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Chap. 2 Fundamentals of Logic

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Proposition Proposition (or statement): an declarative sentence that is either true or false, but not both. e.g. –Margret Mitchell wrote Gone with the Wind. –2+3=6. –What a beautiful evening. –Get up and do your exercise. –Combinatorics is a required course for sophomores. O O O X X

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Negation of Proposition The negation of a proposition p, denoted by ￢ p, is the statement “It is not the case that p”. e.g. –p: Combinatorics is a required course for sophomores. – ￢ p: It is not the case that combinatorics is a required course for sophomores.

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Truth Table of Negation of Proposition p ￢p￢p 1 0 1: true 0: false 0 1

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Compound Statement Compound Statement: a statement that is combined by two or more statements using logic connections, including ⋀ (conjunction), ⋁ (disjunction), ⊻ (exclusive or), → (implication), and ↔ (biconditional).

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Conjunction The conjunction of statements p and q, denoted by p ⋀ q, is the statement “p and q”. e.g. –p: Combinatorics is a required course for sophomores. –q: Margret Mitchell wrote Gone with the Wind. –p ⋀ q : Combinatorics is a required course for sophomores and Margret Mitchell wrote Gone with the Wind.

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Disjunction The disjunction of statements p and q, denoted by p ⋁ q, is the statement “p or q”. e.g. –p: Combinatorics is a required course for sophomores. –q: Margret Mitchell wrote Gone with the Wind. –p ⋁ q : Combinatorics is a required course for sophomores or Margret Mitchell wrote Gone with the Wind.

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Exclusive Or The exclusirve or of statements p and q, denoted by p ⊻ q, is the statement “p or q, but not both”. e.g. –p: Combinatorics is a required course for sophomores. –q: Margret Mitchell wrote Gone with the Wind. –p ⊻ q :–p ⊻ q : Combinatorics is a required course for sophomores or Margret Mitchell wrote Gone with the Wind, but not both.

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Implication The implication of statements p and q, denoted by p → q, is the statement “if p, then q”. e.g. –p: Combinatorics is a required course for sophomores. –q: Margret Mitchell wrote Gone with the Wind. –p → q :–p → q : If combinatorics is a required course for sophomores, then Margret Mitchell wrote Gone with the Wind.

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Biconditional The biconditional of statements p and q, denoted by p ↔ q, is the statement “p if and only if q”. e.g. –p: Combinatorics is a required course for sophomores. –q: Margret Mitchell wrote Gone with the Wind. –p ↔ q :–p ↔ q : Combinatorics is a required course for sophomores if and only if Margret Mitchell wrote Gone with the Wind.

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Truth Table of Conjunction, Disjunction, Exclusive Or, Implication, and Biconditional pq p ⋀ qp ⋀ qp ⋁ qp ⋁ qp ⊻ qp ⊻ qp → qp → qp ↔ qp ↔ q 00 01 10 11 0 1 0 0 1 1 1 0 1 0 1 0 0 1 1 1 0 1 0 1

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Truth Table of Implication p → qp → q p q 100 101 111 p → q means: If p is true, then q is true. If p is false, then q is true or false. Thus,

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Tautology and Contradiction A compound statement is a tautology if it is true for all truth value assignments for its component statements. If a compound statement is false for all such assignment, then it is a contradiction. e.g. p ￢p￢p p ⋀ ￢pp ⋀ ￢p p ⋁ ￢pp ⋁ ￢p 01 01 10 10 0 0 0 0 1 1 1 1 ∴ p ⋀ ￢ p is a p ⋁ ￢ p is a contradiction. tautology.

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Logical Equivalence Two statements p, q are logically equivalent, and we write p ⇔ q, when p ↔ q is a tautology. e.g. pq ￢(p⋀q)￢(p⋀q) ￢p⋁￢q￢p⋁￢q ￢ (p ⋀ q) ↔ ￢ p ⋁￢ q p→qp→q ￢p⋁q￢p⋁qp→q ↔ ￢ p ⋁ q 00 01 10 11 1 1 1 0 1 1 1 0 1 1 1 1 1 1 0 1 1 1 0 1 1 1 1 1 ￢ (p ⋀ q) ⇔ p→q ⇔ ￢p⋁￢q￢p⋁￢q ￢p⋁q￢p⋁q

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The Laws of Logic For any primitive statements p, q, r, any tautology T 0, and any contradiction F 0. – ￢￢ p ⇔ p (Law of Double Negation) – ￢ (p ⋁ q) ⇔ ￢ p ⋀￢ q (DeMorgan’s Law) –p ⋁ q ⇔ q ⋁ p (Commutative Law) –p ⋁ (q ⋁ r) ⇔ (p ⋁ q) ⋁ r (Associative Law) –p ⋁ (q ⋀ r) ⇔ (p ⋁ q) ⋀ (p ⋁ r) (Distributive Law)

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The Laws of Logic (2) For any primitive statements p, q, r, any tautology T 0, and any contradiction F 0. –p ⋁ p ⇔ p (Idempotent Law) –p ⋁ F 0 ⇔ p (Identity Law) –p ⋁￢ p ⇔ T 0 (Inverse Law) –p ⋁ T 0 ⇔ T 0 (Domination Laws) –p ⋁ (p ⋀ q) ⇔ p (Absorption Law)

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Dual of Statement Let s be a statement. If s contains no logical connectives other than ⋀ and ⋁, then the dual of s, denoted s d, is the statement obtained from s by replacing each occurrence of ⋀ and ⋁ by ⋁ and ⋀, respectively, and each occurrence of T 0 and F 0 by F 0 and T 0, respectively. e.g. s: (p ⋀￢ q) ⋁ (r ⋀ T 0 ) s d : (p⋁￢q)⋀(r⋁F0)(p⋁￢q)⋀(r⋁F0)

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Principle of Duality The Principle of Duality. Let s and t be statements that contain no logical connectives other than ⋀ and ⋁. If s ⇔ t, then s d ⇔ t d. e.g. ￢ (p ⋁ q) ⇔ ￢ p ⋀￢ q ￢ (p ⋀ q) ⇔ ￢ p ⋁￢ q (1 st DeMorgan’s Law) (2 nd DeMorgan’s Law)

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The Laws of Logic (3) For any primitive statements p, q, r, any tautology T 0, and any contradiction F 0. – ￢ (p ⋁ q) ⇔ ￢ p ⋀￢ q (1 st DeMorgan’s Law) – ￢ (p ⋀ q) ⇔ ￢ p ⋁￢ q (2 nd DeMorgan’s Law) –p ⋁ q ⇔ q ⋁ p (1 st Communication Law) –p ⋀ q ⇔ q ⋀ p (2 nd Communication Law) –p ⋁ (q ⋁ r) ⇔ (p ⋁ q) ⋁ r (1 st Associative Law) –p ⋀ (q ⋀ r) ⇔ (p ⋀ q) ⋀ r (2 nd Associative Law)

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The Laws of Logic (4) For any primitive statements p, q, r, any tautology T 0, and any contradiction F 0. –p ⋁ (q ⋀ r) ⇔ (p ⋁ q) ⋀ (p ⋁ r) (1 st Distributive Law) –p ⋀ (q ⋁ r) ⇔ (p ⋀ q) ⋁ (p ⋀ r) (2 nd Distributive Law) –p ⋁ p ⇔ p (1 st Idempotent Law) –p ⋀ p ⇔ p (2 nd Idempotent Law) –p ⋁ F 0 ⇔ p (1 st Identity Law) –p ⋀ T 0 ⇔ p (2 nd Identity Law)

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The Laws of Logic (5) For any primitive statements p, q, r, any tautology T 0, and any contradiction F 0. –p ⋁￢ p ⇔ T 0 (1 st Inverse Law) –p ⋀￢ p ⇔ F 0 (2 nd Inverse Law) –p ⋁ T 0 ⇔ T 0 (1 st Domination Laws) –p ⋀ F 0 ⇔ F 0 (2 nd Domination Laws) –p ⋁ (p ⋀ q) ⇔ p (1 st Absorption Law) –p ⋀ (p ⋁ q) ⇔ p (2 nd Absorption Law)

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Substitution Rule 1 Suppose that the compound statement P is a tautology. If p is a primitive statement that appears in P and we replace each occurrence of p by the same statement q, then the resulting compound statement P 1 is also a tautology. e.g. P: ￢ (p ⋁ q)↔( ￢ p ⋀￢ q) is a tautology ∴ P 1 : ￢ [(r ⋀ s) ⋁ q]↔[ ￢ (r ⋀ s) ⋀￢ q] is a tautology

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Substitution Rule 1 (2) e.g. ￢ ￢ (p ∨ q) ⇔ (p ∨ q) e.g. (p ∨ q) ∧ (p ∨￢ q) ⇔ (p ∨ (q ∧￢ q)) ∵ ￢ ￢ p ⇔ p (Law of Double Negation) ∵ p ∨ (q ∧ r) ⇔ (p ∨ q) ∧ (p ∨ r) (1 st Distributive Law)

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Substitution Rule 2 Let P be a compound statement where p is an arbitrary statement that appears in P, and let q be a statement such that q ⇔ p. Suppose that in P we replace one or more occurrences of p by q. Then this replacement yields the compound statement P 1. Under these circumstances P 1 ⇔ P. e.g. P: (p→q)→r, P 1 : ( ￢ p ∨ q)→r ∵ (p→q) ⇔ ￢ p ∨ q ∴ P 1 ⇔ P

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Substitution Rule 2 (2) e.g. ￢ ￢ (p ∨ q) ∧￢ r ⇔ (p ∨ q) ∧￢ r e.g. (p ∨ q) ∧ (r ∧ q) ⇔ (p ∨ q) ∧ (q ∧ r) ∵ ￢ ￢ (p ∨ q) ⇔ (p ∨ q) (Law of Double Negation and Substitution Rule 1) ∵ r ∧ q ⇔ q ∧ r (Commutative Law)

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Example 2.16 Show that (p ∨ q) ∧￢ ( ￢ p ∧ q) ⇔ p (p∨q)∧￢(￢p∧q)(p∨q)∧￢(￢p∧q) ⇔ (p ∨ q) ∧ ( ￢￢ p ∨￢ q) ⇔ (p ∨ q) ∧ (p ∨￢ q) ⇔ p ∨ (q ∧￢ q) ⇔ p ∨ F 0 ⇔ p (2 nd DeMorgan’s Law) (Law of Double Negation) (1 st Distributive Law) (2 nd Inverse Law) (1 st Identity Law)

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Example 2.17 Show that ￢ [ ￢ [(p ∨ q) ∧ r] ∨￢ q] ⇔ q ∧ r ￢ [ ￢ [(p ∨ q) ∧ r] ∨￢ q] ⇔ ￢￢ [(p ∨ q) ∧ r] ∧ ( ￢￢ q) ⇔ [(p ∨ q) ∧ r] ∧ q ⇔ (p ∨ q) ∧ (r ∧ q) ⇔ (p ∨ q) ∧ (q ∧ r) ⇔ [(p ∨ q) ∧ q ] ∧ r) ⇔ q ∧ r (1 st DeMorgan’s Law) (Law of Double Negation) (2 nd Associative Law) (2 nd Commutative Law) (2 nd Associative Law) (2 nd Absorption Law)

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Logical Implication If p, q are arbitrary statements such that p→q is a tautology, then we say that p logically implies q and we write p ⇒ q to denote this situation. e.g. [p ∧ (p→q)] ⇒ q

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Rule of Inference [p ∧ (p→q)] ⇒ q The actual rule will be written in the tabular form p p→q ∴ q

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Example 2.31 Show the following argument is valid ( ￢ p ∨￢ q)→(r ∧ s) r→t ￢ t ∴ p

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Example 2.31 (2) Steps Reasons 1)r→t 2) ￢ t 3) ￢ r 4) ￢ r ∨￢ s 5) ￢ (r ∧ s) 6)( ￢ p ∨￢ q)→(r ∧ s) 7) ￢ ( ￢ p ∨￢ q) 8)p ∧ q 9) ∴ p Premise Steps (1) and (2) and Modus Tollens Step (3) and Rule of Disjunctive Amplification Step (4) and 2 nd DeMorgan’s Laws Steps (6) and (5) and Modus Tollens Step (7), DeMorgan’s Laws, and Law of Double Negation Step (8) and Rule of Conjunctive Simplification

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Example 2.32 Show the following argument is valid by the method of Contradiction ￢ p↔q q→r ￢ r ∴ p

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Example 2.32 (2) Steps Reasons 1) ￢ p↔q 2)( ￢ p→q) ∧ (q→ ￢ p) 3) ￢ p→q 4)q→r 5) ￢ p→r 6) ￢ p 7)r 8) ￢ r 9)r ∧￢ r ( ⇔ F 0 ) 10) ∴ p Premise Step (1) and ￢ p↔q ⇔ ( ￢ p→q) ∧ (q→ ￢ p) Premise (the one assumed) Step (2) and Rule of Conjunctive Simplification Premise Steps (3) and (4) and Law of the Syllogism Steps (5) and (6) and Rule of Detachment Premise Steps (7) and (8) and Rule of Conjunction Steps (6) and (9) and the method of Proof by Contradiction

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Open Statement A declarative sentence is an open statement if –1) it contains one or more variables, and –2) it is not a statement, but –3) it becomes a statement when the variables in it are replaced by certain allowable choices. e.g. –p(x): The number x+2 is an even integer.

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Open Statement (2) e.g. q(x, y): The numbers y+2, x−y, and x+2y are even integers. p(x): x ≥0. q(x): x 2 ≥0.

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Quantifier Here the universe comprises all real numbers. The open statements p(x), q(x), r(x), and s(x) are given by p(x): x≥0, r(x): x 2 −3x−4=0, q(x): x 2 ≥0, s(x): x 2 −3>0. Then –∃ x [p(x) ∧ r(x)] is –∀ x [p(x)→q(x)] is –∀ x [q(x)→s(x)] is (∃: Existential Quantifier) (∀: Universal Quantifier) true false

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Table 2.21 StatementWhen Is It True? When Is It False? ∃ x p(x) ∀ x p(x) ∃ x ￢ p(x) ∀ x ￢ p(x) For some a in the universe, p(a) is true. For every a in the universe, p(a) is false. For every a in the universe, p(a) is true. For some a in the universe, p(a) is false. For every a in the universe, p(a) is true. For every a in the universe, p(a) is false. For some a in the universe, p(a) is true.

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Logical Equivalence of Open Statements Let p(x), q(x) be open statements defined for a given universe. The open statements p(x) and q(x) are called (logically) equivalent, and we write ∀ x [p(x) ⇔ q(x)] when the biconditional p(a) ↔ q(a) is true for each replacement a from the universe (that is, p(a) ⇔ q(a) for each a in the universe).

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Logical Equivalence of Open Statements (2) For the universe of all triangles in the plane, let p(x), q(x) denote the open statements: p(x): x is equiangular, q(x): x is equilateral. ∵ p(a) ↔ q(a) is true for every triangle a in the plane ∴ ∀ x [p(x) ⇔ q(x)] (p(x) and q(x) are logically equivalent)

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Logical Implication of Open Statements Let p(x), q(x) be open statements defined for a given universe. If the implication p(a)→ q(a) is true for each a in the universe (that is, p(a) ⇒ q(a) for each a in the universe), then we write ∀ x [p(x) ⇒ q(x)] and say that p(x) logically implies q(x).

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Logical Implication of Open Statements (2) For the universe of all triangles in the plane, let p(x), q(x) denote the open statements: p(x): x is equiangular, q(x): x is equilateral. ∵ p(a) → q(a) is true for every triangle a in the plane ∴ ∀ x [p(x) ⇒ q(x)] (p(x) logically implies q(x))

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Logical Equivalence and Logical Implication for Qualifier Statement ∃ x [p(x) ∧ q(x)] ⇒ [ ∃ x p(x) ∧∃ x q(x)] 1.Suppose ∃ x [p(x) ∧ q(x)] is true. 2.p(a) ∧ q(a) is true for some a. 3.p(a) is true and q(a) is true for some a. 4.∃ x p(x) is true and ∃ x q(x) is true. 5.[ ∃ x p(x) ∧∃ x q(x)] is true.

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Negating Statement with One Quantifier ￢ [ ∀ x p(x)] ⇒ ∃ x ￢ p(x) 1.Suppose ￢ [ ∀ x p(x)] is true. 2.∀ x p(x) is false. 3.p(a) is false for some a. 4. ￢ p(a) is true for some a. 5.∃ x ￢ p(x) is true.

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Negating Statement with One Quantifier (2) ∃ x ￢ p(x) ⇒ ￢ [ ∀ x p(x)] 1.Suppose ∃ x ￢ p(x) is true. 2. ￢ p(a) is true for some a. 3.p(a) is false for some a. 4.∀ x p(x) is false. 5. ￢ [ ∀ x p(x)] is true

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See Table 2.21

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Example 2.49 Let p(x, y), q(x, y), and r(x, y) represent three open statements, with replacements for the variables x, y chosen from some prescribed universe(s). What is the negation of the following statement? ∀ x ∃ y [(p(x, y) ∧ q(x, y))→r(x, y)]

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Example 2.49 (2) ￢ [ ∀ x ∃ y [(p(x, y) ∧ q(x, y))→r(x, y)]] ⇔ ∃ x [ ￢ [ ∃ y [(p(x, y) ∧ q(x, y))→r(x, y)]]] ⇔ ∃ x ∀ y ￢ [(p(x, y) ∧ q(x, y))→r(x, y)] ⇔ ∃ x ∀ y ￢ [ ￢ [p(x, y) ∧ q(x, y)] ∨ r(x, y)] ⇔ ∃ x ∀ y [ ￢￢ [p(x, y) ∧ q(x, y)] ∧￢ r(x, y)] ⇔ ∃ x ∀ y [p(x, y) ∧ q(x, y) ∧￢ r(x, y)].

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Rule of Universal Specification If an open statement becomes true for all replacements by the members in a given universe, then that open statement is true for each specific individual member in that universe. (A bit more symbolically—if p(x) is an open statement for a given universe, and if ∀ x p(x) is true, then p(a) is true for each a in the universe.)

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Example 2.53 For the universe of all people, consider the open statements m(x): x is a mathematics professor, c(x): x has studied calculus. Now consider the following argument. –All mathematics professors have studied calculus. –Leona is a mathematics professor. –Therefore Leona has studied calculus. ∀ x [m(x)→c(x)] m(Leona) ∴ c(Leona)

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Example 2.53 (2) If we let l represent this particular woman (in our universe) named Leona, then we can rewrite this argument in symbolic form as ∀ x [m(x)→c(x)] m(l) ∴ c(l)

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Example 2.53 (3) Steps Reasons 1)∀ x [m(x)→c(x)] 2)m(l)→c(l) 3)m(l) 4)∴ c(l) Premise Step (1) and Rule of Universal Specification Premise Steps (2) and (3) and the Rule of Detachment

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Rule of Universal Generalization If an open statement p(x) is proved to be true when x is replaced by any arbitrarily chosen element c from our universe, then the universally quantified statement ∀ x p(x) is true. Furthermore, the rule extends beyond a single variable. So if, for example, we have an open statement q(x, y) that is proved to be true when x and y are replaced by arbitrarily chosen elements from the same universe, or their own respective universes, then the universally quantified statement ∀ x ∀ y q(x, y) [or, ∀ x,y q(x, y)] is true. Similar results hold for the cases of three or more variables.

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Example 2.56 Show the following argument is ∀ x [p(x) ∨ q(x)] ∀ x [( ￢ p(x) ∧ q(x))→r(x)] ∴ ∀ x [ ￢ r(x)→ p(x)]

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Example 2.56 (2) Steps Reasons 1)∀ x [p(x) ∨ q(x)] 2)p(c) ∨ q(c) 3)∀ x [( ￢ p(x) ∧ q(x))→r(x)] 4)[( ￢ p(c) ∧ q(c))→r(c)] 5) ￢ r(c)→ ￢ ( ￢ p(c) ∧ q(c)) 6) ￢ r(c)→[p(c) ∨￢ q(c)] 7) ￢ r(c) 8)p(c) ∨￢ q(c) 9)[p(c) ∨ q(c)] ∧ [p(c) ∨￢ q(c)] 10)p(c) ∨ [q(c) ∧￢ q(c)] 11)p(c) ∨ F 0 12)p(c) 13)∀ x [ ￢ r(x)→ p(x)] Premise Step (1) and Rule of Universal Specification Step (5), 2 nd DeMorgan’s Law, and Law of Double Negation Premise Step (3) and Rule of Universal Specification Step (4) and s →t ⇔ ￢ t → ￢ s Premise (assumed) Steps (7) and (6) and Modus Ponens Steps (2) and (8) and Rule of Conjunction Step (9) and 1 st Distributive Law Step (10) and 2 nd Inverse Law Steps (7) and (11) and Rule of Universal Generalization Step (11) and 1 st Indentity Law

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