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Chapter 1: (Part 2): The Foundations: Logic and Proofs

Chapter 1: (Part 2): The Foundations: Logic and Proofs. Propositional Equivalence (Section 1.2) Predicates & Quantifiers (Section 1.3). Propositional Equivalences (1.2). A tautology is a proposition which is always true . Classic Example: P V  P

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Chapter 1: (Part 2): The Foundations: Logic and Proofs

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  1. Chapter 1: (Part 2): The Foundations: Logic and Proofs Propositional Equivalence (Section 1.2) Predicates & Quantifiers (Section 1.3)

  2. Propositional Equivalences (1.2) • A tautology is a proposition which is always true . Classic Example: P V P • A contradiction is a proposition which is always false . Classic Example: P P • A contingency is a proposition which neither a tautology nor a contradiction. Example: (P V Q)  R CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  3. Propositional Equivalences (1.2) (cont.) • Two propositions P and Q are logically equivalent if P  Q is a tautology. We write: P  Q CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  4. Propositional Equivalences (1.2) (cont.) • Example: (P  Q)  (Q  P)  (P  Q) • Proof: • The left side and the right side must have the same truth values independent of the truth value of the component propositions. • To show a proposition is not a tautology: use an abbreviated truth table • try to find a counter example or to disprove the assertion. • search for a case where the proposition is false CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  5. Propositional Equivalences (1.2) (cont.) • Case 1:Try left side false, right side true Left side false: only one of PQ or Q P need be false. 1a. Assume PQ = F. Then P = T , Q = F. But then right side PQ = F. Wrong guess. 1b. Try Q P = F. Then Q = T, P = F. Then PQ = F. Another wrong guess. CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  6. Propositional Equivalences (1.2) • Case 2.Try left side true, right side false If right side is false, P and Q cannot have the same truth value. 2a. Assume P =T, Q = F. Then PQ = F and the conjunction must be false so the left side cannot be true in this case. Another wrong guess. 2b. Assume Q = T, P = F. Again the left side cannot be true. We have exhausted all possibilities and not found a counterexample. The two propositions must be logically equivalent. Note: Because of this equivalence, if and only if or iff is also stated as is a necessary and sufficient condition for. CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  7. CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  8. Note: equivalent expressions can always be substituted for each other in a more complex expression - useful for simplification. CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  9. Propositional Equivalences (1.2) (cont.) • Normal or Canonical Forms • Unique representations of a proposition • Examples: Construct a simple proposition of two variables which is true only when • P is true and Q is false: P Q • P is true and Q is true: P  Q • P is true and Q is false or P is true and Q is true:(P Q) V (P  Q) CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  10. Propositional Equivalences (1.2) (cont.) • A disjunction of conjunctions where • every variable or its negation is represented once in each conjunction (a minterm) • each minterms appears only once Disjunctive Normal Form (DNF) • Important in switching theory, simplification in the design of circuits. CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  11. Propositional Equivalences (1.2) (cont.) • Method: To find the minterms of the DNF. • Use the rows of the truth table where the proposition is 1 or True • If a zero appears under a variable, use the negation of the propositional variable in the minterm • If a one appears, use the propositional variable. CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  12. Propositional Equivalences (1.2) (cont.) • Example: Find the DNF of (P V Q) R CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  13. Propositional Equivalences (1.2) (cont.) • There are 5 cases where the proposition is true, hence 5 minterms. Rows 1,2,3, 5 and 7 produce the following disjunction of minterms: (P V Q) R  (P Q  R) V (P Q  R) V (P  Q R) V (P  Q  R) V (P  Q  R) • Note that you get a Conjunctive Normal Form (CNF) if you negate a DNF and use DeMorgan’s Laws. CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  14. Predicates & Quantifiers (1.3) • A generalization of propositions - propositional functions or predicates: propositions which contain variables • Predicates become propositions once every variable is bound- by • assigning it a value from the Universe of Discourse U or • quantifying it CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  15. Predicates & Quantifiers (1.3) (cont.) • Examples: • Let U = Z, the integers = {. . . -2, -1, 0 , 1, 2, 3, . . .} • P(x): x > 0 is the predicate. It has no truth value until the variable x is bound. • Examples of propositions where x is assigned a value: • P(-3) is false, • P(0) is false, • P(3) is true. • The collection of integers for which P(x) is true are the positive integers. CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  16. Predicates & Quantifiers (1.3) (cont.) • P(y) V P(0) is not a proposition. The variable y has not been bound. However, P(3) V P(0) is a proposition which is true. • Let R be the three-variable predicate R(x, y z): x + y = z • Find the truth value of R(2, -1, 5), R(3, 4, 7), R(x, 3, z) CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  17. Predicates & Quantifiers (1.3) (cont.) • Quantifiers • Universal P(x) is true for every x in the universe of discourse. Notation: universal quantifier x P(x) ‘For all x, P(x)’, ‘For every x, P(x)’ The variable x is bound by the universal quantifier producing a proposition. CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  18. Predicates & Quantifiers (1.3) (cont.) • Example: U = {1, 2, 3} x P(x)  P(1)  P(2)  P(3) CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  19. Predicates & Quantifiers (1.3) (cont.) • Quantifiers (cont.) • Existential • P(x) is true for some x in the universe of discourse. Notation: existential quantifier x P(x) ‘There is an x such that P(x),’‘For some x, P(x)’, ‘For at least one x, P(x)’, ‘I can find an x such that P(x).’ Example: U={1,2,3} x P(x)  P(1) V P(2) V P(3) CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  20. Predicates & Quantifiers (1.3) (cont.) • Quantifiers (cont.) • Unique Existential P(x) is true for one and only one x in the universe of discourse. Notation: unique existential quantifier !x P(x) ‘There is a unique x such that P(x),’‘There is one and only one x such that P(x),’‘One can find only one x such that P(x).’ CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  21. Predicates & Quantifiers (1.3) (cont.) • Example: U = {1, 2, 3, 4} How many minterms are in the DNF? CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  22. Predicates & Quantifiers (1.3) (cont.) REMEMBER! A predicate is not a proposition until all variables have been bound either by quantification or assignment of a value! CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  23. Predicates & Quantifiers (1.3) (cont.) • Equivalences involving the negation operator (x P(x ))  x P(x) (x P(x))  x P(x) • Distributing a negation operator across a quantifier changes a universal to an existential and vice versa. • (x P(x))  (P(x1)  P(x2)  …  P(xn)) P(x1) V P(x2) V … V P(xn) x P(x) CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  24. Predicates & Quantifiers (1.3) (cont.) • Multiple Quantifiers: read left to right . . . • Example: Let U = R, the real numbers, P(x,y): xy= 0 x y P(x, y) x y P(x, y) x y P(x, y) x y P(x, y) The only one that is false is the first one. What’s about the case when P(x,y) is the predicate x/y=1? CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  25. Predicates & Quantifiers (1.3) (cont.) • Multiple Quantifiers: read left to right . . . • Example: Let U = {1,2,3}. Find an expression equivalent to x y P(x, y) where the variables are bound by substitution instead: Expand from inside out or outside in. Outside in: y P(1, y) y P(2, y) y P(3, y) [P(1,1) V P(1,2) V P(1,3)]  [P(2,1) V P(2,2) V P(2,3)]  [P(3,1) V P(3,2) V P(3,3)] CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  26. Predicates & Quantifiers (1.3) (cont.) • Converting from English (Can be very difficult!) “Every student in this class has studied calculus”transformed into:“For every student in this class, that student has studied calculus” C(x): “x has studied calculus” x C(x) This is one way of converting from English! CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  27. Predicates & Quantifiers (1.3) (cont.) • Multiple Quantifiers: read left to right . . . (cont.) • Example: F(x): x is a fleegle S(x): x is a snurd T(x): x is a thingamabob U={fleegles, snurds, thingamabobs} (Note: the equivalent form using the existential quantifier is also given) CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  28. Predicates & Quantifiers (1.3) (cont.) • Everything is a fleegle x F( x)   (x F(x)) • Nothing is a snurd. x  S(x)   (x S( x)) • All fleegles are snurds. x [F(x)S(x)]  x [F(x) V S(x)]  x  [F(x)  S(x)]   (x [F(x) V S(x)]) CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  29. Predicates & Quantifiers (1.3) (cont.) • Some fleegles are thingamabobs. x [F(x)  T(x)] (x [F(x) V T(x)]) • No snurd is a thingamabob. x [S(x) T(x)]  (x [S(x )  T(x)]) • If any fleegle is a snurd then it's also a thingamabob x [(F(x)  S(x))  T(x)]  (x [F(x)  S(x)  T( x)]) CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  30. Predicates & Quantifiers (1.3) (cont.) • Extra Definitions: • An assertion involving predicates is valid if it is true for every universe of discourse. • An assertion involving predicates is satisfiable if there is a universe and an interpretation for which the assertion is true. Else it is unsatisfiable. • The scope of a quantifier is the part of an assertion in which variables are bound by the quantifier CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  31. Predicates & Quantifiers (1.3) (cont.) • Examples: Valid: x S(x) [x S( x)] Not valid but satisfiable: x [F(x)  T(x)] Not satisfiable: x [F(x)  F(x)] Scope: x [F(x) V S( x)] vs. x [F(x)] V x [S(x)] CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  32. Predicates & Quantifiers (1.3) (cont.) • Dangerous situations: • Commutativity of quantifiers x y P(x, y) y x P( x, y)? YES! x y P(x, y)  y x P(x, y)? NO! DIFFERENT MEANING! • Distributivity of quantifiers over operators x [P(x)  Q(x)] x P( x) x Q( x)? YES! x [P( x)  Q( x)] [x P(x)  x Q( x)]? NO! CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  33. Sets (1.6) • A set is a collection or group of objects or elements or members. (Cantor 1895) • A set is said to contain its elements. • There must be an underlying universal set U, either specifically stated or understood. CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  34. Sets (1.6) (cont.) • Notation: • list the elements between braces: S = {a, b, c, d}={b, c, a, d, d} (Note: listing an object more than once does not change the set. Ordering means nothing.) • specification by predicates: S= {x| P(x)}, S contains all the elements from U which make the predicate P true. • brace notation with ellipses: S = { . . . , -3, -2, -1}, the negative integers. CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  35. Sets (1.6) (cont.) • Common Universal Sets • R = reals • N = natural numbers = {0,1, 2, 3, . . . }, the counting numbers • Z = all integers = {. . , -3, -2, -1, 0, 1, 2, 3, 4, . .} • Z+ is the set of positive integers • Notation: x is a member of S or x is an element of S: x  S. x is not an element of S: x  S. CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  36. Sets (1.6) (cont.) • Subsets • Definition: The set A is a subset of the set B, denoted A  B, iff x [x  A  x  B] • Definition: The void set, the null set, the empty set, denoted , is the set with no members. Note: the assertion x  is always false. Hence x [x  x  B] is always true(vacuously). Therefore,  is a subset of every set. Note: A set B is always a subset of itself. CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  37. Sets (1.6) (cont.) • Definition: If A  B but A  B the we say A is a proper subset of B, denoted A  B (in some texts). • Definition: The set of all subset of a set A, denoted P(A), is called the power set of A. • Example: If A = {a, b} then P(A) = {, {a}, {b}, {a,b}} CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  38. Sets (1.6) (cont.) • Definition: The number of (distinct) elements in A, denoted |A|, is called the cardinality of A. If the cardinality is a natural number (in N), then the set is called finite, else infinite. • Example: A = {a, b}, |{a, b}| = 2, |P({a, b})| = 4. A is finite and so is P(A). Useful Fact: |A|=n implies |P(A)| = 2n CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  39. Sets (1.6) (cont.) • N is infinite since |N| is not a natural number. It is called a transfinite cardinal number. • Note: Sets can be both members and subsets of other sets. • Example: A = {,{}}. A has two elements and hence four subsets: , {}, {{}}. {,{}} Note that  is both a member of A and a subset of A! • Russell's paradox: Let S be the set of all sets which are not members of themselves. Is S a member of itself? • Another paradox: Henry is a barber who shaves all people who do not shave themselves. Does Henry shave himself? CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

  40. Sets (1.6) (cont.) • Definition: The Cartesian product of A with B, denoted A x B, is the set of ordered pairs {<a, b> | a  A  b  B} Notation: Note: The Cartesian product of anything with  is . (why?) • Example: A = {a,b}, B = {1, 2, 3} AxB = {<a, 1>, <a, 2>, <a, 3>, <b, 1>, <b, 2>, <b, 3>} What is BxA? AxBxA? • If |A| = m and |B| = n, what is |AxB|? CS 210, Ch.1 (part 2): The foundations: Logic & Proof, Sets, and Functions

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