Discrete Mathematics

Predicates - the universal quantifier 11/28/2015 Suppose P(x) is a predicate on some universe of discourse. Ex. B(x) = “x is carrying a backpack,” x is set of CSER1209 students. The universal quantifier of P(x) is the proposition: “P(x) is true for all x in the universe of discourse.” We write it  x P(x), and say “for all x, P(x)”  x P(x) is TRUE if P(x) is true for every single x.  x P(x) is FALSE if there is an x for which P(x) is false.

Predicates - the universal quantifier 11/28/2015 Let A = {1, 2, 3, 4, 5}. Determine the truth value of each of the following statements: (a) ( ∀ x ∈ A)(x + 3 < 10) (b) ( ∀ x ∈ A)(x + 3 ≤ 7) (a) True. For every number in A satisfies x + 3 < 10. (b) False. For if x = 5, then x + 3 is not less than or equal 7. In other words, 5 is not a solution to the given condition.

Predicates - the existential quantifier 11/28/2015 Another way of changing a predicate into a proposition. Suppose P(x) is a predicate on some universe of discourse. Ex. C(x) = “x has a candy bar,” x is set of cser1209 students. The existential quantifier of P(x) is the proposition: “P(x) is true for some x in the universe of discourse.” We write it  x P(x), and say “for some x, P(x)”  x P(x) is TRUE if there is an x for which P(x) is true.  x P(x) is FALSE if P(x) is false for every single x.

Predicates - the existential quantifier 11/28/2015 Let A = {1, 2, 3, 4, 5}. Determine the truth value of each of the following statements: (a) ( ∃ x ∈ A)(x + 3 = 10) (b) ( ∃ x ∈ A)(x + 3 < 5) (a) False. For no number in A is a solution to x + 3 = 10. (b) True. For if x0 = 1, then x0 + 3 < 5, i.e., 1 is a solution.

Predicates - more examples 11/28/2015 L(x) = “x is a lion.” F(x) = “x is fierce.” C(x) = “x drinks coffee.” All lions are fierce. Some lions don’t drink coffee. Some fierce creatures don’t drink coffee. Universe of discourse is all creatures.  x (L(x)  F(x))  x (L(x)   C(x))  x (F(x)   C(x))

Predicates - more examples 11/28/2015 B(x) = “x is a hummingbird.” L(x) = “x is a large bird.” H(x) = “x lives on honey.” R(x) = “x is richly colored.” All hummingbirds are richly colored. No large birds live on honey. Birds that do not live on honey are poorly colored. Universe of discourse is all creatures.  x (B(x)  R(x))  x (L(x)  H(x))  x (  H(x)   R(x))

Example Translate each of these statements into logical expressions using predicates, quantifiers, and logical connectives. a) No one is perfect. b) Not everyone is perfect. c) All your friends are perfect. d) At least one of your friends is perfect. e) Not everybody is your friend or someone is not perfect.  x  P(x)  x P (x)  x(F(x)  P (x))  x(F(x)  P (x))(  x F(x)  (  x  P(x)) Let P (x) be "x is perfect"; let F(x) be "x is your friend";

Predicates - quantifier negation 11/28/2015 Not all large birds live on honey.  x P(x) means “P(x) is true for every x.” What about  x P(x) ? Not [“P(x) is true for every x.”] “There is an x for which P(x) is not true.”  x  P(x) So,  x P(x) is the same as  x  P(x).  x (L(x)  H(x))  x  (L(x)  H(x))

Predicates - quantifier negation 11/28/2015 No large birds live on honey.  x P(x) means “P(x) is true for some x.” What about  x P(x) ? Not [“P(x) is true for some x.”] “P(x) is not true for all x.”  x  P(x) So,  x P(x) is the same as  x  P(x).  x (L(x)  H(x))  x  (L(x)  H(x))

Predicates - quantifier negation 11/28/2015 So,  x P(x) is the same as  x  P(x). So,  x P(x) is the same as  x  P(x). General rule: to negate a quantifier, move negation to the right, changing quantifiers as you go.

Predicates - quantifier negation 11/28/2015 Negate each of the following statements: (a) ∃ x ∀ y P(x, y) (b) ∀ x ∀ y P(x, y) (c) ∃ y ∃ x ∀ z P(x, y, z) Use ￢∀ x p(x) ≡ ∃ x ￢ p(x) and ￢∃ x p(x) ≡ ∀ x ￢ p(x): (a) ￢ ( ∃ x ∀ y, p(x, y)) ≡ ∀ x ∃ y ￢ p(x, y) (b) ￢ ( ∀ x ∀ y, p(x, y)) ≡ ∃ x ∃ y ￢ p(x, y) (c) ￢ ( ∃ y ∃ x ∀ z, p(x, y, z)) ≡ ∀ y ∀ x ∃ z ￢ p(x, y, z)

Predicates - free and bound variables 11/28/2015 A variable is bound if it is known or quantified. Otherwise, it is free. Examples: P(x)x is free P(5)x is bound to 5  x P(x)x is bound by quantifier Reminder: in a proposition, all variables must be bound.

Predicates - the meaning of multiple quantifiers 11/28/2015  x  y P(x,y)  x  y P(x,y)  x  y P(x,y)  x  y P(x,y) P(x,y) true for all x, y pairs. For every value of x we can find a (possibly different) y so that P(x,y) is true. P(x,y) true for at least one x, y pair. There is at least one x for which P(x,y) is always true. quantification order is not commutative. Suppose P(x,y) = “x’s favorite class is y.”

Predicates - multiple quantifiers 11/28/2015 To bind many variables, use many quantifiers! Example: P(x,y) = “x > y”  x P(x,y)  x  y P(x,y)  x  y P(x,y)  x P(x,3) a)True proposition b)False proposition c)Not a proposition d)No clue c)b)a)b)

Proofs - how do you know? 11/28/2015 The following statements are true: If I am Mila, then I am a great swimmer. I am Mila. What do we know to be true? I am a great swimmer! How do we know it?

Proofs - how do you know? 11/28/2015 A theorem is a statement that can be shown to be true. A proof is the means of doing so. Axiom, postulates, hypotheses and previously proven theorems. Rules of inference Proof

Proofs - how do you know? 11/28/2015 The following statements are true: If I am Mila, then I am a great swimmer. I am Mila. What do we know to be true? I am a great swimmer! What rule of inference can we use to justify it?

Proofs - Modus Ponens 11/28/2015 I am Mila. If I am Mila, then I am a great swimmer.  I am a great swimmer! p p  q  q Tautology: (p  (p  q))  q Inference Rule: Modus Ponens

Proofs - Modus Tollens 11/28/2015 I am not a great skater. If I am Erik, then I am a great skater.  I am not Erik!  q p  q  p p Tautology: (  q  (p  q))   p Inference Rule: Modus Tollens

Proofs - Addition 11/28/2015 I am not a great skater.  I am not a great skater or I am tall. p  p  q Tautology: p  (p  q) Inference Rule: Addition

Proofs - Simplification 11/28/2015 you are sleepy and I am not a great skater.  you are sleepy. p  q  p Tautology: (p  q)  p Inference Rule: Simplification

Proofs - Disjunctive Syllogism 11/28/2015 I am a great eater or I am a great skater. I am not a great skater.  I am a great eater! p  q  q  p Tautology: ((p  q)   q)  p Inference Rule: Disjunctive Syllogism

Proofs - Hypothetical Syllogism 11/28/2015 If you are an athlete, you are always hungry. If you are always hungry, you have a snickers in your backpack.  If you are an athlete, you have a snickers in your backpack. p  q q  r  p  r Tautology: ((p  q)  (q  r))  (p  r) Inference Rule: Hypothetical Syllogism

Proofs - A little quiz… 11/28/2015 Amy is a computer science major.  Amy is a math major or a computer science major. Addition If Ernie is a math major then Ernie is geeky. Ernie is not geeky!  Ernie is not a math major. Modus Tollens

Proofs - A little proof… 11/28/2015 Here ’ s what you know: Ellen is a math major or a CS major. If Ellen does not like discrete math, she is not a CS major. If Ellen likes discrete math, she is smart. Ellen is not a math major. Can you conclude Ellen is smart? M  C  D   C D  S  M

Proofs - A little proof… 11/28/ M  CGiven 2.  D   CGiven 3. D  SGiven 4.  MGiven 5. CDS (1,4) 6. DMT (2,5) 7. SMP (3,6) Ellen is smart!

Proofs - A little proof…(Another approach) 11/28/ M  CGiven 2.  D   CGiven 3. D  SGiven 4.  MGiven 5. CDisjunctive Syllogism (1,4) 6. C  DContrapositive of 2 7. C  S Hypothetical Syllogism (6,3) 8. SModus Ponens (5,7) Ellen is smart!

Challenge.. Here ’ s what you know: "It is not sunny this afternoon and it is colder than yesterday" "We will go swimming only if it is sunny" "If we do not go swimming, then we will take a canoe trip" "If we take a canoe trip, then we will be home by sunset“ Can you conclude "We will be home by sunset." p be the proposition "It is sunny this afternoon," q the proposition "It is colder than yesterday," r the proposition "We will go swimming," s the proposition "We will take a canoe trip," and t the proposition "We will be home by sunset."  p  qr  p  r  ss  tt

Challenge.. Here’s what you know: "If you send me an message, then I will finish writing the program," "If you do not send me an message, then I will go to sleep early," "If I go to sleep early, then I will wake up feeling refreshed" Can you conclude "If I do not finish writing the program, then I will wake up feeling refreshed." p be the proposition "You send me an message," q the proposition "I will finish writing the program," r the proposition "I will go to sleep early," and s the proposition "I will wake up feeling refreshed." p  qp  q  p  rr  s  q  s

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