Universal Conditional Statements s2

CmSc 175 Discrete Mathematics Study Guide for Unit Exam 2Predicate Logic, Valid and Invalid Arguments in Predicate Logic Proofs, Mathematical Induction, Set, Boolean Algebra1. Be able to translate universally and existentially quantified statements in predicate logic and find their negation2. Be able to recognize valid and invalid arguments in predicate logic, determine the inference rule applied and the types of errors.3. Know how to prove statements using direct proofs and mathematical induction.4. Know set identities5. Know the formal definitions of set operations: union, intersection, difference and complement.6. Be able to compute union, complement, intersection of discrete sets and of sets of real numbers7. Know how to simplify set expressions and to show that two expressions are equivalent using set identities and the formal definitions of set operations.8. Know how to find the powerset of a given set, the Cartesian product of two sets, and to determine if a given set of subsets of some set is a partition or not.9. Know the axioms of Boolean Algebra 10. Be able to prove statements in Boolean algebra, e.g. x + x = xSample problems with solutions are given below. See also problems in homework assignments and quizzes1 1. Be able to translate universally and existentially quantified statements in predicate logic and find their negationMen are mortal x, man(x)  mortal(x) Negation: x, man(x)  ~mortal(x) Some men are immortalSome cheaters sit at the back row x, cheater(x)  sit_in_the_back_row (x) Negation: x, cheater(x)  sit_in_the_back_row (x) (x) No cheaters sit in the back rowNegations of quantified statements Babies cry. Some babies don’t cryNo textbook is cheap Some textbooks are cheapSome students study in the library No students study in the librarySome students don’t take introductory classes All students take introductory classes 2. Be able to recognize valid and invalid arguments in predicate logic, determine the inference rule applied and the types of errors.All honest people pay their taxes. Darth is not honest Therefore Darth does not pay his taxes Invalid, Inverse errorSenior students must take a capstone class Anna is a senior student Therefore Anna must take a capstone class. Valid, Modus PonensAll teachers occasionally make mistakes. No gods make mistakes Therefore no teachers are gods Valid Hypothetical Syllogism. Explanation: x, teacher(x)  make_mistakes(x) (1) x, god(x)  ~make_mistakes(x) (2) x, make_mistakes (x)  ~ god (x) (2’ contrapositive of (2))2 Now combine (1) and (2’) with HS, we get: x, teacher(x)  ~ god (x)All teachers occasionally make mistakes. Peter makes mistakes Therefore Peter is a teacher Invalid converse errorAll teachers occasionally make mistakes. Peter does not make mistakes Therefore Peter is not a teacher Valid Modus Tollens3. Know how to prove statements using direct proofs and mathematical induction.3.1. Prove that the sum of two odd numbers is an even number. Proof: Let x and y be two odd numbers. By the definition of odd numbers, there is some p integer such that x = 2p -1, and there is some q integer such that y = 2q – 1. x + y = 2p – 1 + 2q – 1 = 2p + 2q – 2 = 2(p + q – 1) Let s = p + q – 1 Then, x + y = 2s x + y has the representation of an even number, therefore it is even.3.2. Let S(n) = 1 + 2 + 3 + 4 + …. + n, n = 1, 2, 3, …. Prove by mathematical induction the statement P(n): S(n) = n(n+1)/2 for all integers n  1 a. Base step Show that P(1) is true. P(1): S(1) = 1*(1 + 1)/2 = 1 By the definition of S(n), S(1) = 1 Therefore P(1) is true b. Inductive step Show that P(k) P(k+1) is true P(k): S(k) = k(k+1)/2 P(k+1): S(k+1) = (k+1)(k+2)/2Assume that P(k) is true. We have to show that P(k+1) is also true. S(k+1) = S(k) + (k+1) = k(k+1)/2 + (k+1) = = k(k+1)/2 + 2(k+1)/23 = ((k+1)(k+2))/2 Therefore P(k+1) is true, therefore P(k) P(k+1) is true. By the principle of math induction, P(n): S(n) = n(n+1)/2 is true for all integers n  14. Know set identitiesSee Lesson 115. Know the formal definitions of set operations: union, intersection, difference and complement.A  B = {x | (x є A )  (x є B)}A  B = {x | (x є A ) V (x є B)}A – B = {x | (x є A ) V (x  B)}~A = {x | (x  A ) }6. Be able to compute union, complement, intersection of discrete sets and of sets of real numbersLet A = {x | -1 < x < 6 V 10  x 15} B = {x | 4 < x  12}A  B = { x | 4 < x < 6 V 10  x  12}A  B = { x | -1 < x 15 }~A = { x | x  -1 V 6  x < 10 V 15 < x} ~B = { x | x  4 V 12 < x}~A  ~B = {x | x  -1 V 15 < x} ~A  ~B = {x | x  4 V 6  x < 10 V 12 < x} 7. Know how to simplify set expressions and to show that two expressions are equivalent using set identities and the formal definitions of set operations.Show that (A  B)  B = B 4 a. Proof using set identities(A  B)  B = (A  B)  ( B  U) by identity law = (A  B)  ( B  (A  ~A)) by complementation law = (A  B)  (B  A)  (B  ~A) by distributive law = (B  A)  (B  A)  (B  ~A) by commutative laws = (B  A)  (B  ~A) by idempotent laws (i.e. A  A = A) = B  (A  ~A) by distributive law = B  U by complementation law = B by identity law b. Proof using the formal definitions(A  B)  B = {x | ((x є A )  (x є B)) V (x є B)} = = {x | (x є B)} by the absorption laws in propositional logic (P V (P Λ Q) ≡ P Here P is x є B, Q is x є A8. Know how to find the power set of a given set, the Cartesian product of two sets, and to determine if a given set of subsets of some set is a partition or not. a. Power set A = {1,2} P (A) = {, {1}, {2}, {1,2}} Note: The power set of a set with n elements has 2n elementsA =  P (A) = {} A = {P (A) = {, {}, {}, {, } } b. Cartesian products A = {1,2}, B = {a, b, c}, C = {3} A x B = {(1,a), (1,b), (1,c), (2,a), (2,b), (2,c)} B x A = {(a,1), (a,2), (b,1), (b,2), (c,1), (c,2)} A x C = {(1,3), (2,3)} The number of pairs in the Cartesian product is the product of the number of elements of each set. c. PartitionsA = {1,2,3,4,5}, Which of the following sets are not partitions? Explain{{1,2},{3,5},{4}} Partition {{1,2,3},{4,5}} Partition {1,2},{4,5}} Not a partition, element 3 is missing5 {{1,2},{2,3,4,5}} Not a partition, element 2 is repeated, i.e. the sets {1,2} and {2,3,4,5} are not disjoint9. Know the axioms of Boolean Algebra See Lesson 1310. Be able to prove statements in Boolean algebra, e.g. x + x = xProve x = x + x x = x + 0 by identity axiom = x + x . ~x by complements axiom = (x + x) . (x + ~x) by distributive axiom = (x + x) . 1 by complements axiom = x + x by identity axiom6

Universal Conditional Statements s2

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