ics 253-01 logic & sets (an overview) week 1

2/17/2008 Sultan Almuhammadi 1

ICS 253-01

Logic & Sets(An Overview)

Week 1


Keywords (1):

Proposition ConjunctionDisjunctionNegation Compound propositionTruth TableLogically equivalence


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


Example:

p ^ q is a “compound” proposition (logical expression)

The value of this proposition (expression) depends on the values of p and q.


Truth table: p ^ q

p q p ^ q

T T T

T F F

F T F

F F F


Truth table: p v q

p q p v q

T T T

T F T

F T T

F F F


Keywords (2)

Conditional proposition:

if p then q p q

Biconditional proposition: p if and only if q

p q


Keywords (2)Conditional proposition: if p then q (read: p implies q )

p q


(write: p iff q for short)

p q


Keywords (2)Conditional proposition: if p then q (read: p implies q )

p q


(write: p iff q for short)

p q

p -> q

p <--> q


Truth Tables: if_then and iff

p q p -> q p <-> q

T T T T

T F F F

F T T F

F F T T


Truth Tables (Example)

p q ~p ~p v q p -> q

T T ?

T F ?

F T ?

F F ?


Truth Tables


T T F ?

T F F ?

F T T ?

F F T ?


Truth Tables


T T F T ?

T F F F ?

F T T T ?

F F T T ?


Truth Tables


T T F T T

T F F F F

F T T T T

F F T T T


Truth Tables


T T F T T

T F F F F

F T T T T

F F T T T

Logically equivalent

(~p v q) = (p -> q)


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.


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


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?


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


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.


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


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.


Set Operations- Intersection

- A B = { x | x A and xB}- Union

- A B = { x | x A or xB}- Complement

- E’ = Ē = { x | x E }


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, bZ 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}


Examples:- Which of the following is true?

- 2 N- 2 Z- 2 N Z- x N x Z- N Z Z Z


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)


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)


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


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)


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)


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: ?


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: ?

ics 253-01 logic & sets (an overview) week 1

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