数学381 - 中考1

1) Prove that premises[ p n q v r] and [r->s] together imply conclusion [p v s] 2) Conclude from "Jasmine is skiing or it isn't snowing" and "Bart is playing hockey or it is snowing" that either J is skiing or B is playing hockey 3) Show that (p ∧ q) → (p ∨ q) is a tautology using successive substitution

1) Premises: i. (pnq)vr 三 (pvr)n(qvr) ← distributive rule ii. r->s 三 !r v s ← * Steps: i. p n qvr <-- premise, so assume true ii. (pvr)n(qvr) <-- logically equivalent with i., so true iii. pvr <-- simplification from ii., bc both must be true! iv. r->s <-- premise so true v. !rvs <-- logically equivalent with iv., so true --> can be p OR r, sub p for r vi. pvs <-- resolution syllogism with iii. and v. 2) p= J skiing; !s=isn't snowing; q=B playing hockey; s=snowing Steps: i. p v !s ii. q v s iii. p v q <-- resolution! 3) i. (p ∧ q) → (p ∨ q) ≡ ¬(p ∧ q) ∨ (p ∨ q) --> by Example 3, table 7.1 ii. ≡ (¬p ∨ ¬q) ∨ (p ∨ q) --> by the first De Morgan law (换 n-->V); distr ! iii. ≡ (¬p ∨ p) ∨ (¬q ∨ q) --> by the associative and commutative laws for disjunction (就是grouping嘿嘿) iv. ≡ T ∨ T --> by Example 1 and the commutative law for disjunction *v. ≡ T* --> T ∨ T = T

1) "It is below freezing." Therefore it is either below freezing or raining. Which inference rule is this? 2) Show that from the premises Npnq, r---> p, Nr--> s, and s--> t, we can conclude t. 3) Prove that formal argument having premises: p-->q, Np-->r, r-->s and conclusion Nq-->s is valid

1) Addition rule 2) Prove valid argument (applying inference rules; premises all assumed to be true): Npnq r---> p Nr--> s s--> t ==> t Steps - using inference rules (8): i. Assume premise 1 is true --> Npnq ii. Apply inference rule - simplification of (i.) --> Np iii. Assume premise 2 is true --> r--> p iv. Use Modus Tallons on (ii.) and (iii.)! --> Nr v. Assume premise 3 is true --> Nr--> s vi. Modus Poneas on iv. and v. --> s vii. Assume premise 4 is true --> s--> t viii. Modus Poneas on vi. and vii. --> t! 3) p-->q Np-->r r-->s Conclusion Nq-->s ASSUME ALL PREMISES ARE TRUE: i. Contrapositive of p-->q --> Nq-->Np ii. Np-->r iii. Hypothetical Syllogism of i. and ii. --> Nq-->r iv. r-->s premise is true vi. Hypothetical syllogism of iii. and iv. --> *Nq-->s*

1) Find contrapositive, converse, and inverse of conditional statement: The home team wins whenever it's raining 2) Define (symbols+how to use in sentence) following logical connectives: i. Conjunction ii. Disjunction iii. Exclusive/or iv. Implication v. Biconditional Operator vi. Negation 3) Translate english sentence into logical expression: you can access the internet from campus only if you are a compsci major or you are NOT a freshman 4) a) Prove that pv(qnr) and (pvq)n(pvr) are logically equivalent. This is distributive law of disjunction over conjunction b) Prove that NOT(p-->q) and p n NOTq are logically equivalent WITHOUT using truth table! 5) What are the 10 logical equivalence laws?

1) Conditional: (p)If raining --> (q)home team wins Contrapositive (反+不）: If the home team does NOT win, then it is NOT raining --> 唯有此等于原句！ Converse: If the home team wins, then it is raining Inverse: If it's not raining, than the home team won't win 2) i. Conjunction (and - in CONJUNCTION with=and) ± = 加减 ii. Disjunction (or) */ = 成除 iii. Exclusive/or (only one=T) iv. Implication (if --> then) v. Biconditional Operator (p if and ONLY if q) --> p truth value必=q truth value vi. Negation (NOTp, NOTq, 云云) 3) a=internet from campus b=compsci major c=freshman *a--> (b v (NOTc))* 4) a) Construct truth table for these compound propositions to prove pv(qnr)=(pvq)n(pvr)--> 使用distributive property而得相等滴: --> n=and=±=BOTH must=T, 否则=F！ --> v=or=*/=至少一个为T便等于T！ P: TTTTFFFF Q: TTFFTTFF R: TFTFTFTF qnr: TFFFTFFF pv(qnr): TTTTTFFF pvq: TTTTTTFF pvr: TTTTTFTF (pvq)n(pvr): TTTTTFFF *《因为俩公式是一样的，证明它们有一样的真理数！》* b) NOT(p-->q) = p n NOTq i. NOT(p-->q) 三 NOT(NOTpvq) --> conditional-disjunction equivalence ii. NOT(NOTpvq)三NOT((NOTp) n NOTq) --> 2nd de morgan law --> NOT(p conj q)=NOTp disj NOTq --> NOT(pnq)=NOTp v NOTq, 还有反过来 iii. 三*p n NOTq* --> double negation law = NOT(NOTp) = p! 5) i. Identity law - --> p n T 三 p --> p v F 三 p ii. Domination law - --> p v T 三 T --> p n F 三 F iii. Idempotent law - --> p v p 三 p --> p n p 三 p iv. Double negation law - --> NOT(NOTp)三q v p v. Commutative law - --> p v q 三 q v p --> p n q 三 q n p vi. Associative law - --> (pvq)vr 三 pv(qvr) --> (pnq)nr 三pn(qnr) vii. Distributive law (就是基本数学哈哈) - --> pv(qnr)三(pvq)n(pvr) --> pn(qvr)三(pnq)v(pnr) viii. De Morgan's law - --> NOT(pnq)三NOTp v NOTq --> NOT(pvq)三NOTp n NOTq viv. Absorption law - --> pv(pnq)三p --> pn(pvq)三p vv. Negation law - --> pvNOTp三T --> pnNOTp三F

1) Prove 2^0.5 is irrational. 2) What is contingency?

1) Key proof to understand: Prove 2^0.5 is irrational. Proof: assume it's false, and find a contradiction. (assume !p; showed r n !r was true - r=certain pair integers=/=even = contradiction) 记住：Recall class lemma: rational number r for integers p,q, r=p/q where p and q cannot BOTH be even. i. We assume 2^0.5 is rational, with integers p, q such that 2^0.5=p/q ii. p/q=2^0.5 = p^2/q^2=2 --> p^2=2q^2 --> Given q is integer, p^2 is even which means p must be even (lemma: x^2 is odd=x is odd, and vice versa) iii. Thus, p^2=2k, where k is some integer --> (2k)^2=2q^2 = 4k^2=2q^2 = q^2=2k^2 iv. Given k is integer, q^2 is even which means q must be even...but p and q cannot BOTH be even, as this is contradiction of rational number lemma. Thus, by assuming that 2^0.5 is rational, we run into the contradiction that p and q are BOTH even, which means 2^0.5 MUST be irrational! 口 By contradiction, we assume that 2^0.5 is actually rational. Using lemma, we may write 2^0.5=p/q where p and q are nonzero integers which are NOT both even Squaring both sides, we get that 2=(p^2/q^2) OR p^2=2q^2 This means p^2 HAS to be even REMEMBER: we proved that n odd → n^2 is odd (01/29 Wed) So, we conlcude that p itself MUST be even since p^2=even p=2k (even) for some keZ (for some integer(Z) k) Substituting back, we get tat 2q^2=p^2 = (2k)^2 = 4k^2 Divide both sides by 2: q^2=2k^2 Same argument as above: q^2 must be even, so q must be even! THIS CONTRADICTS definition of p and q → found a contradiction so our proof is concluded. This shows that 2^0.5=irrational canNOT be false. 口 2) Contingency = 不总是/不总非之推论（proposition）

1) What does the statement ∀xN(x) mean if N(x) is "Computer x is connected to the network" and the domain consists of all computers on campus? 2) What is the truth value of ∀x(x^2 ≥ x) if the domain consists of all real numbers? b) What is the truth value of this statement if the domain consists of all integers? 3) Let Q(x) denote the statement "x = x + 1." What is the truth value of the quantification ∃xQ(x), where the domain consists of all real numbers? 4) What are the negations of the statements (a) "There is an honest politician" and (b) "All Americans eat cheeseburgers"? Write out the quantification of 是与非 5) Show that ¬∀x(P(x) → Q(x)) and ∃x(P(x) ∧ ¬Q(x)) are logically equivalent. 6) "For every person x, if person x is a student in this class, then x has studied calculus" --> express in logical expression 7) "There is a person x having the properties that x is a student in this class and x has visited Mexico." --> express in logical expression

1) The statement ∀xN(x) = for every computer x on campus, that computer x is connected to the network. This statement can be expressed in English as "Every computer on campus is connected to the network." --> V = every x IS 2) Truth value = T a) For every real number x, x^2 >= x --> FALSE! Real numbers=include decimals; 0.5^2 < 0.5 b) For every integer x, x^2 >= x --> TRUE! Integers=没有decimals, 所以真理 3) FALSE! Q(x) is false for every real number x --> existential quantification of Q(x)=∃xQ(x) is FALSE! 4) a) H(x) = x is honest There is an honest politician = there EXISTS honest politician in the domain=all politicians, such that H(x) or x is honest --> *∃xH(x)* Negation? There does NOT exist even ONE honest politician in the domain=all politicians, such that --> ALL POLITICIANS are dishonest! --> *Negation = Vx!(H(x)* --> for ALL x, !H(x)--> H(x) is false = there are no honest politicians --> Disprove there exists one x=T within D, have to prove that all x=F within D! --> 此更难 b) C(x) = eat cheeseburgers ALL eat hamburgers = for all x, C(x) is true OR for all x within domain=all amerikkkans, they eat cheeseburgers C(x) --> *VxC(x)* Negation? There exists ONE amerikkan who does NOT eat cheeseburger, thus negating statement that ALL eat cheeseburgers --> *∃x!C(x)* --> Disprove that all x=T within D, just have to prove that ONE x=F within D! --》轻而易举似的 5) 【Show that ¬∀x(P(x) → Q(x)) and ∃x(P(x) ∧ ¬Q(x)) are logically equivalent.】 Proof: i. !Vx(P(x)-->Q(x)) ii. ∃x!(!P(x) v Q(x)) - table 7 equivalence; demorgans iii. ∃x(P(x) n !Q(x)) - demorgan's 完成！口 6) S(x) = student in class C(x) = calculus Vx = for every person x =*Vx(S(x) → C(x))* 7) ∃x=person x having properties/exists person x with element... S(x)=student in class M(x)=visited MX n=AND =*∃x (S(x) n M(x))*

1) Suppose r = (p v (NOTq)) → (p ∧ q). Calculate truth table for r. 2) Determine whether each of the compound propositions (p ∨ ¬q) ∧ (q ∨ ¬r) ∧(r ∨ ¬p), (p ∨ q ∨ r) ∧ (¬p ∨ ¬q ∨ ¬r), and (p ∨ ¬q) ∧ (q ∨ ¬r) ∧ (r ∨ ¬p) ∧ (p ∨ q ∨ r) ∧ (¬p ∨ ¬q ∨ ¬r) is satisfiable.

1) Truth Table (P, Q, all possible combos of T/F): i. P: TTFF ii. Q: TFTF iii. NOTq: FTFT iv. p ∧ q: TFFF --> 俩都一样=T v. (p v (NOTq)): TTFT --> [等于(P true OR Q false OR both the same! (but canNOT be p=F+q=T!))] -->Disjunction (OR) of first three columns vi. R: TFTF --> Conjunction (AND) of [p ∧ q] 与 [(p v (NOTq))] --> Analyze (iv) and (v) ⇒ 一样=T！ 2) Instead of using a truth table to solve this problem, we will reason about truth values. Note that (p ∨ ¬q) ∧ (q ∨ ¬r) ∧ (r ∨ ¬p) is true when the three variables p, q, and r have the same truth value (see Exercise 42 of Section 1.1). Hence, it is satisfiable as there is at least one assignment of truth values for p, q, and r that makes it true. - Similarly, note that (p ∨ q ∨ r) ∧ (¬p ∨ ¬q ∨ ¬r) is true when at least one of p, q, and r is true and at least one is false (see Exercise 43 of Section 1.1). - Hence, (p ∨ q ∨ r) ∧ (¬p ∨ ¬q ∨ ¬r) is satisfiable, as there is at least one assignment of truth values for p, q, and r that makes it true. - Finally, note that for (p ∨ ¬q) ∧ (q ∨ ¬r) ∧ (r ∨ ¬p) ∧ (p ∨ q ∨ r) ∧ (¬p ∨ ¬q ∨ ¬r) to be true, (p ∨ ¬q) ∧ (q ∨ ¬r) ∧ (r ∨ ¬p) and (p ∨ q ∨ r) ∧ (¬p ∨ ¬q ∨ ¬r) must both be true. For the first to be true, the three variables must have the same truth values, and for the second to be true, at least one of the three variables must be true and at least one must be false. However, these conditions are contradictory. From these observations we conclude that no assignment of truth values to p, q, and r makes (p ∨ ¬q) ∧ (q ∨ ¬r) ∧ (r ∨ ¬p) ∧ (p ∨ q ∨ r) ∧ (¬p ∨ ¬q ∨ ¬r) true. Hence, it is unsatisfiable.

作业1 - 第二部: (1.3) 1ace, 3a, 4a, 5-6, 7ac, 8ac, 17a, 18-19, 24-25 (using proof by truth table), 35-36 1. Use truth tables to verify equivalences: a) pnT三p c) pnF三F e) pvp三p 3. Use truth tables to verify the commutative laws a) p ∨ q ≡ q ∨ p. 4. Use truth tables to verify the associative laws a) (p ∨ q) ∨ r ≡ p ∨ (q ∨ r). 5. Use a truth table to verify the distributive law p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r). 6. Use a truth table to verify the first De Morgan law ¬(p ∧ q) ≡ ¬p ∨ ¬q. 7. Use De Morgan's laws to find the negation of each of the following statements. a) Jan is rich and happy. c) Mei walks or takes the bus to class. 8. Use De Morgan's laws to find the negation of each of the following statements. a) Kwame will take a job in industry or go to graduate school. c) James is young and strong. 18. Determine whether (¬p ∧ (p → q)) → ¬q is a tautology. 19. Determine whether (¬q ∧ (p → q)) → ¬p is a tautology. Show that two compound propositions are logically equivalent. To do this, either show that both sides are true, or that both sides are false, for exactly the same combinations of truth values of the propositional variables in these expressions (whichever is easier). --> USE TRUTH TABLES 24. Show that ¬(p ⊕ q) and p ↔ q are logically equivalent. 25. Show that ¬(p ↔ q) and ¬p ↔ q are logically equivalent. 35. Show that (p → q) → r and p → (q → r) are not logically equivalent. 36. Show that (p ∧ q) → r and (p → r) ∧ (q → r) are not logically equivalent.

1) a) pnT三p --> p n T 三 p 证明: --Truth Table (i. identity law): i. p: TF ii. T: TT iii. *PnT: TF! = p: TF* [--> p v F 三 p --Truth Table: i. p: TF ii. F: FF iii. *PvF: T! = p: TF*] c) pnF三F --Truth Table (domination law): i. p: TF ii. F: FF iii. *PnF: FF! =/= p:TF* --> 必须俩皆是T!!!!!! e) pvp三p --Truth Table (domination law): i. p: TF ii. P: FT iii. *PvP: TT!* 3) a) p ∨ q ≡ q ∨ p --p ∨ q: i. p: TTFF ii. q: TFTF iii. pvq: TTTF --q ∨ p: i. p: TTFF ii. q: TFTF iii. qvp: TTTF *--> 相等！* 4) (p ∨ q) ∨ r ≡ p ∨ (q ∨ r) --(p ∨ q) ∨ r: i. p: TTTTFFFF ii. q: TFTFFFTT iii. r: TTFFFTFT iv. pvq: TTTTFFTT v. *(pvq)vr: TTTTFTTT* --p ∨ (q ∨ r): i. p: TTTTFFFF ii. q: TFTFFFTT iii. r: TTFFFTFT iv. qvr: TTTFFTTT v. *pv(qvr): TTTTFTTT* 5) p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r) (distributive law) - n=俩都是T - v=至少一个T便行 --p ∧ (q ∨ r): i. p: TTTTFFFF ii. q: TFTFFFTT iii. r: TTFFFTFT iv. qvr: TTTFFTTT v. *pn(qvr): TTTFFFFF* --(p ∧ q) ∨ (p ∧ r): i. p: TTTTFFFF ii. q: TFTFFFTT iii. r: TTFFFTFT iv. pnq: TFTFFFFF v. pnr : TTFFFFFF vi. *(pnq)v(pnr): TTTFFFFF, equal to each other=proves equivalence* --> 相等！ 6) ¬(p ∧ q) ≡ ¬p ∨ ¬q. (de morgan) --¬(p ∧ q): i. p: TTFF ii. q: TFTF iii. pnq: TFFF iv. *!(pnq): FTTT* --¬p ∨ ¬q: i. !p: FFTT (反着来) ii. !q: FTFT iii. *!pv!q: FTTT, equal to each other=proves equivalence* --> 相等！ 7. Use De Morgan's laws to find the negation of each of the following statements. a) Jan is rich and happy. --> (i) NOT(p n q) 三 NOTp v NOTq --> *Jan is NOT rich or NOT happy!* c) Mei walks or takes the bus to class. --> (ii) NOT(p v q) 三 NOTp n NOTq --> *妹子哈哈哈 Mei does not walk to class and does not take the bus to class.* 8. Use De Morgan's laws to find the negation of each of the following statements. a) Kwame will take a job in industry or go to graduate school. --> (ii) NOT(p v q) 三 NOTp n NOTq --> *Kwame will NOT take a job in industry and NOT go to graduate* school. c) James is young and strong. --> (i) NOT(p n q) 三 NOTp v NOTq --> *James is NOT young or NOT strong.* 18. Determine whether (¬p ∧ (p → q)) → ¬q is a tautology. --> If p=false and p-->q=true, then q=false to satisfy --> Modus Poneas rule = (p n (p-->q)) --> q --> *是 It is tautology, always true!* Truth Table: p: TTFF !p: FFTT q: TFTF !q: FTFT p->q: TFTT ¬p∧(p → q): FFTT *¬p∧(p → q)->!q: TTFT; NOT a tautology as it contains an F!* 19. Determine whether (¬q ∧ (p → q)) → ¬p is a tautology. --> Modus Tollens: If q=false AND p-->q is true, then p=false --> *YES, this is tautology - ALWAYS TRUE!* Truth Table: p: TTFF !p: FFTT q: TFTF !q: FTFT p->q: TFTT ¬q∧(p→q): FFTF *¬q∧(p→q)→¬p: TTTT!* Show that two compound propositions are logically equivalent. To do this, either show that both sides are true, or that both sides are false, for exactly the same combinations of truth values of the propositional variables in these expressions (whichever is easier). --> USE TRUTH TABLES 24. Show that ¬(p ⊕ q) and p ↔ q are logically equivalent. i. ¬(p ⊕ q) truth table: p: TTFF q: TFTF p ⊕ q: FTTF --> (唯有一个能=T) *¬(p ⊕ q): TFFT* ii. p ↔ q truth table: p: TTFF q: TFTF *p ↔ q: TFFT, the same! proves logical equivalence* --> (p=T IF AND ONLY IF q=T! --> pq必是同样的) 25. Show that ¬(p ↔ q) and ¬p ↔ q are logically equivalent. i. ¬(p ↔ q) truth table: p: TTFF q: TFTF p ↔ q: TFFT --> (pq必是同样的) *¬(p ↔ q): FTTF* ii. ¬p ↔ q truth table: p: TTFF !p: FFTT q: TFTF *¬p ↔ q: FTTF, the same! proves logical equivalence* --> (pq必是同样的) 35. Show that (p → q) → r and p → (q → r) are NOT logically equivalent. --> If p, then q=true, then r is true --> If p=true, then (if q, then r=true) --> = 俩必须是同样地 Make truth table, evaluate conditionals individually: p: TTTTFFFF q: TFTFFFTT r: TTFFFTFT p->q: TFTFTTTT --> =Np v q (只要不是p n Nq就行) *(p → q) → r: TTFTFTFT* p: TTTTFFFF q: TFTFFFTT r: TTFFFTFT q→r: TTFTTTFT *p → (q → r): TTFTTTTT* --> 不相等！ 36. Show that (p ∧ q) → r and (p → r) ∧ (q → r) are not logically equivalent. p: TTTTFFFF q: TFTFFFTT r: TTFFFTFT p n q: TFTFFFFF *(p n q) → r: TTFTTTTT* p: TTTTFFFF q: TFTFFFTT r: TTFFFTFT p → r: TTFFTTTT q → r: TTFTTTFT *(p → r) ∧ (q → r): TTFFTTFT, NOT equal!* --> 不相等！

作业2 (第二部): 1.6: 1-6, 9bc, 10af, 11, 12 1. Find the argument form for the following argument and determine whether it is valid. Can we conclude that the conclusion is true if the premises are true? If Socrates is human, then Socrates is mortal. Socrates is human. ∴ Socrates is mortal. 2. Find the argument form for the following argument and determine whether it is valid. Can we conclude that the conclusion is true if the premises are true? If George does not have eight legs, then he is not a spider. George is a spider. ∴ George has eight legs. 3. What rule of inference is used in each of these arguments? a) Alice is a mathematics major. Therefore, Alice is either a mathematics major or a computer science major. b) Jerry is a mathematics major and a computer science major. Therefore, Jerry is a mathematics major. c) If it is rainy, then the pool will be closed. It is rainy. Therefore, the pool is closed. d) If it snows today, the university will close. The university is not closed today. Therefore, it did not snow today. e) If I go swimming, then I will stay in the sun too long. If I stay in the sun too long, then I will sunburn. Therefore, if I go swimming, then I will sunburn. 4. What rule of inference is used in each of these arguments? a) Kangaroos live in Australia and are marsupials. Therefore, kangaroos are marsupials. b) It is either hotter than 100 degrees today or the pollution is dangerous. It is less than 100 degrees outside today. Therefore, the pollution is dangerous. c) Linda is an excellent swimmer. If Linda is an excellent swimmer, then she can work as a lifeguard. Therefore, Linda can work as a lifeguard. d) Steve will work at a computer company this summer. Therefore, this summer Steve will work at a computer company or he will be a beach bum. e) If I work all night on this homework, then I can answer all the exercises. If I answer all the exercises, I will understand the material. Therefore, if I work all night on this homework, then I will understand the material. 5. Use rules of inference to show that the hypotheses "Randy works hard," "If Randy works hard, then he is a dull boy," and "If Randy is a dull boy, then he will not get the job" imply the conclusion "Randy will not get the job." 6. Use rules of inference to show that the hypotheses "If it does not rain or if it is not foggy, then the sailing race will be held and the lifesaving demonstration will go on," "If the sailing race is held, then the trophy will be awarded," and "The trophy was not awarded" imply the conclusion "It rained." 9. For each of these collections of premises, what relevant conclusion or conclusions can be drawn? Explain the rules of inference used to obtain each conclusion from the premises. b) "If I eat spicy foods, then I have strange dreams." "I have strange dreams if there is thunder while I sleep." "I did not have strange dreams." c) "I am either clever or lucky." "I am not lucky." "If I am lucky, then I will win the lottery." 10. For each of these sets of premises, what relevant conclusion or conclusions can be drawn? Explain the rules of inference used to obtain each conclusion from the premises. a) "If I play hockey, then I am sore the next day." "I use the whirlpool if I am sore." "I did not use the whirlpool." f ) "I am either dreaming or hallucinating." "I am not dreaming." "If I am hallucinating, I see elephants running down the road." 11. Show that the argument form with premises p1, p2,... , pn and conclusion q → r is valid if the argument form with premises p1, p2,... , pn, q, and conclusion r is valid.

1. Find the argument form for the following argument and determine whether it is valid. Can we conclude that the conclusion is true if the premises are true? If Socrates is human, then Socrates is mortal. Socrates is human. ∴ Socrates is mortal. *Modus Poneas!=TRUE* p->q p q 2. Find the argument form for the following argument and determine whether it is valid. Can we conclude that the conclusion is true if the premises are true? If George does not have eight legs, then he is not a spider. George is a spider. ∴ George has eight legs. p=george has 8 legs q=george is a spider !p--> !q q p --> *modus tollens = TRUE!* 3. What rule of inference is used in each of these arguments? a) Alice is a mathematics major. Therefore, Alice is either a mathematics major or a computer science major. p pvq --> *addition rule!* b) Jerry is a mathematics major and a computer science major. Therefore, Jerry is a mathematics major. pnq p --> *simplification rule!* c) If it is rainy, then the pool will be closed. It is rainy. Therefore, the pool is closed. p->q p q --> *modus poneas!* d) If it snows today, the university will close. The university is not closed today. Therefore, it did not snow today. p->q !q !p --> *modus tollens!* e) If I go swimming, then I will stay in the sun too long. If I stay in the sun too long, then I will sunburn. Therefore, if I go swimming, then I will sunburn. p->q q->r p->r --> *hypothetical syllogism* 4. What rule of inference is used in each of these arguments? a) Kangaroos live in Australia and are marsupials. Therefore, kangaroos are marsupials. pnq q --> *simplification rule* b) It is either hotter than 100 degrees today or the pollution is dangerous. It is less than 100 degrees outside today. Therefore, the pollution is dangerous. pvq 《-- EITHER hotter OR pollution !p q --> *Disjunctive syllogism.* c) Linda is an excellent swimmer. If Linda is an excellent swimmer, then she can work as a lifeguard. Therefore, Linda can work as a lifeguard. p p->q q --> *modus poneas!* d) Steve will work at a computer company this summer. Therefore, this summer Steve will work at a computer company or he will be a beach bum. p pvq --> *addition rule* e) If I work all night on this homework, then I can answer all the exercises. If I answer all the exercises, I will understand the material. Therefore, if I work all night on this homework, then I will understand the material. p->q q->r p->r --> *hypothetical syllogism* 5. Use rules of inference to show that the hypotheses "Randy works hard," "If Randy works hard, then he is a dull boy," and "If Randy is a dull boy, then he will not get the job" imply the conclusion "Randy will not get the job." p p->q q->r p->r --> *hypothetical syllogism* r --> *modus poneas rule* ???6. Use rules of inference to show that the hypotheses "If it does not rain(p) or if it is not foggy(q), then the sailing race will be held(r) and the lifesaving demonstration will go on(s)," "If the sailing race is held(r), then the trophy will be awarded(t)," and "The trophy was not awarded(t)" imply the conclusion "It rained(p)." p = rains q = It is foggy r = Sailing race is held s = Lifesaving demonstration will go on t = Trophy is awarded P1: !pv!q -> rns P2: r-->t P3: !t To prove: !p r--> t !t !r = modus tollens !pv!q -> rns !(rns) = disj syllogism pvq = modus tollens p = conjunction --> !r ==> 2->3 Modus Tollens --> !(rns)->!(!pv!q) ==> equivalency of p->q 三 !q->!p --> !rv!s -> pnq ==> demorgan's law --> !rv!s ==> addition --> pnq ==> modus ponens --> p ==》 Thus, it rained! --> Using Modus Tollens, De Morgan's Laws, Addition, Modus Ponens, and simplification 9. For each of these collections of premises, what relevant conclusion or conclusions can be drawn? Explain the rules of inference used to obtain each conclusion from the premises. b) "If I eat spicy foods(p), then I have strange dreams(q)." "I have strange dreams(q) if there is thunder while I sleep(r)." "I did not have strange dreams(!q)." p->q r->q !q --> *!p = modus tollens* --> *!r = modus tollens* c) "I am either clever or lucky." "I am not lucky." "If I am lucky, then I will win the lottery." pvq !q <-- canNOT apply modus tollens just bc hyp=F; !p v q 🉑️) q->r --> *p = disjunc syllogism* 10. For each of these sets of premises, what relevant conclusion or conclusions can be drawn? Explain the rules of inference used to obtain each conclusion from the premises. a) "If I play hockey(p), then I am sore the next day(q)," "I use the whirlpool(s) if I am sore." "I did not use the whirlpool." p->q q->s --> p->s (hyp syllogism) !q --> *!s (modus tollens since !q and !q->!s=T)* f ) "I am either dreaming or hallucinating." "I am not dreaming." "If I am hallucinating, I see elephants running down the road." pvq !p --> q = disjunctive syllogism q->r --> *r = modus tollens (backward)* 11. Show that the argument form with premises p1, p2,... , pn and conclusion q → r is valid if the argument form with premises p1, p2,... , pn, q, and conclusion r is valid.

作业1 - 第一部: (1.1) 1,2,5,6,10-13,16,18-25,29-30 1. Which of these sentences are propositions? What are the truth values of those that are propositions? a) Boston is the capital of Massachusetts. b) Miami is the capital of Florida c) 2 + 3 = 5. d) 5+7 = 10 e) x + 2 = 11. f) Answer this question. 2. Which of these are propositions? What are the truth values of those that are propositions? a) Do not pass go. b) What time is it? c) There are no black flies in Maine. d) 4+x=5 e) The moon is made of green cheese. f) 2^n >= 100 5. What is the negation of each of these propositions? a) Mei has an MP3 player. c) 2 + 1 = 3. 6. What is the negation of each of these propositions? a) Jennifer and Teja are friends. c) Abby sent more than 100 text messages yesterday. 10. Let p and q be the propositions --p: I bought a lottery ticket this week. --q: I won the million dollar jackpot. Express each of these propositions as an English sen- tence. a) ¬p c) p → q e) p ↔ q g) ¬p ∧ ¬q 11. Let p and q be the propositions "Swimming at the New Jersey shore is allowed" and "Sharks have been spotted near the shore," respectively. Express each of these com- pound propositions as an English sentence. a) ¬q c) ¬p ∨ q e) ¬q → p g) p ↔ ¬q 12. Let p and q be the propositions "The election is decided" and "The votes have been counted," respectively. Express each of these compound propositions as an English sen- tence. a) ¬p c) ¬p ∧ q e) ¬q → ¬p g) p ↔ q 13. Let p and q be the propositions --p: It is below freezing. --q: It is snowing. Write these propositions using p and q and logical connectives (including negations). a) It is below freezing and snowing. c) It is not below freezing and it is not snowing. e) If it is below freezing, it is also snowing. g) That it is below freezing is necessary and sufficient for it to be snowing. 16. Let p, q, and r be the propositions --p: You get an A on the final exam. --q: You do every exercise in this book. --r: You get an A in this class. Write these propositions using p, q, and r and logical con- nectives (including negations): a) You get an A in this class, but you do not do every exercise in this book. c) To get an A in this class, it is necessary for you to get an A on the final. e) Getting an A on the final and doing every exercise in this book is sufficient for getting an A in this class. 18. Determine whether these biconditionals are true or false. a) 2 + 2 = 4 if and only if 1 + 1 = 2. c) 1 + 1 = 3 if and only if monkeys can fly. 19. Determine whether each of these conditional statements is true or false. a) If 1 + 1 = 2, then 2 + 2 = 5. c) If 1 + 1 = 3, then 2 + 2 = 5. 20. Determine whether each of these conditional statements is true or false. a) If 1 + 1 = 3, then unicorns exist. c) If 1 + 1 = 2, then dogs can fly. 21. For each of these sentences, determine whether an inclusive or, or an exclusive or, is intended. Explain your answer. a) Coffee or tea comes with dinner. c) The prerequisite for the course is a course in number theory or a course in cryptography. 22. For each of these sentences, determine whether an inclusive or, or an exclusive or, is intended. Explain your answer. a) Experience with C++ or Java is required. c) To enter the country you need a passport or a voter registration card. 23. For each of these sentences, state what the sentence means if the logical connective or is an inclusive or (that is, a disjunction) versus an exclusive or. Which of these meanings of or do you think is intended? a) To take discrete mathematics, you must have taken calculus or a course in computer science. c) Dinner for two includes two items from column A or three items from column B. **24. Write each of these statements in the form "if p, then q" in English. [Hint: Refer to the list of common ways to ex- press conditional statements provided in this section.] a) It is necessary to wash the boss's car to get promoted. c) A sufficient condition for the warranty to be good is that you bought the computer less than a year ago. e) You can access the website only if you pay a subscription fee. g) Carol gets seasick whenever she is on a boat. 25. Write each of these statements in the form "if p, then q" in English. [Hint: Refer to the list of common ways to express conditional statements.] a) It snows whenever the wind blows from the northeast. c) That the Pistons win the championship implies that they beat the Lakers. e) To get tenure as a professor, it is sufficient to be world famous. g) Your guarantee is good only if you bought your CD player less than 90 days ago. 29. State the converse, contrapositive, and inverse of each of these conditional statements. a) If it snows today, I will ski tomorrow. c) A positive integer is a prime only if it has no divisors other than 1 and itself. 30. State the converse, contrapositive, and inverse of each of these conditional statements. a) If it snows tonight, then I will stay at home. c) When I stay up late, it is necessary that I sleep until noon.

1. Propositions=truth value: a=T, b=T, c=T, d=F 2. Propositions=truth value: c=F, e=F 5. 反差: a) Mei does not have an MP3 player c) 2+1 != 3 6. 反差: a) Jennifer and Teja are not friends c) Abby did not send more than 100 txt msges yesterday 10. p=买lottery ticket这周；q=赢了100万美元jackpot a) I did not buy a lottery ticket this week c) If I bought a lottery ticket this week, then I'd win the million dollar jackpot e) I'll buy lottery ticket this week IF AND ONLY IF i'll win million dollar jackpot g) I did not buy a lottery ticket this week AND I did not win the million dollar jackpot 11. p=🉑️在NJ岸游泳；q=鲨鱼在岸附近被看见 a) I did not buy a lottery ticket this week c) Swimming is NOT allowed at NJ shore OR sharks were spotted near the shore, OR BOTH are true! e) If sharks have NOT been spotted near shore, then swimming at NJ shore is allowed g) Swimming allowed on NJ shore IF AND ONLY IF no sharks were spotted near shore 12. p=总选已被决定；q=总选投票已被数 a) The election has not been decided c) The election is NOT decided and the votes have been counted e) If the votes have NOT been counted, then the election has NOT been decided g) The election has been decided IF AND ONLY IF votes have been counted 13. p=below freezing；q=snowing a) p ∧ q c) ¬p ∧ ¬q e) p → q g) p ↔ q --> "necessary and sufficient"=biconditional! 16. p=A on final exam q=every exercise in book r=A in this class a) r n NOTq --> but=AND (conjunction) c) r->p --> q is necessary condition for p --> To get an A in this class(r), it is necessary for you to get an A on the final(p). e) (pnq) → q --> sufficient = conditional CONDITIONAL： Truth table (p与q一样 或者 p=T): P: TTFF Q: TFTF P->Q: TFTT 18) a) T c) T --> F=F --> T! (p和q须得含于同样真理数) 19) a) F --> T+F=F c) T --> F+F+T 20) a) T --> p: 1+1=3 --> q:unicorns exist --> p=F, q=F ===> p-->q=T (F+F=T) c) F --> T+F=F 21) Exclusive/or=唯有一个=T; Inclusive/or=至少1个T，不允许有俩F a) Exclusive/or --> 要不选咖啡，要不选茶 - 不能俩都选哦 c) Inclusive/or --> 俩课程其中都行，若俩都有那更好！ 22) a) Inclusive/or --> 俩技术其中都行，若俩都有那更好 c) Inclusive/or --> 俩签证其中都行，若俩都有那更好 23) Inclusive/or (disjunction) 比 exclusive/or; intention? a) IN: You can take calculus OR compsci OR both, to take discrete maths EX: You can take EITHER calc OR compsci, to take discrete maths Intention: IN c) IN: Dinner for 2 includes 2 items-A, OR 3 items-B, OR both! EX: Dinner for 2 includes EITHER 2 items-A, OR 3 items-B Intention: EX 24) a) If you get promoted, then you washed the boss's car --> p is necessary for q (p=wash boss's car) c) If you bought computer less than year ago, then warranty is good/fulfilled --> Sufficient condition for q is p = p-->q e) If you can access the website, then you pay a subscription fee --> q if p = p-->q g) If Carol is on a boat, then she will get seasick --> q when p = p-->q 25) a) If the wind blows from NE, then it snows --> q when p c) If the pistons win championship, then 她们已 beat the lakers --> p implies q e) If you are world famous, then you will get tenured as professor --> Sufficient condition for q is p g) If you bought ur CD player <90天前, then guarantee is good --> q only if p 29) a) i. Converse: If I ski tmr, it will snow today ii. Contrapositive (C+I): I will NOT ski tmr, if it does NOT snow today iii. Inverse: If it does NOT snow today, will NOT ski tmr c) i. Converse: If positive integer has no divisers other than 1以及本数,then positive integer is prime ii. Contrapositive (C+I): If positive integer HAS divisers other than 1以及本数,then positive integer is NOT prime iii. Inverse: If positive integer is NOT prime, then it HAS divisors other than 1以及本数 30) a) i. Converse: If I stay at home, then it snows tonight ii. Contrapositive (C+I): If I do NOT stay at home, then it does NOT snow tonight iii. Inverse: If it does NOT snow tonight, then i will NOT stay home c) i. Converse: If I sleep until noon, then I stayed up late ii. Contrapositive (C+I): If I DON'T sleep until noon, then i DIDN'T stay up late iii. Inverse: If i DON'T stay up late, then i DON'T sleep until noon

作业4: 1.7: 1,2,4-7,10,12 (use proof by contraposition),13,14,17-20,28,29,82-84(supplemental) 1. Use a direct proof to show that the sum of two odd integers is even. 2. Use a direct proof to show that the sum of two even integers is even. 4. Show that the additive inverse, or negative, of an even number is an even number using a direct proof. 5. Prove that if m + n and n + p are even integers, where m, n, and p are integers, then m + p is even. What kind of proof did you use? 6. Use a direct proof to show that the product of two odd numbers is odd. 7. Use a direct proof to show that every odd integer is the difference of two squares. [Hint: Find the difference of the squares of k + 1 and k where k is a positive integer.] 10. Use a direct proof to show that the product of two rational numbers is rational. 12. Prove or disprove that the product of a nonzero rational number and an irrational number is irrational. (use proof by contraposition) 13. Prove that if x is irrational, then 1∕x is irrational. 14. Prove that if x is rational and x ≠ 0, then 1∕x is rational 17. Use a proof by contraposition to show that if x + y ≥ 2, where x and y are real numbers, then x ≥ 1 or y ≥ 1. 18. Prove that if m and n are integers and mn is even, then m is even or n is even. 19. Show that if n is an integer and n3 + 5 is odd, then n is even using a) a proof by contraposition. b) a proof by contradiction. 20. Prove that if n is an integer and 3n + 2 is even, then n is even using a) a proof by contraposition. b) a proof by contradiction. 28. Prove that if n is a positive integer, then n is even if and only if 7n + 4 is even. 29. Prove that if n is a positive integer, then n is odd if and only if 5n + 6 is odd. Suppl：https://drive.google.com/open?id=1zi7BNB8jICwfwQTEBNlvbgxvv_ws0eVa

1. Use a direct proof to show that the sum of two odd integers is even. To prove: sum of two odd integers = even Recall: - Odd: integer k such that 2k+1 - Even: integer k such that 2k Proof: i. Let x and y be odd integers. Then there exists integers such that x=2k+1; y=2z+1. ii. Thus, x+y = (2k+1)+(2z+1) = 2k+2z+2 = 2(k+z+1) iii. Since k and z are integers, k+z+1 is also an integer and x+y is even (given that product of 2 and some integer is even, as proven in lecture), based on properties for integers and odd/even numbers above. 口 2. Use a direct proof to show that the sum of two even integers is even. Recall: - Odd: integer k such that 2k+1 - Even: integer k such that 2k Proof: i. Let x and y be even integers. Then there exists integers such that x=2k; y=2z ii. Thus, x+y=2k+2z=2(k+z) iii. Since k and z are integers, 2(k+z) is also an integer. Recalling that 2(an integer) is even (it can be divided by 2), then 2(k+z)=x+y is even. 口 4. Show that the additive inverse, or negative, of an even number is an even number using a direct proof. Proof: i. Let x be an even real number. Then there exists real number such that x=2k. ii. Thus, the additive inverse of x is -x = -2k = 2(-k), making j=-k this can be rewritten as 2j iii. Since x is a real number, 2k is also a real number, -x and -2k are also real numbers as real numbers includes negative numbers. Knowing that an even number is divisible by 2, then 2j or 2(-k) is even. 口 5. Prove that if m + n and n + p are even integers, where m, n, and p are integers, then m + p is even. What kind of proof did you use? Direct Proof: i. Suppose (m+n) and (n+p) are even integers. Then there exists integers x and y such that 2x=(m+n) and 2y=(n+p). ii. Now we find the sum of the integers: m+n+n+p = 2x+2y=2 --> m+2n+p = 2(x+y) iii. We're interested in m+p, thus subtract 2n from both sides --> m+p = 2(x+y-n) iv. Since x, y, n are integers, then x+y-n is also an integer and 2(x+y-n) is an even integer. Thus, m+p is even (given property for even integers that some even integer x, x=2k). 口 6. Use a direct proof to show that the product of two odd numbers is odd. Proof: i. Suppose there are 2 odd integers x and y. Then there exists integers k and z such that x=2k+1 and y=2z+1. ii. Then the product of xy=(2k+1)(2z+1)=4kz+2k+2z+1 = 2(2kz+k+z+1) = 2(2kz+k+z)+1. iii. Since k and z are integers, 2(2kz+k+z)+1 = xy is an integer. Thus, xy is odd. 口 7. Use a direct proof to show that every odd integer is the difference of two squares. [Hint: Find the difference of the squares of k + 1 and k where k is a positive integer.] Proof: i. Let k+1 and k be positive integers. Then (k+1)^2 and k^2 are positive integers. ii. Find their diff: (k+1)^2-(k^2) = (k+1)(k+1) - (k^2) = k^2+2k+1 - k^2 = 2k+1 iii. Since k+1 and k are positive integers, and 2k+1 defines any odd integer, this completes the proof. 口 10. Use a direct proof to show that the product of two rational numbers is rational. - Recall: rational number is some fraction p/q Proof: i. Suppose there exists two rational numbers x and y. Then suppose there exist integers p,q,r,s such that x=p/q and y=r/s ii. Thus, xy=(p/q)(r/s) = pr/qs iii. Given that p,r,q,s are integers, then pr and qs are integers (product of 2 integers is also an integer). Thus, pr/qs is a rational number, and xy is proven to be a rational number. 口 12. Prove or disprove that the product of a nonzero rational number (p) and an irrational number (q) is irrational (r). (use proof by contraposition) - p=nonzero rational number x - q=irrational number y - r=irrational product xy pnq--> r Proof: = !r --> !(pnq) = !r --> !p v !q --> EITHER !p=T or !q=T (in this case we assume !p=F; !q=T) = !r n p--> !q --> Prove (rational xy)(nonzero rational x) = rational y Contraposition Proof (!rnp-->!q): i. Assume rational xy is made up of integers p,q such that xy=p/q; assume nonzero rational x made of integers r,s such that x=r/s. ii. We multiply rational xy and nonzero rational x, (p/q)(r/s)=pr/qs iii. Given p,q,r,s are integers, pr/qs is a rational number—it follows from our contrapositive assumption xy is rational and x is rational and nonzero, that y=pr/qs is rational! This effectively proves that product of nonzero rational number and irrational number is irrational! 口 [Contradiction Proof: i. Assume !q: the product of nonzero rational number x (such that x=p/q) and irrational number y = RATIONAL. --> p,q,r,s=nonzero integers ii. Then, xy=r/s iii. Since x=p/q, then (p/q)y = r/s ==> y=(q/p)(r/s) = qr/ps --> y=qr/ps iv. Since p,q,r,s are nonzero integers and qr and ps are also nonzero integers, this would imply that y is a rational number --> which is in contradiction of the given fact that y is irrational. This proves the original claim, that xy=irrational! 口] 13. Prove that if x is irrational, then 1∕x is irrational. Contradiction (!q) Proof: i. We assume 1/x is a rational number and y,z are nonzero integers such that 1/x=y/z. --> x=/=0 as 1/x has to be rational! ii. 1/x=y/z --> x=z/y iii. Since z,y are nonzero integers then z/y=x is rational...BUT this contradicts the fact that x is given as irrational, thus proving that the assumption that 1/x is rational is false! Thus if 1/x cannot be a rational number, it follows that it must be irrational. 口 14. Prove that if x is rational and x ≠ 0, then 1∕x is rational. Direct proof: i. Suppose that x is a nonzero rational number. Then y,z are nonzero integers such that x=y/z. ii. x=y/z --> 1/x=1/(y/z) = 1/x=z/y iii. Since z,y are nonzero integers then 1/x=z/y is rational (also x is nonzero rational). Thus this proves that 1/x is rational number! 口 17. Use a proof by contraposition to show that if x + y ≥ 2, where x and y are real numbers, then x ≥ 1 or y ≥ 1. Proof: i. Assume !q: x<1 and y<1 (and are real numbers) ii. Add two inequalities x+y<1+1 --> x+y<2 iii. Thus, we have shown that if x+y>=2, then x>=1 or y>=1 by proving its contrapositive x+y<2 then x<1 and y<1 true! 口 18. Prove that if m and n are integers and mn is even, then m is even or n is even. pnq--> rvs pnq--> !r n !s Contrapositive Proof: i. Assume m and n are odd integers. Then, k and z are integers such that m=2k+1 and n=2z+1. ii. mn=(2k+1)(2z+1) --> 4kz+2k+2z+1 --> 2(2kz+k+z)+1 iii. Knowing that k,z are integers, we've proven that 2(2kz+k+z)+1 is odd. Thus, we've proven via contraposition that if m and n are odd, then mn is odd (2k+1) which implies that the contrapositive (if mn is even, then m and n are even) is true! 口 19. Show that if n is an integer and n^3 + 5 is odd, then n is even using a) a proof by contraposition. Proof: i. Assume n is an odd integer, and that k is an integer such that n=2k+1 ii. Thus, n^3+5 = (2k+1)^3 + 5 = (2k+1)(2k+1)(2k+1) + 5 = 4k^2+4k+1(2k+1) + 5 = 8k^3 + 12k^2 + 6k + 6 = 2(4k^3+6k^2 + 3k + 3) --> Replace (4k^3+6k^2 + 3k + 3)=l, then n^3+5 = 2l! iii. Given that k is integer, 2(4k^3+6k^2 + 3k + 3) is even, thus proving that if n is odd, then n^3+5 is even --> proving the contrapositive that if n^3 + 5=odd, then n is even! 口 b) a proof by contradiction. Proof (hyp=T;concl=F): i. Given n^3 + 5 is odd, we assume that n is odd, and k is integer such that n=2k+1. ii. Then n^3+5 = (2k+1)^3 + 5 = (2k+1)(2k+1)(2k+1) + 5 = 4k^2+4k+1(2k+1) + 5 = 8k^3 + 12k^2 + 6k + 6 = 2(4k^3+6k^2 + 3k + 3) iii. Replace (4k^3+6k^2 + 3k + 3)=l, then n^3+5 = 2l = n^3+5 is ... even? iv. Assuming n=odd, this leads us to contradiction that that n^3+5 is even AND odd given our earlier assertion. This is a contradiction! Thus "n is odd" must be a false statement; n is even is proven. 口 20. Prove that if n is an integer and 3n + 2 is even, then n is even using a) a proof by contraposition. Proof: i. Assume that n is odd integer, and k is integer such that n=2k+1. ii. Then 3(2k+1)+2 = 6k+3+2 = 6k+4+1 = 2(3k+2)+1 iii. Since k is integer, then 3k+2 is integer and 2(3k+2)+1 is odd, which proves through contraposition that if n is odd then 3n+2 is odd --> this proves that if n is even, then 3n+2 is even! 口 b) a proof by contradiction (hyp is true but conclusion is false). Proof: i. Given that 3n+2 is even, we assume n is (not even) odd, and k is integer such that n=2k+1 ii. Then n=3(2k+1)+2 = 6k+3+2 = 2(3k+4)+1 iii. Since k is integer, then 2(3k+4)+1=3n+2 is odd but ALSO even。。。which is contradiction! Thus, assuming "n is odd" leads us to contradiction which proves "n is odd" FALSE, proving "n is even" is true! 口 28. Prove that if n is a positive integer, then n is even if and only if 7n + 4 is even. Direct proof (if n is even, then 7n+4 is even): i. Assume n is even integer and k is integer such that n=2k ii. Then 7(2k)+4 = 14k+4 = 2(7k+2) iii. Given k is integer, then 2(7k+2) is even and thus 7n+4 is even, proving that if n is even then 7n+4 is even! 口 Contraposition Proof (if 7n+4 is even, then if n is even) (!q-->!p): i. Assume n is ODD integer and k is integer such that k=2k+1 ii. Then, 7(2k+1)+4 = 14k+7+4 = 2(7k+5)+1 iii. Given k is integer, then 2(7k+5)+1 = 7n+4 is odd, which follows from assertion that n is ODD. This proves through contraposition "if 7n+4 is even then n is even" true! 口 29. Prove that if n is a positive integer, then n is odd if and only if 5n + 6 is odd. Direct proof: n is odd --> 5n+6 is odd i. Suppose n is odd integer, with k = integer such that n=2k+1 ii. Then, 5(2k+1)+6 = 10k+5+6 = 10k+10+1 = 2(5k+5)+1 iii. From the above, we see that (given k is integer) 2(5k+5)+1 is odd, meaning 5n+6 is odd, if n is odd. This is directly proves that if n is odd 5n+6 is odd! 口 Contraposition proof: 5n+6 is odd--> n is odd; !(n odd)-->!(5n+6 odd) i. Suppose n is EVEN integer, with k = integer such that n=2k ii. Then, 5(2k)+6 = 10k+6 = 2(5k+3) iii. From the above, we see that (given k is integer) 2(5k+3) is EVEN, meaning 5n+6 is even, if n is even. This proves the contrapositive of 5n+6 is odd, n is odd, proving that if 5n+6 is odd, then n is odd! 口 Proof: 5n+6 is odd--> n is odd 82. [n] is greatest integer of n; [n]=rounded down of n! a) n0/2 b) n0/2; Z --> 因为k rounded DOWN! c) Z; n0/2 d) n0/2 e) Z f) n0/2 g) n0/2 h) n0 i) 2k+1 84) Pf: i. Let n be an integer, given that p-->q is logically equivalent to !pvq and from lemma 1.7.2 VneZ, !P(n)-->Q(n), that !p-->q is logically equivalent p v q. ii. Then we can apply universal instantiation to VneZ, !P(n)-->Q(n) to get !P(n)--> Q(n) which is logically equivalent to !(!P(n)) v Q(n) which is logically equivalent P(n) v Q(n). We then use universal generalization to get VneZ P(n) v Q(n), that n is either even or odd for every n. 口

作业3: 1.4: 32, 33, 46, 52 1.5: 1ac, 2ac, 3be, 6be, 9bdfgh, 10bdfgh, 13cd, 14cd, 19ad, 20ad, 24b, 25b, 27ad, 28ad, 31ad, 32ad, 36de, 37ad, 41, 42, 45, 46 1.6: 7, 8, 9ade, 13bd, 14bd, 15ad, 16ad, 17, 18, 19a, 20a, 23, 24 https://docs.google.com/document/d/1JmLjjx_nHxFdR21jFt_nUMgwpQ3mXoYNWL16QeqouLA/edit

1.4: 32) a) P(x)=dog has fleas --> VxP(x) --> N: ∃x!P(x) = There is one dog that does NOT have fleas! b) P(x)=horse can add --> ∃xP(x) --> N: Vx!P(x) = ALL of horses canNOT add! c) P(x)=koala can climb --> VxP(x) --> N: ∃x!P(x) = There is one koala that canNOT climb! d) P(x)=monkey can speak French. --> Vx!P(x) --> N: ∃xP(x) = There is one monkey that can speak french! e) P(x)=pig can swim; Q(x)=catch fish. --> ∃x[P(x) n Q(x)] --> N: Vx[!P(x) v !Q(x)] = NONE of pigs can swim nor catch fish! or ALL pigs can not swim or not catch fish! 33) 46) ∀x[P(x) ↔ Q(x)] = ∀xP(x) ↔ ∀xQ(x) --> 相等么？ Pf: Consider a domain of discourse D with two elements, a and b, such that P(a) is true and Q(a) is false, while P(b) is false and Q(b) is true. - 前者: P(a), Q(b) THEN P(b), Q(b) - 后者: P(a), P(b) THEN Q(a), Q(b) Then the propositions ∀x P(x) is false, since P(b) is. Similarly, since Q(a) is false then so is ∀x Q(x). Because both propositions are false, the biconditional ∀x P(x) ↔ ∀x Q(x) is true. However, the biconditional P(a) ↔ Q(a) is false since the two prositions have different truth values. (The same can be said for P(b) ↔ Q(b)).) Consequently, the statement ∀x (P(x) ↔ Q(x)) is false. Since ∀x (P(x) ↔ Q(x)) is false for this domain D while ∀x P(x) ↔ ∀x Q(x) is true, the two statements are not logically equivalent. 52) 显明 ∀xP(x) ∨ ∀xQ(x) =/= ∀x[P(x) ∨ Q(x)] Pf: Suppose Px=odd; Qx=even x=domain of all positive integers i. ∀xP(x) ∨ ∀xQ(x) =分开的！--> every x is odd OR every x is even --> F i. ∀x[P(x) ∨ Q(x)] =连一起的！ --> every x is EITHER odd or even! --> T *From proof, it is clear that ∀xP(x) ∨ ∀xQ(x) =/= ∀x[P(x) ∨ Q(x)], as the former is false (as it is not true, within domain of positive integers, that every x is odd or every x is even) and the latter is true (it is true that every x is either odd or even).* 1.5: 2) domain=real numbers (包括decimals) a) There exists a real number x such that for every y, xy is equal to y. c) For every real number x and every real number y, there exists a real number z such that y+z is equal to x. 6) C(x,y)=x enrolled in y Domain x=all students in school Domain y=all classes given at school b) There is a student in school who is enrolled in MATH 695. c) There is a class given at school that Carol Sitea is enrolled in. 9) b) Vx∃yL(x,y) d) VxVy!L(x,y) --> 记得，"！"得放到推测之前！ --> = !∃xVyL(x,y) f) ∃yVx!L(x,y) 10) F(x,y)=x can fool y Domain for x and y=all people in world --> somebody=∃ --> everybody=V b) VyF(Evelyn,y) d) !∃xVyF(x,y) --> Nobody=contrary to one existing, all are false! = !∃x --> 也能写出 Vx∃y!F(x,y) --> demorgans=flip! 颠倒黑白嘿嘿 f) !∃xF(x,Fred) n F(x,Jerry) --> x=nobody g) i. Nancy can fool two ppl a and b AND --> ∃a∃b[F(Nancy,a) n F(Nancy,b)] ii. a and b=separate ppl AND (not equal) --> a=/=b iii. All people Nancy fools (c) = c=a or c=b --> Vc[F(Nancy, c)--> (c=a)v(c=b)] (for every c, 需要括号) *∃a∃b[[F(Nancy,a) n F(Nancy,b)] n [a=/=b] n [Vc[F(Nancy,c) --> (c=a v c=b)]]]* h) i. There is a person(y) everyone(x) fools --> ∃yVxF(x,y) ii. ALL people(a) fooled by everyone(b) MUST = y --> Va(VbF(b,a)--> a=y) *∃y[VxF(x,y) n Va((VbF(b,a)) --> a=y)]* 14) c) P(x,y)=student x in this class visited location y - (x,y)=x visited y, x did y --> but=AND! ∃x[P(x,Alaska) n !P(x,Hawaii)] d) P(x,y)=student x in this class have programming language y *Vx∃yP(x,y)* --> ∃=for some x in Domain, P(x)=T; for at least one x, exists element x in domain such that P(x)=T等等 20) a) 理所当然无详细特出=V - For every integer x and y, if x AND y are negative THEN their product integer z is positive. --> integers=不需要P若不是predicate --> *VxVy[(x<0 n y<0) --> xy>0]* --> 不是 VxP(x<0) n VyP(y<0) --> 这个等于every integer x is negative and every integer y is negative, 全然不同！！！！！！ d) For every integer x and y, their total absolute value sum does NOT exceed(equalorgreaterthan) sum of the separate absolute values of x and y --> *VxVy[|(x+y)| <= |x|+|y|]* 24) ∀x∀y((x ≠ 0) ∧ (y ≠ 0) ↔ (xy ≠ 0)) --> Domain=all real nmbrs *For every two real numbers x and y, x and y do NOT equal 0 IF AND ONLY IF their product does not equal 0* 28) Domain=all real numbers a) ∀x∃y(x^2 = y) = T --> For every real number there exists number that is its square = 永远T！不需多想！ d) ∃x∃y(x + y ≠ y + x) = F --> Commutative property holds for all real numbers! 41) x,y,z=real numbers - x(yz)=(xy)z --> *VxVyVz[x(yz)=(xy)z]* 42) x,y,z=real numbers --> add universal quantifiers such that VxVyVz=always holds - x(y+z)=xy+yz - (x+y)z=xz+yz --> Distributive property ALWAYS holds! --> *VxVyVz[x(y+z)=xy+yz]* --> *VxVyVz[(x+y)z=xz+yz]* 46) ∃x∀y(x ≤ y^2) = there exists x in domain <= square of EVERY value in domain a) positive real numbers (>0) (decimals) = F! --> For every y, y^2>= some x --> ZERO is only number that is <= square of every # in domain --> ZERO not included --> F! b) integers = T! --> x<= EVERY squared # in domain--> NO decimals=true! --> (-5)<=(-5)^2 --> x=09⃣️🉑️ c) nonzero real numbers = T! --> x as -1 <= smaller than EVERY squared # y^2 in domain! --> T! 1.6: 8) *modus tollens, double negation, universal instantiation* P(x)=x is human; Q(x)=x is island - No man is an island=for all x, if x is human then x is NOT island! - 后果: !P(Manhattan) i. Vx(P(x) --> !Q(x)) = premise no man is island ii. Q(manhattan) = premise manhattan is island iii. P(manhattan) --> !Q(manhattan) = universal instantiation of (i) --> VxP(x)=T, then P(c)=TRUE PREDICATE! --> Vx[P(x)--> Q(x)] = P(c)--> Q(c) (若每个都是T，那当然把某个"c"提出来也会吃真的！) iv. ![!Q(manhattan)] = double negation (ii) (negation of the conclusions=negation of hyp!) --> setting up for MODUS TOLLENS! - p-->q; !q=!p to be true! v. *!P(manhattan)*! --> modus tollens (iv, iii) 9) P(x)=take x day off Q(x)=rains S(x)=snows t=tues; r=thurs P1: Vx(P(x)--> Q(x) v S(x)) P2: P(t) v P(r) P3: ![Q(t) v S(t)] P4: !S(r) i. Univ instantiation of P1 --> P(t)--> Q(t) v S(t) --> P(r)--> Q(r) v S(r) ii. P4 and (i) = modus tollens --> !P(t) iii. Disjunctive syllogism !P(t) and P2 --> P(r) iv. Modus Pollens P(r) and (i) --> Q(r) v S(r) v. Disjunctive syllogism P4: !S(r) and Q(r) v S(r) --> *Q(r)* 14) b) *universal generalization(put it in), hypothetical syllogism(清除中间), universal instantiation(take it out)* Pf: c=arbitrary # r(x)=x is roommate d(x)=x has taken course in discrete maths a(x)=x can take algorithm course 后果: Vx(r(x) --> a(x)) i. Vx(r(x)--> d(x)) = premise --> Each of five roommates, Melissa, Aaron, Ralph, Veneesha, and Keeshawn, has taken a course in discrete mathematics. ii. r(c)--> d(c) = universal instantiation (i) iii. Vx(d(x)--> a(x)) = premise --> Every student who has taken a course in discrete mathematics can take a course in algorithms. iv. d(c)--> a(c) = universal instantiation (iii) v. r(c)--> a(c) = hypothetical syllogism (除去中间宿主）(ii,iv) --> Therefore, all five roommates can take a course in algorithms next year. vi. Vx(r(x)--> a(x)) = universal generalization (v) d) *exist gen, exist inst, modus ponens, conjunction, simplification, univ instant* c(x)=in the class f(x)=been to france l(x)=visits the louvre 后果: ∃x(c(x) n l(x)) i. ∃x(c(x) n f(x)) = premise --> someone in class AND been to france --> There is someone in this class who has been to France. ii. c(c) n f(c) = existential instantiation iii. c(c) = simplification (ii) - ONLY to conjunction n iv. f(c) = simplification (ii) - ONLY to conjunction n v. V(x)(f(x)--> l(x)) --> Everyone who goes to France visits the Louvre. vi. f(c)--> l(c) = universal instantiation vii. l(c) = modus ponens (iv, vi) viii. c(c) n l(c) = conjunction (iii, vii) viv. ∃x(c(x) n l(x)) = existential generalization（塞进去!) --> Therefore, someone in this class has visited the Louvre. 16) a) 结果: ∃x!u(M) = *CORRECT, using modus tollens(negationv)+univ instant* u(x)=enrolled in uni d(x)=lived in dorm d(M)=mia lived in dorm --> 自domain之例 i. Vx(u(x)--> d(x)) = premise --> Vx(u(x) v !d(x)) --> Everyone enrolled in the university has lived in a dormitory. ii. u(M)--> d(M) = univ instant iii. !d(M) = premise --> Mia has never lived in a dormitory. iv. !u(M) = modus tollens (!u-->!d=T!) (ii,iii) --> Therefore, Mia is not enrolled in the university. d) *CORRECT, using univ insant, modus ponens(p=positivev)* All lobstermen set at least a dozen traps. Hamilton is a lobsterman. Therefore, Hamilton sets at least a dozen traps. l(x)=lobsterman t(x)=set dozen+ traps l(H)=hamilton --> 自domain之例(平常用c，于此用H!) 结果: t(H) i. Vx(l(x)--> t(x)) = premise --> If you're lobsterman, you've set at least a dozen traps ii. l(H)--> t(H) = univ instant iii. l(H) = premise iv. t(H) = modus ponens (ii,iii) 18) Let S(x, y) be "x is shorter than y." i. ∃sS(s, Max) = premise ii. S(Max, Max) --> *where does this come from? (i) says 存在以至于s is shorter than max, NOT anything about max being shorter* --> *entity cannot differ from itself==>inherently contradictory, 凭空捏造啊虚滴!* iii. ∃xS(x, x), existential generalization --> so that someone is shorter than himself. 20) a) If x is a positive real number, then x^2 is a positive real number. Therefore, if a^2 is positive, where a is a real number, then a is a positive real number. --> *INVALID! fallacy of affirming the conclusion (只能用MP于推测然后结论！) (-x)^2 can also=positive number --> (-5)^2=25, but -5=negative!* 24) 1. ∀x(P(x) ∨ Q(x)) —Premise 2. P(c) ∨ Q(c) —Universal instantiation from (1) 3. P(c) —Simplification from (2) --> *can only use simplification for CONJUNCTION!* (因为cnd=c和d皆是T所以能自其中随便抽，若v=或者的话就不行喽！） 4. ∀xP(x) —Universal generalization from (3) 5. Q(c) —Simplification from (2) --> *can only use simplification for CONJUNCTION!* 6. ∀xQ(x) —Universal generalization from (5) 7. ∀x(P(x) ∨ ∀xQ(x)) —Conjunction from (4) and (6) --> *CONJUNCTION是n, 84 v!*

作业2 (第一部): 1.3: 11a, 12a, 15a, 16a, using SS: 20, 25-27 11. Show that each of these conditional statements is a tautology by using truth tables. a) (p ∧ q) → p 12. Show that each of these conditional statements is a tautology by using truth tables. a) [¬p ∧ (p ∨ q)] → q --> EXAMPLE 8 Show that (p ∧ q) → (p ∨ q) is a tautology. Solution: To show that this statement is a tautology, we will use logical equivalences to demonstrate that it is logically equivalent to T. (Note: This could also be done using a truth table.) i. (p ∧ q) → (p ∨ q) ≡ ¬(p ∧ q) ∨ (p ∨ q) --> by Example 3 ii. ≡ (¬p ∨ ¬q) ∨ (p ∨ q) --> by the first De Morgan law iii. ≡ (¬p ∨ p) ∨ (¬q ∨ q) --> by the associative and commutative laws for disjunction iv. ≡ T ∨ T --> by Example 1 and the commutative law for disjunction v. ≡ T --> by the domination law 15. Show that each conditional statement in Exercise 11 is a tautology by applying a chain of logical identities as in Example 8. (Do not use truth tables.) a) (p ∧ q) → p 16. Show that each conditional statement in Exercise 12 is a tautology by applying a chain of logical identities as in Example 8. (Do not use truth tables.) a) [¬p ∧ (p ∨ q)] → q USING SUCCESSIVE SUBSTITUTION: 20. Show that p ↔ q and (p ∧ q) ∨ (¬p ∧ ¬q) are logically equivalent. 25. Show that ¬(p ↔ q) and ¬p ↔ q are logically equivalent. 26. Show that (p → q) ∧ (p → r) and p → (q ∧ r) are logically equivalent. 27. Show that (p → r) ∧ (q → r) and (p ∨ q) → r are logically equivalent

11. Show that each of these conditional statements is a tautology by using truth tables. a) (p ∧ q) → p p: TTFF q: TFTF pnq: TFFF *(pnq)->p: TTTT* 12. Show that each of these conditional statements is a tautology by using truth tables. a) [¬p ∧ (p ∨ q)] → q p: TTFF !p: FFTT q: TFTF !q: FTFT pvq: TTTF !pn(pvq): FFTF --> n=俩必须都是T *!pn(pvq)->q: TTTT* --> EXAMPLE 8 Show that (p ∧ q) → (p ∨ q) is a tautology. Solution: To show that this statement is a tautology, we will use logical equivalences to demonstrate that it is logically equivalent to T. (Note: This could also be done using a truth table.) i. (p ∧ q) → (p ∨ q) ≡ ¬(p ∧ q) ∨ (p ∨ q) --> by Example 3 ii. ≡ (¬p ∨ ¬q) ∨ (p ∨ q) --> by the first De Morgan law iii. ≡ (¬p ∨ p) ∨ (¬q ∨ q) --> by the associative and commutative laws for disjunction iv. ≡ T ∨ T --> by Example 1 and the commutative law for disjunction v. *≡ T * --> by the domination law 15. Show that each conditional statement in Exercise 11 is a tautology by applying a chain of logical identities as in Example 8. (Do not use truth tables.) a) (p ∧ q) → p ≡!(pnq) v p --> 7.1 ≡(!p v !q) v p --> demorgan's rule ≡(!p v p) v !q --> associative rule ≡T v T *≡T!* 16. Show that each conditional statement in Exercise 12 is a tautology by applying a chain of logical identities as in Example 8. (Do not use truth tables.) a) [¬p ∧ (p ∨ q)] → q ≡![!p n (p v q)] v q ≡[p n !(p v q)] v q --> demorgan's rule ≡[p v (!p n !q)] v q --> demorgan's rule again ≡(p v !p) n (p v !q) v q ≡T n (p v !q) v q ≡T n [p v (!q v q)] --> T n x = x (identity law) ≡pv (!q v q) ≡p v T --> !q v q = T (domination law) *≡T!* USING SUCCESSIVE SUBSTITUTION: 20. Show that p ↔ q and (p ∧ q) ∨ (¬p ∧ ¬q) are logically equivalent. p↔q (p->q)n(q->p) (!pvq)n(!qvp) --> 7.1 *(!pn!q)v(pnq)≡(!pn!q)v(pnq)* --> associative rule 25. Show that ¬(p ↔ q) and ¬p ↔ q are logically equivalent. ¬(p↔q) ![(!p v q) n (!q v p)] (p n !q) v (q n !p) --> demorgan's law *(pvq) n (!qv!p) ≡ (p v q)n(!q v !p)* ≡ ¬p↔q --> associative law 26. Show that (p → q) ∧ (p → r) and p → (q ∧ r) are logically equivalent. (p→q)n(p→r) ≡? p→(q n r) (!p v q)n(!p v r) ≡ !p v (q n r) *(!p v q)n(!p v r) ≡ (!p v q)n(!p v r)* --> dist law! 27. Show that (p → r) ∧ (q → r) and (p ∨ q) → r are logically equivalent (p→r) n (q→r) ≡ (p v q)→r (!pvr) n (!qvr) ≡ !(p v q) v r (!pvr) n (!qvr) ≡ (!p n !q) v r --> de morgans law (!p n !q) n rvr ≡ (!p n !q) v r *(!p n !q) n r ≡ (!p n !q) v r*

作业2: 第三部 1.4: 2, 3, 5, 6, 9ad, 10ad, 11-16 (part d only), 17b, 18c, 19d, 20a, 23b, 24c, 25ad, 28ad, 30a, 31a, 37, 38, 55a, 56 (得交13道题) 2. Let P(x) be the statement "The word x contains the letter a." What are these truth values? a) P(orange) b) P(lemon) c) P(true) d) P(false) 3. Let Q(x, y) denote the statement "x is the capital of y." What are these truth values? a) Q(Denver, Colorado) b) Q(Detroit, Michigan) c) Q(Massachusetts, Boston) d) Q(New York, New York) 5. Let P(x) be the statement "x spends more than five hours every weekday in class," where the domain for x consists of all students. Express each of these quantifications in English. a) ∃xP(x) b) ∀xP(x) c) ∃x ¬P(x) d) ∀x ¬P(x) 6. Let N(x) be the statement "x has visited North Dakota," where the domain consists of the students in your school. Express each of these quantifications in English. a) ∃xN(x) b) ∀xN(x) c) ¬∃xN(x) d) ∃x¬N(x) e) ¬∀xN(x) f ) ∀x¬N(x) 9. Let P(x) be the statement "x can speak Russian" and let Q(x) be the statement "x knows the computer language C++." Express each of these sentences in terms of P(x), Q(x), quantifiers, and logical connectives. The domain for quantifiers consists of all students at your school. a) There is a student at your school who can speak Russian and who knows C++ d) No student at your school can speak Russian or knows C++. 10. Let C(x) be the statement "x has a cat," let D(x) be the statement "x has a dog," and let F(x) be the statement "x has a ferret." Express each of these statements in terms of C(x), D(x), F(x), quantifiers, and logical connectives. Let the domain consist of all students in your class. a) A student in your class has a cat, a dog, and a ferret. d) No student in your class has a cat, a dog, and a ferret. 11. Let P(x) be the statement "x = x2." If the domain consists of the integers, what are these truth values? d) P(−1) 12. Let Q(x) be the statement "x + 1 > 2x." If the domain consists of all integers, what are these truth values? d) ∃xQ(x) 13. Determine the truth value of each of these statements if the domain consists of all integers. d) ∀n(3n ≤ 4n) 14. Determine the truth value of each of these statements if the domain consists of all real numbers. d) ∀x(2x > x) 15. Determine the truth value of each of these statements if the domain for all variables consists of all integers. d) ∃n(n2 < 0) 16. Determine the truth value of each of these statements if the domain of each variable consists of all real numbers. d) ∀x(x2 ≠ x) 17. Suppose that the domain of the propositional function P(x) consists of the integers 0, 1, 2, 3, and 4. Write out each of these propositions using disjunctions, conjunctions, and negations. b) ∀xP(x) **18. Suppose that the domain of the propositional function P(x) consists of the integers −2, −1, 0, 1, and 2. Write out each of these propositions using disjunctions, conjunctions, and negations. c) ∃x¬P(x) **19. Suppose that the domain of the propositional function P(x) consists of the integers 1, 2, 3, 4, and 5. Express these statements without using quantifiers, instead using only negations, disjunctions, and conjunctions. d) ¬∀xP(x) 20. Suppose that the domain of the propositional function P(x) consists of −5, −3, −1, 1, 3, and 5. Express these statements without using quantifiers, instead using only negations, disjunctions, and conjunctions. a) ∃xP(x) 23. Translate in two ways each of these statements into logical expressions using predicates, quantifiers, and logical connectives. First, let the domain consist of the students in your class and second, let it consist of all people. b) Everyone in your class is friendly 24. Translate in two ways each of these statements into logical expressions using predicates, quantifiers, and logical connectives. First, let the domain consist of the students in your class and second, let it consist of all people. c) There is a person in your class who cannot swim. 25. Translate each of these statements into logical expressions using predicates, quantifiers, and logical connectives. a) No one is perfect. d) At least one of your friends is perfect. 28. Translate each of these statements into logical expressions using predicates, quantifiers, and logical connectives. a) Something is not in the correct place. d) Nothing is in the correct place and is in excellent condition. 30. Suppose the domain of the propositional function P(x, y) consists of pairs x and y, where x is 1, 2, or 3 and y is 1, 2, or 3. Write out these propositions using disjunctions and conjunctions. a) ∃x P(x, 3) 31. Suppose that the domain of Q(x, y, z) consists of triples x, y, z, where x = 0, 1, or 2, y = 0 or 1, and z = 0 or 1. Write out these propositions using disjunctions and conjunctions. a) ∀yQ(0, y, 0) 37. Find a counterexample, if possible, to these universally quantified statements, where the domain for all variables consists of all integers. a) ∀x(x2 ≥ x) b) ∀x(x > 0 ∨ x < 0) c) ∀x(x = 1) 38. Find a counterexample, if possible, to these universally quantified statements, where the domain for all variables consists of all real numbers. a) ∀x(x2 ≠ x) b) ∀x(x2 ≠ 2) c) ∀x(|x| > 0) 55. What are the truth values of these statements? a) ∃!xP(x) → ∃xP(x) **56. Write out ∃!xP(x), where the domain consists of the integers 1, 2, and 3, in terms of negations, conjunctions, and disjunctions.

2. Let P(x) be the statement "The word x contains the letter a." What are these truth values? a) P(orange) = T b) P(lemon) = F c) P(true) = F d) P(false) = T 3. Let Q(x, y) denote the statement "x is the capital of y." What are these truth values? a) Q(Denver, Colorado) = T b) Q(Detroit, Michigan) = T c) Q(Massachusetts, Boston) = F d) Q(New York, New York) = F 5. Let P(x) be the statement "x spends more than five hours every weekday in class," where the domain for x consists of all students. Express each of these quantifications in English. a) ∃xP(x) = There exists a student who spends more than 5 hours每个weekday in class b) ∀xP(x) = Every student spends more than five hours每个weekday in class c) ∃x ¬P(x) = There exists a student who DOESN'T spend 5+ hours每个weekday in class --> Negation of (b) d) ∀x ¬P(x) = NO students spend more than 5 hours每个weekday in class --> Negation of (a) 6. Let N(x) be the statement "x has visited North Dakota," where the domain consists of the students in your school. Express each of these quantifications in English. a) ∃xN(x) = There is a student who has visited ND b) ∀xN(x) = All students have visited ND c) ¬∃xN(x) = NO students have visited ND --> Negation = Vx!N(x) d) ∃x¬N(x) = There is a student who has NOT visited ND e) ¬∀xN(x) = There is a student who has NOT visited ND --> Negation = ∃x!N(x) f) ∀x¬N(x) = NO students have visited ND 9. Let P(x) be the statement "x can speak Russian" and let Q(x) be the statement "x knows the computer language C++." Express each of these sentences in terms of P(x), Q(x), quantifiers, and logical connectives. The domain for quantifiers consists of all students at your school. a) There is a student at your school who can speak Russian and who knows C++ --> P(x)=russian; Q(x)=C++ --> *∃x(P(x) n Q(x))* d) No student at your school can speak Russian OR knows C++. -记住：或与和的区别不是虚的！！！说"或"有缘故，仔细看题目！ --> *Vx!(P(x) v Q(x))* --> (=!∃x等等等) 10. Let C(x) be the statement "x has a cat," let D(x) be the statement "x has a dog," and let F(x) be the statement "x has a ferret." Express each of these statements in terms of C(x), D(x), F(x), quantifiers, and logical connectives. Let the domain consist of all students in your class. a) A student in your class has a cat, a dog, and a ferret. --> ∃x(C(x) n D(x) n F(x)) d) No student in your class has a cat, a dog, and a ferret. --> Vx!(C(x) n D(x) n F(x)) (=!∃x等等等) 11. Let P(x) be the statement "x = x^2." If the domain consists of the integers, what are these truth values? d) P(−1) = FALSE 12. Let Q(x) be the statement "x + 1 > 2x." If the domain consists of all integers(无decimals), what are these truth values? d) ∃xQ(x) = ∃x(x + 1 > 2x) --> one x for which x+1>2x --> TRUE! x=-1 --> -1+1 > 2(-1) --> 0 > -2 13. Determine the truth value of each of these statements if the domain consists of all integers. d) ∀n(3n ≤ 4n) = for all n, (3n <= 4n) --> FALSE! n=-1 --> -3 >= -4 14. Determine the truth value of each of these statements if the domain consists of all real numbers(有decimals). d) ∀x(2x > x) = For all x, (2x>x) --> FALSE! x=-1 --> -2 < -1 15. Determine the truth value of each of these statements if the domain for all variables consists of all integers(没decimals). d) ∃n(n^2 < 0) = there is one n for which n2<0 --> 非！但凡被......^2的都》0！ 16. Determine the truth value of each of these statements if the domain of each variable consists of all real numbers(decimals). d) ∀x(x^2 ≠ x) = for every x, x^2=/=x --> 非! 1^2 = 1! 17. Suppose that the domain of the propositional function P(x) consists of the integers 0, 1, 2, 3, and 4. Write out each of these propositions using disjunctions, conjunctions, and negations. b) ∀xP(x) = for every x, P(x) --> AND --> *(P(0) n P(1) n P(2) n P(3) n P(4))* 18. Suppose that the domain of the propositional function P(x) consists of the integers −2, −1, 0, 1, and 2. Write out each of these propositions using disjunctions, conjunctions, and negations. c) ∃x!P(x) = !VxP(x) = exists value x such that !P(x)=TRUE --> !Vx(P(-2) n P(-1) n P(0) n P(1) n P(2)) = --> *!P(-2) v !P(-1) v !P(0) v !P(1) v !P(2)* 19. Suppose that the domain of the propositional function P(x) consists of the integers 1, 2, 3, 4, and 5. Express these statements without using quantifiers, instead using only negations, disjunctions, and conjunctions. d) ¬∀xP(x) = negation is ∃x!(P(x)) = There exists value such that !P(x)=true. （因为是V所以=所有=AND！） --> *!P(1) v !P(2) v !P(3) v !P(4) v !P(5)* 20. Suppose that the domain of the propositional function P(x) consists of −5, −3, −1, 1, 3, and 5. Express these statements without using quantifiers, instead using only negations, disjunctions, and conjunctions. a) ∃xP(x) = for one x, P(x)=T (this OR that, bc there exists at least one!) --> P(-5) v P(-3) v P(-1) v P(1) v P(3) v P(5) 23. Translate in two ways each of these statements into logical expressions using predicates, quantifiers, and logical connectives. First, let the domain consist of the students in your class and second, let it consist of all people. b) Everyone in your class is friendly F(x)=all my class students friendly S(x)=students are in my class --> VxF(x) --> Vx(S(x) --> F(x)) - For all x, if they are student, then they are friendly (因为是所有) - If existential (someone in class is friendly) 就简单滴 = ∃x(F(x) n S(x)) --> 只用 if/then 于说 ALL bc it implies FOR ALL x, IF they are x, then they are y! 24. Translate in two ways each of these statements into logical expressions using predicates, quantifiers, and logical connectives. First, let the domain consist of the students in your class and second, let it consist of all people. c) There is a person in your class who cannot swim. W(x)=student can swim S(x)=students are in my class --> ∃x!S(x) --> ∃x(S(x) n !W(x)) - There is one x wherein=student in my class AND cannot swim 25. Translate each of these statements into logical expressions using predicates, quantifiers, and logical connectives. a) No one is perfect. = Vx!P(x) d) At least one of your friends is perfect. = ∃x(F(x) n P(x)) --> If my friend, 那至少有一个是完美滴 --> F(x)=friend; P(x)=perfect 28. Translate each of these statements into logical expressions using predicates, quantifiers, and logical connectives. a) Something is not in the correct place. = ∃x!C(x) --> C(x)=correct place d) Nothing (not one thing) is in the correct place NOR in excellent condition. = !∃x(C(x) n E(x)) --> C(x)=correct place --> E(x)=excellent condition 30. Suppose the domain of the propositional function P(x, y) consists of pairs x and y, where x is 1, 2, or 3 and y is 1, 2, or 3. Write out these propositions using disjunctions and conjunctions. a) ∃x P(x, 3) = there exists x such that y=3 --> P(1,3) v P(2,3) v P(3,3) 31. Suppose that the domain of Q(x, y, z) consists of triples x, y, z, where x = 0, 1, or 2, y = 0 or 1, and z = 0 or 1. Write out these propositions using disjunctions and conjunctions. a) ∀yQ(0, y, 0) = all y ==> 0,y,0 --> ALL y values for which 0,y,0 = true --> AND! --> Q(0,0,0) n Q(0,1,0) - 记住: NO quantifiers when they ask for just disjunctions/conjunctions! 37. Find a counterexample, if possible, to these universally quantified statements, where the domain for all variables consists of all integers(无decimals). a) ∀x(x^2 ≥ x) --> 🈚️反驳：不含有decimal，所以算是。 b) ∀x(x>0 ∨ x<0) --> x=0 --> counterexample! c) ∀x(x = 1) --> x=2 --> 2=/=1 38. Find a counterexample, if possible, to these universally quantified statements, where the domain for all variables consists of all real numbers(有decimals). a) ∀x(x^2 ≠ x) --> 1^2=1! b) ∀x(x^2 ≠ 2) --> x=(2^0.5) --> (2^0.5)^2 = 2! c) ∀x(|x| > 0) --> x=0 55. What are the truth values of these statements? a) ∃!xP(x) → ∃xP(x) = If for exactly ONE x, P(x)=T --> then there is an x, P(x)=T --> T! ??56. Write out ∃!xP(x), where the domain consists of the integers 1, 2, and 3, in terms of negations, conjunctions, and disjunctions. --> Exactly ONE x--> P(x) --> [P(1) v P(2) v P(3)] n ![P(1) n P(2)] n ![P(2) n P(3)] n ![P(1) n P(3)] - 不懂

Rules of Inference for Quantified Statements Universal instantiation: If VxP(x) is true, then P(c) or any particular specialization of this = TRUE predicate NOT the same as saying it's a tautology Really, this means in propositional logic that "The statement VxP(x) → P(c) is TRUE whenever "c" belongs to the domain! Universal generalization: If P(x) for arbitrary "c", then you can deduce the proposition that VxP(x) is TRUE Existential instantiation: If ∃xP(x) is true proposition, then some element c in domain P(c) is TRUE Existential generalization: P(c) is true, then ∃xP(x) is TRUE!

Rules of Inference for Quantified Statements Universal instantiation - 解放所以能更方便措: Given VxP(c): VxP(x) = P(c) Vx(P(x)-->Q(x)) = P(c)-->Q(c) 基本上把里头的东西掏出来=univinst Universal generalization: Given P(c): P(c)=VxP(x) P(c)-->Q(c) = Vx(P(x)-->Q(x)) 基本上把东西塞回去=univgen Existential instantiation - 解放所以能更方便措: If ∃xP(x) is true proposition, then some element c in domain P(c) is TRUE Existential generalization: P(c) is true, then ∃xP(x) is TRUE!

a: What are Inference Rules (八)? b: What is p-->q equal to?

a: Modus Poneas: valid inference rule (p→ q AND p imply q) (a+b=c is true) p→ q p = q --> INCORRECT USE OF MODUS PONEAS: Take the following two premises: Assuming "2^0.5 > 3/2" and "If 2^0.5 > 3/2", then (2^0.5)^2 > (3/2)^2. So conclude thant 2 > (9/4) Valid inference proceduce...where did we go wrong? INCORRECT inference! → p is FALSE in this case! 2) Modus Tollens: if q is false and p→ q is true, than p is false (a+b=c) Nq p→ q = Np 3) Hypothetical Syllogism: (a+b=c is true) assuming a and b are true p→ q q→ r = p→ r 4) Disjunctive Syllogism: (a+b=c is true) assuming a and b are true Pvq Np = Q 5) Addition: (a+b=c is true) assuming a and b are true P = Pvq 6) Simplification: (a+b=c is true) assuming a and b are true Pnq = P 7) Conjunction: (a+b=c is true) assuming a and b are true P Q = Pnq 8) Resolution Syllogism: (a+b=c is true) assuming a and b are true Pvq Np v r = Qvr is true b: p-->q = !p v q --> 只要不是TF --> 都是T！

Memorize: tables 6, 7.1, PROVE using successive substitution (basic algebra): p ←> q 三 (p-->q) n (q→ p) (table 8.1)

三(!p v q) n (!q v p) ← table 7.1 三(!pvq) n !q v (!pvq) n p ← distributativity 三(!p n !q) v (q n !q) v (!p n p) v (qnp) 三!p n !q v F v F v pnq ← negation law 三!p n !q v (pnq) ← identity law 三(pnq) v (!p n !q) ← commutativity