ϕ ϕ ϕ ϕ 7 / 39
8 Intuitive Meaning of Propositional Logic formula reading is true if... P P the proposition represented by P is true 假, falsity, bottom never true P Q P 與 Q, conjunction P is true, and Q is also true P Q P 或 Q, disjunction P is true, or Q is true, or both P Q P 就 Q, implication it is not the case that P is true and Q is false 8 / 39
9 Example Formulas P, Q P P Q P; interpreted as (P Q) P not as P (Q P) P P Q; interpreted as P (P Q) not as (P P) Q Operator Precedence binds tighter than ; tighter than 9 / 39
10 Example Formulas (ctd.) P Q R; interpreted as (P Q) R not as P (Q R) P Q R; interpreted as (P Q) R not as P (Q R) P P Q; interpreted as P (P Q) not as (P P) Q Operator Associativity and associate to the left, to the right 10 / 39
11 Defined Connectives and Syntactic Equality other connectives can be defined in terms of the basic ones: ϕ := ϕ P: 非 P, negation ϕ ψ := (ϕ ψ) (ψ ϕ) P Q: P 若且唯若 Q, equivalence, bi-implication := : 真, truth, top and are not real connectives, but only abbreviations; e.g., P P (the same formula) Precedence and Associativity binds tighter than ; less tight than, associates to the left 11 / 39
12 Truth Value Semantics in general, to know whether a formula ϕ is true, we need to know whether its propositional letters are true need a truth value assignment (interpretation) I : R B, where B := {T, F} given an interpretation I, define the semantics ϕ I of a formula ϕ: 1. for r R: r I := I (r). 2. I := F. 3. ϕ ψ I := T if ϕ I = T and ψ I = T, else ϕ ψ I := F. 4. ϕ ψ I := F if ϕ I = F and ψ I = F, else ϕ ψ I := T. 5. ϕ ψ I := F if ϕ I = T and ψ I = F, else ϕ ψ I := T. 12 / 39
13 Truth Value Semantics in general, to know whether a formula ϕ is true, we need to know whether its propositional letters are true need a truth value assignment (interpretation) I : R B, where B := {T, F} given an interpretation I, define the semantics ϕ I of a formula ϕ: 1. for r R: r I := I (r). 2. I := F. 3. ϕ ψ I := T if ϕ I = T and ψ I = T, else ϕ ψ I := F. 4. ϕ ψ I := F if ϕ I = F and ψ I = F, else ϕ ψ I := T. 5. ϕ ψ I := F if ϕ I = T and ψ I = F, else ϕ ψ I := T. 12 / 39
14 Derived Truth Values Lemma For any interpretation I and formulas ϕ and ψ we have I = T. ϕ I = T if ϕ I = F, else ϕ I = F. ϕ ψ I = T if ϕ I = ψ I, else ϕ ψ I = F. 13 / 39
15 Validity and Satisfiability I = ϕ ( I is a model for ϕ ): ϕ I = T ϕ is satisfiable: there is I with I = ϕ ϕ is valid: for all I we have I = ϕ ϕ ψ ( ϕ entails ψ ): whenever I = ϕ also I = ψ ϕ ψ ( ϕ and ψ are equivalent ): both ϕ ψ and ψ ϕ 14 / 39
16 Examples P P is valid P P is satisfiable P Q P Q 15 / 39
17 Properties of Validity and Satisfiability Theorem ϕ is valid iff ϕ is unsatisfiable iff ϕ ; furthermore, ϕ ψ iff ϕ I = ψ I for every interpretation I. Example: ϕ ϕ 16 / 39
18 Notational Caveat ϕ ψ and ϕ ψ do not mean the same! is syntactic equality; different ways of writing, same formula (P Q) R P Q R P Q (R ) is semantic equality; different formulas, same semantics P P, but not P P 17 / 39
19 Propositional Letters in a Formula define set PL(ϕ) of propositional letters that occur in a formula ϕ: PL(r) = {r}, for every r R PL( ) = PL(ϕ ψ) = PL(ϕ ψ) = PL(ϕ ψ) = PL(ϕ) PL(ψ) Example PL(((P Q) P) P) = PL( ) = 18 / 39
20 Propositional Letters in a Formula define set PL(ϕ) of propositional letters that occur in a formula ϕ: PL(r) = {r}, for every r R PL( ) = PL(ϕ ψ) = PL(ϕ ψ) = PL(ϕ ψ) = PL(ϕ) PL(ψ) Example PL(((P Q) P) P) = {P, Q} PL( ) = 18 / 39
21 Propositional Letters in a Formula define set PL(ϕ) of propositional letters that occur in a formula ϕ: PL(r) = {r}, for every r R PL( ) = PL(ϕ ψ) = PL(ϕ ψ) = PL(ϕ ψ) = PL(ϕ) PL(ψ) Example PL(((P Q) P) P) = {P, Q} PL( ) = 18 / 39
22 Agreement Lemma Lemma For every formula ϕ and interpretations I 1, I 2 such that I 1 (P) = I 2 (P) for every P PL(ϕ), we have ϕ I1 = ϕ I2 Propositional letters that don t occur in a formula do not matter when determining its semantics. 19 / 39
23 Truth Tabling for every formula ϕ, PL(ϕ) is finite, say PL(ϕ) = n every one of these n variables could be either true or false; this gives 2 n combinations to know whether ϕ is valid, we only need to try them all out! 20 / 39
24 Example We can use truth tables to show: = P Q P Q (P Q) P Q P Q P Q P Q P Q P Q P Q P Q (P Q) F F F T T F T T P Q P Q 21 / 39
25 Example We can use truth tables to show: = P Q P Q (P Q) P Q P Q P Q P Q P Q P Q P Q P Q (P Q) F F F F T F T F F T T T P Q P Q 21 / 39
26 Example We can use truth tables to show: = P Q P Q (P Q) P Q P Q P Q P Q P Q P Q P Q P Q (P Q) F F F F F T F T T F F T T T T T P Q P Q 21 / 39
27 Example We can use truth tables to show: = P Q P Q (P Q) P Q P Q P Q P Q P Q P Q P Q P Q (P Q) F F F F T F T F T F T F F T F T T T T T P Q P Q 21 / 39
28 Example We can use truth tables to show: = P Q P Q (P Q) P Q P Q P Q P Q P Q P Q P Q P Q (P Q) F F F F T T F T F T F F T F F T F F T T T T T T P Q P Q 21 / 39
29 Example We can use truth tables to show: = P Q P Q (P Q) P Q P Q P Q P Q P Q P Q P Q P Q (P Q) F F F F T T T F T F T F F T T F F T F F T T T T T T T T P Q P Q 21 / 39
30 Some Properties of Equivalence Equivalence is reflexive: ϕ ϕ symmetric: if ϕ ψ then ψ ϕ transitive: if ϕ ψ and ψ χ then ϕ χ Connection between and = ϕ ψ iff ϕ ψ 22 / 39
31 Substitution for Propositional Letters We substitute a formula ϑ for all occurrences of r R in a formula ϕ, written as ϕ[ϑ/r] as follows: if ϕ is some r R, then ϕ[ϑ/r] := ϑ if r = r ϕ[ϑ/r] := r otherwise [ϑ/r] := (ψ χ)[ϑ/r] := ψ[ϑ/r] χ[ϑ/r] (ψ χ)[ϑ/r] := ψ[ϑ/r] χ[ϑ/r] (ψ χ)[ϑ/r] := ψ[ϑ/r] χ[ϑ/r]. 23 / 39
32 Tautologies for free! Substitution in Tautologies If = ϕ then = ϕ[ψ/r]. Once we have shown that = P P, we know that = ϕ ϕ for any ϕ. If ϕ 1 ϕ 2, then also ϕ 1 [ψ/r] ϕ 2 [ψ/r]. 24 / 39
33 Properties of Substitution (II) Leibniz Law If ψ 1 ψ 2 then ϕ[ψ 1 /r] ϕ[ψ 2 /r]. ϕ (ψ χ) ϕ ( ψ χ) We can eliminate from any formula! 25 / 39
34 Truth Functions truth function of arity n: function from B n to B formulas give rise to truth functions: for r 1,..., r n R and x 1,..., x n B, define interpretation { xi r = r I r1:=x 1,...,r n:=x n (r) := i F else for formula ϕ with PL(ϕ) = {r 1,..., r n } define truth function f ϕ : B n B by f ϕ (x 1,..., x n ) := ϕ Ir1 :=x 1,...,rn:=xn { T if x1 = x Example: f P Q (x 1, x 2 ) = 2 = T F else 26 / 39
35 Functional Completeness Functional Completeness A set O of operators is functionally complete if, for every f : B n B, there is a formula ϕ f using only operators from O such that f ϕf = f. {,,, } is functionally complete so are {,, } and {, } but {, } is not 27 / 39
36 Functional Completeness (ctd.) P Q R f F F F F F F T F F T F F F T T T T F F F T F T T T T F T T T T T ϕ f := P Q R P Q R P Q R P Q R 28 / 39
37 Functional Completeness (ctd.) P Q R f F F F F F F T F F T F F F T T T T F F F T F T T T T F T T T T T ϕ f := P Q R P Q R P Q R P Q R 28 / 39
38 Functional Completeness (ctd.) P Q R f F F F F F F T F F T F F F T T T T F F F T F T T T T F T T T T T ϕ f := P Q R P Q R P Q R P Q R 28 / 39
39 Functional Completeness (ctd.) P Q R f F F F F F F T F F T F F F T T T T F F F T F T T T T F T T T T T ϕ f := P Q R P Q R P Q R P Q R 28 / 39
40 Functional Completeness (ctd.) P Q R f F F F F F F T F F T F F F T T T T F F F T F T T T T F T T T T T ϕ f := P Q R P Q R P Q R P Q R 28 / 39
41 Calculational Logic 演算邏輯 29 / 39
42 Principles of Calculational Logic calculational logic is a deductive system for propositional logic not a new logic idea: calculate with formulas to establish their truth avoid case distinctions, truth tables make use of a set of laws: tautologies of the form ϕ ψ replace equivalent formulas 30 / 39
43 Example Derivation Laws: Associativity of : ((P Q) R) (P (Q R)) Symmetry of : P Q Q P Unfolding : P P Derivation: (P Q) 31 / 39
44 Example Derivation Laws: Associativity of : ((P Q) R) (P (Q R)) Symmetry of : P Q Q P Unfolding : (P Q) ((P Q) ) Derivation: (P Q) 31 / 39
45 Laws: Example Derivation Associativity of : ((P Q) R) (P (Q R)) Symmetry of : P Q Q P Unfolding : (P Q) ((P Q) ) Derivation: (P Q) { Unfolding } (P Q) 31 / 39
46 Laws: Example Derivation Associativity of : ((P Q) ) (P (Q )) Symmetry of : P Q Q P Unfolding : P P Derivation: (P Q) { Unfolding } (P Q) 31 / 39
47 Laws: Example Derivation Associativity of : ((P Q) ) (P (Q )) Symmetry of : P Q Q P Unfolding : P P Derivation: (P Q) { Unfolding } (P Q) { Associativity of } P (Q ) 31 / 39
48 Laws: Example Derivation Associativity of : ((P Q) R) (P (Q R)) Symmetry of : P Q Q P Unfolding : Q Q Derivation: (P Q) { Unfolding } (P Q) { Associativity of } P (Q ) 31 / 39
49 Laws: Example Derivation Associativity of : ((P Q) R) (P (Q R)) Symmetry of : P Q Q P Unfolding : Q Q Derivation: (P Q) { Unfolding } (P Q) { Associativity of } P (Q ) { Folding } P Q 31 / 39
50 Laws: Example Derivation Associativity of : ((P Q) R) (P (Q R)) Symmetry of : P Q Q P Unfolding : P P Derivation: (P Q) { Unfolding } P Q { Folding } P Q By transitivity: (P Q) P Q 31 / 39
51 Laws for and Unfolding : P P Idempotence of : P P P Symmetry of : P Q Q P Associativity of : P (Q R) (P Q) R Distributivity of : P (Q R) P Q P R Excluded Middle: P P Examples: P P, = P 32 / 39
52 Law for Golden Rule: P Q P Q P Q P Q P Q P Q P Q P Q P Q P Q P Q (P Q) F F F F T T T F T F T F F T T F F T F F T T T T T T T T P Q is true iff P Q and P Q have the same truth value. Example: P (P Q) P 33 / 39
53 Law for Golden Rule: P Q P Q P Q P Q P Q P Q P Q P Q P Q P Q P Q (P Q) F F F F T T T F T F T F F T T F F T F F T T T T T T T T P Q is true iff P Q and P Q have the same truth value. Example: P (P Q) P 33 / 39
54 Law for and Substitution Law Unfolding : P Q Q P Q Substitution: (P Q) ϕ[p/r] (P Q) ϕ[q/r] Example: P P, P (P Q) P Q 34 / 39
55 The Island of Knights and Knaves ( 騎士與惡棍之島 ) on an island, there are two kinds of inhabitants: knights and knaves knives always speak the truth, knaves always lie assume inhabitant A says: If you ask B, he will say he is a knight. what can we infer about A? what about B? 35 / 39
56 Knights and Knaves in Propositional Logic propositional letters represent identity of inhabitants let A mean A is a knight ; then A is A is a knave statements about who is a knight or knave become propositional formulas assume A says ϕ: if A is a knight, then ϕ is true if A is a knave, then ϕ is false So whenever A says ϕ, we have A ϕ! A says: B says he is a knight. is A B B 36 / 39
57 Statements about Knights and Knaves A says he is a knight. A says he is a knave. A and B are of the same kind. A says: I am of the same kind as B. 37 / 39
58 Statements about Knights and Knaves A says he is a knight. A A A says he is a knave. A and B are of the same kind. A says: I am of the same kind as B. 37 / 39
59 Statements about Knights and Knaves A says he is a knight. A A A says he is a knave. A A A and B are of the same kind. A says: I am of the same kind as B. 37 / 39
60 Statements about Knights and Knaves A says he is a knight. A A A says he is a knave. A A A and B are of the same kind. A B A says: I am of the same kind as B. 37 / 39
61 Statements about Knights and Knaves A says he is a knight. A A A says he is a knave. A A A and B are of the same kind. A B A says: I am of the same kind as B. A (A B) 37 / 39
62 Finding a Treasure on the Island Legend has it there is a treasure on the island. We want to find out whether that is true. Let the propositional letter T stand for there is a treasure. Assume we meet inhabitant A. What question Q should we ask him to find out whether T is true? A answers Q with yes : A Q A answers Q with yes iff there is a treasure: A Q T this is the same as Q (A T ) So we should ask Does there is a treasure on this island equivale that you are a knight? 38 / 39
63 Finding a Treasure on the Island Legend has it there is a treasure on the island. We want to find out whether that is true. Let the propositional letter T stand for there is a treasure. Assume we meet inhabitant A. What question Q should we ask him to find out whether T is true? A answers Q with yes : A Q A answers Q with yes iff there is a treasure: A Q T this is the same as Q (A T ) So we should ask Does there is a treasure on this island equivale that you are a knight? 38 / 39
64 Finding a Treasure on the Island Legend has it there is a treasure on the island. We want to find out whether that is true. Let the propositional letter T stand for there is a treasure. Assume we meet inhabitant A. What question Q should we ask him to find out whether T is true? A answers Q with yes : A Q A answers Q with yes iff there is a treasure: A Q T this is the same as Q (A T ) So we should ask Does there is a treasure on this island equivale that you are a knight? 38 / 39
65 Finding a Treasure on the Island Legend has it there is a treasure on the island. We want to find out whether that is true. Let the propositional letter T stand for there is a treasure. Assume we meet inhabitant A. What question Q should we ask him to find out whether T is true? A answers Q with yes : A Q A answers Q with yes iff there is a treasure: A Q T this is the same as Q (A T ) So we should ask Does there is a treasure on this island equivale that you are a knight? 38 / 39
66 Some More Logic Puzzles Assume A says any of the following things; what can you deduce about A and B? If I am a knight, then so is B. If B is a knight, then so am I. If I am a knave, then B is a knight. If I am a knight, then B is a knave. If B is a knave, then I am a knave. B says one of us is a knight. 39 / 39
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