2/17/2008Sultan Almuhammadi1 ICS Logic & Sets (An Overview) Week 1

2/17/2008Sultan Almuhammadi2 Keywords (1): Proposition Conjunction Disjunction Negation Compound proposition Truth Table Logically equivalence

2/17/2008Sultan Almuhammadi3 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

2/17/2008Sultan Almuhammadi4 Example: p ^ q is a “compound” proposition (logical expression) The value of this proposition (expression) depends on the values of p and q.

2/17/2008Sultan Almuhammadi5 Truth table: p ^ q pqp ^ q TTT TFF FTF FFF

2/17/2008Sultan Almuhammadi6 Truth table: p v q pqp v q TTT TFT FTT FFF

2/17/2008Sultan Almuhammadi7 Keywords (2) Conditional proposition: if p then q pq Biconditional proposition: p if and only if q pq

2/17/2008Sultan Almuhammadi8 Keywords (2) Conditional proposition: if p then q (read: p implies q ) pq Biconditional proposition: p if and only if q (write: p iff q for short) pq

2/17/2008Sultan Almuhammadi9 Keywords (2) Conditional proposition: if p then q (read: p implies q ) pq Biconditional proposition: p if and only if q (write: p iff q for short) pq p -> q p q

2/17/2008Sultan Almuhammadi10 Truth Tables: if_then and iff pqp -> qp q TTTT TFFF FTTF FFTT

2/17/2008Sultan Almuhammadi11 Truth Tables (Example) pq~p~p v qp -> q TT? TF? FT? FF?

2/17/2008Sultan Almuhammadi12 Truth Tables pq~p~p v qp -> q TTF? TFF? FTT? FFT?

2/17/2008Sultan Almuhammadi13 Truth Tables pq~p~p v qp -> q TTFT? TFFF? FTTT? FFTT?

2/17/2008Sultan Almuhammadi14 Truth Tables pq~p~p v qp -> q TTFTT TFFFF FTTTT FFTTT

2/17/2008Sultan Almuhammadi15 Truth Tables pq~p~p v qp -> q TTFTT TFFFF FTTTT FFTTT Logically equivalent (~p v q) = (p -> q)

2/17/2008Sultan Almuhammadi16 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.

2/17/2008Sultan Almuhammadi17 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 

2/17/2008Sultan Almuhammadi18 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?

2/17/2008Sultan Almuhammadi19 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

2/17/2008Sultan Almuhammadi20 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.

2/17/2008Sultan Almuhammadi21 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

2/17/2008Sultan Almuhammadi22 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.

2/17/2008Sultan Almuhammadi23 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 }

2/17/2008Sultan Almuhammadi24 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}

2/17/2008Sultan Almuhammadi25 Examples: - Which of the following is true? - 2  N - 2  Z - 2  N  Z - x  N  x  Z - N  Z -   Z -   Z

2/17/2008Sultan Almuhammadi26 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)

2/17/2008Sultan Almuhammadi27 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)

2/17/2008Sultan Almuhammadi28 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  x  y (x > y)(domain of discourse is Q) - False, for y = x  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

2/17/2008Sultan Almuhammadi29 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)

2/17/2008Sultan Almuhammadi30 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)

2/17/2008Sultan Almuhammadi31 Exer 2: Domain = all cs253 students p(x, q) = student x solved question q Write the following in symbolic notation: A.Everybody got full mark. B.Nobody got full mark. C.Negation of A. (Not everybody got full mark). D.Negation of B. E.There was a hard question nobody solved it. F.Negation of E. Solution: ?

2/17/2008Sultan Almuhammadi32 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: ?