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ICS 253-01 Logic & Sets (An Overview) Week 1

ICS 253-01 Logic & Sets (An Overview) Week 1. Keywords (1):. Proposition Conjunction Disjunction Negation Compound proposition Truth Table Logically equivalence. Getting started. Proposition: true/false statement cannot be both at the same time e.g. Today is Monday

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ICS 253-01 Logic & Sets (An Overview) Week 1

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  1. ICS 253-01 Logic & Sets (An Overview) Week 1 Sultan Almuhammadi

  2. Keywords (1): Proposition Conjunction Disjunction Negation Compound proposition Truth Table Logically equivalence Sultan Almuhammadi

  3. Getting started • Proposition: • true/false statement • cannot be both at the same time • e.g. Today is Monday • Conjunction: (and) • p ^ q • Disjunction: (or) • p v q • Negation: (not) • ~p Sultan Almuhammadi

  4. Example: p ^ q is a “compound” proposition (logical expression) The value of this proposition (expression) depends on the values of p and q. Sultan Almuhammadi

  5. Truth table: p ^ q Sultan Almuhammadi

  6. Truth table: p v q Sultan Almuhammadi

  7. Keywords (2) Conditional proposition: if p then q p q Biconditional proposition: p if and only if q p q Sultan Almuhammadi

  8. Keywords (2) Conditional proposition: if p then q (read: p implies q ) p q Biconditional proposition: p if and only if q (write: p iff q for short) p q Sultan Almuhammadi

  9. Keywords (2) Conditional proposition: if p then q (read: p implies q ) p q p -> q p <--> q Biconditional proposition: p if and only if q (write: p iff q for short) p q Sultan Almuhammadi

  10. Truth Tables: if_then and iff Sultan Almuhammadi

  11. Truth Tables (Example) Sultan Almuhammadi

  12. Truth Tables Sultan Almuhammadi

  13. Truth Tables Sultan Almuhammadi

  14. Truth Tables Sultan Almuhammadi

  15. Logically equivalent (~p v q) = (p -> q) Truth Tables Sultan Almuhammadi

  16. Exer 1. (p -> q) = (~p v q) • If it is Wednesday, John has discussion. • It is not Wed, or John has discussion. • If you don’t study hard, you will fail. • Study hard or you fail. • I have a meeting on Friday. • I have a meeting today, or it is not Friday. Sultan Almuhammadi

  17. Logical Equivalence: p  q • (p -> q )  (~p v q ) • ~( p v q )  ~p ^ ~q [De Morgan’s] • ~( p ^ q )  ~p v ~q [De Morgan’s] • (p <--> q)  (p -> q) ^ (q -> p) • (p <--> q)  (~p v q) ^ (~q v p) • Remember: I use = for  Sultan Almuhammadi

  18. Binary Logic • Binary Logic • p has one of two values: True / False • p cannot have both. • Values can be {T,F}, {0,1}, {High, Low} • n-ary Logic • p has one of n valuse: e.g. 1, 2, …, n • Conjunction, disjunction, and negation are defined over these n values. • Sounds weird? Sultan Almuhammadi

  19. Keywords (3): • Propositional logic • First order logic • Domain of discourse • Sets • Natural Numbers (N) • Integers (Z) • Rational Numbers (Q) • Irrational Numbers ( Q’ ) • Real Numbers (R) • Prime numbers Sultan Almuhammadi

  20. First Order Logic • Propositional logic • E.g. p ^ q -> r • First order logic • E.g. p(x) ^ q(y) • x and y are from some Domain of discourse. • The value of p(x) depends on x. Sultan Almuhammadi

  21. Sets • Notations: • Upper case letters: A, B X, Y • Elements can be listed in braces {…} • E.g. A = {1, 2, 5} • Elements can be described: • E.g. B = set of all even numbers. • Or B = {x | x is an even number} • E.g. C = {a : a is even and 1< a <10} • The size of set A is denoted by |A| • E.g. |A| = 3 , |B| = ∞ , |C| = 4 Sultan Almuhammadi

  22. Sets • Membership • x  A // x belongs to A, • y  A • Subsets • A  B // can be equal • B  N // proper-subset • The Empty Set: denoted by  = { } • The Universe: U = set of all elements. Sultan Almuhammadi

  23. Set Operations • Intersection • A  B = { x | x  A and xB} • Union • A  B = { x | x  A or xB} • Complement • E’ = Ē = { x | x  E } Sultan Almuhammadi

  24. Number Systems • Integers: • Z = { …, -3, -2, -1, 0, 1, 2, 3, …} • Z+ = {1, 2, 3, …} (positive integers) • Natural Numbers: • N = {0, 1, 2, 3, … } (nonnegative integers) • Rational Numbers: • Q = {a/b | a, bZ and b  0} • Irrational Numbers: • Q’ = { x | x  R and x  Q} • Prime numbers: • {x | x  N and x is divisible by 1 and x only} Sultan Almuhammadi

  25. Examples: • Which of the following is true? • 2  N • 2  Z • 2  N  Z • x  N  x  Z • N  Z •   Z •   Z Sultan Almuhammadi

  26. Warm up: • Domain of discourse (the domain) • Set {x | x is prime} • Set {x | x is even and x is prime} • For all x, P(x) x P(x) • For some x, P(x) x P(x) • Eg. • x, x > 1 (domain = N) • x, x > 1 (domain = N) • x, y, x > y (domain = N) Sultan Almuhammadi

  27. Exer 1: • x y (x > y) (domain of discourse is R) • x y (x > y) (domain of discourse is N) • x y (x < y) (domain of discourse is Z) • x y (x > y) (domain of discourse is Q) • x y (x < y) (domain of discourse is N) • x y (x < y) (domain of discourse is Z) Sultan Almuhammadi

  28. Exer 1: Solution • x y (x > y) (domain of discourse is R) • False, for x = 1, y = 2 • x y (x > y) (domain of discourse is N) • True, for x = 2, y = 1 • x y (x < y) (domain of discourse is Z) • True, for y = x + 1 • x y (x > y) (domain of discourse is Q) • False, for y = x + 1 • x y (x < y) (domain of discourse is N) • False, for y = x • x y (x < y) (domain of discourse is Z) • False, for y = x - 1 Sultan Almuhammadi

  29. Negation: • E.g. • Domain: set of all horses • p(x) : x is a black horse • q(x): x is a white horse • Universal quantifier • x p(x) /* all horses are black */ • Negation: ~ x p(x) = x ~p(x) • Existential quantifier • x p(x) • Negation: ~ x p(x) = x ~p(x) Sultan Almuhammadi

  30. Negation: • e.g.1. • x y(x > y) • = x [ y(x > y) ] • Negation: ~ x [ y (x > y) ] • = x ~ y (x > y) • = x y ~ ( x > y) • = x y ( x ≤ y) • e.g.2. • x yz (x < z) ^ (z < y) • Negation: ~ x yz (x < z) ^ (z < y) • = x y z ~ [ (x < z) ^ (z < y) ] • = x y z (x ≥ z) V (z ≥ y) Sultan Almuhammadi

  31. Exer 2: • Domain = all cs253 students • p(x, q) = student x solved question q • Write the following in symbolic notation: • Everybody got full mark. • Nobody got full mark. • Negation of A. (Not everybody got full mark). • Negation of B. • There was a hard question nobody solved it. • Negation of E. Solution: ? Sultan Almuhammadi

  32. Quiz (A1) (A1: for practice only, not counted) • Domain = all cs253 students • p(x, q) = student x solved question q • Write the following in symbolic notation: “There is exactly one student who got full mark.” Solution: ? Sultan Almuhammadi

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