离散数学习题及答案01

CIS 607: Mathematical Basis for ComputingSOLUTIONS OF HOMEWORK 1Homework 1 -- Propositional Logic, Predicate Logic and ProofsDue Date: February 14, 2017In this homework, you will answer the following questions. Prepare a pdf file for your solutions and upload that pdf file into the blackboard system.Q1)Let p and q be the propositionsp : I bought a lottery ticket this week.q : I won the million dollar jackpot.Express each of these propositions as an English sentence.∙p →qo If I bought a lottery ticket this week, then I won the million dollar jackpot on Friday.∙￢p ∨(p ∧q)o Either I did not buy a lottery ticket this week, or else I did buy one and won the million dollar jackpot on Friday.∙￢p →￢qo If I did not buy a lottery ticket this week, then I did not win the million dollar jackpot on Friday.Q2)Let p and q be the propositionsp : It is below freezing.q : It is snowing.Write these propositions using p and q and logical connectives (including negations).∙It is not below freezing and it is not snowing.o(￢p ∧￢q)∙If it is below freezing, it is also snowing.o p →q∙Either it is below freezing or it is snowing, but it is not snowing if it is below freezing.o(p ∨ q) ∧ ( p →￢q)Q3)Construct a truth table for each of these compound propositions.∙p →(￢q ∨r)∙(p →q) ∧(￢p →r)∙(￢p ↔￢q) ↔(q ↔r)∙Show that ￢(p ↔q) and p ↔￢q are logically equivalent.o p q ￢(p ↔q) p ↔￢qo T T F Fo T F T To F T T To F F F F∙Show that (p →q) →r and p →(q →r) are not logically equivalent.o p q r (p →q) →r p →(q →r)o F F F F TQ5)Determine the truth value of each of these statements if the domain consists of all integers.∙∀n(n + 1 > n)o T∙∃n(2n = 3n)o T (when n=0)∙∃n(n = −n)o T (when n=0)∙∀n(3n ≤4n)o F (when n is a negative integer)Q6)Suppose that the domain of the propositional function P(x) consists of the integers 1, 2, and 3. Write out each of these propositions using disjunctions, conjunctions, and negations.∙∃xP(x)o P(1) ∨P(2) ∨P(3)∙∀xP(x)o P(1) ∧P(2) ∧P(3)∙￢∃xP(x)o￢ (P(1) ∨P(2) ∨P(3))∙￢∀xP(x)o￢ (P(1) ∧P(2) ∧P(3))Express the negations of each of these statements so that all negation symbols immediately precede predicates.∙∀x∃y∀zT (x, y, z)o∃x∀y∃z￢T (x, y, z)∙∀x∃yP(x, y) ∨∀x∃yQ(x, y)o∃x∀y￢P(x, y) ∧∃x∀y￢Q(x, y)∙∀x∃y(P(x, y) ∧∃zR(x, y, z))o∃x∀y(￢P(x, y) ∨∀z￢R(x, y, z))∙∀x∃y(P(x, y) →Q(x, y))o∃x∀y(P(x, y) ∧￢Q(x, y))Q8)∙Determine whether ∀x(P(x) →Q(x)) and ∀xP(x) →∀xQ(x) are logically equivalent. Justify your answer.o NOTo P: is even number, Q: is odd numbero∀x P(x) →∀xQ(x) is true since ∀xP(x) is false and ∀xQ(x) is falseo But ∀x(P(x) →Q(x)) is false∙Determine whether ∀x(P(x) ↔Q(x)) and ∀x P(x) ↔∀xQ(x) are logically equivalent. Justify your answer.o NOTo P: is even number, Q: is odd numbero∀x P(x) ↔∀xQ(x) is true since ∀xP(x) is false and ∀xQ(x) is falseo But ∀x(P(x) ↔Q(x)) is false∙Show that ∃x(P(x) ∨Q(x)) and ∃xP(x) ∨∃xQ(x) are logically equivalent.o if ∃x(P(x) ∨Q(x)) is trueo P(a) ∨Q(a) is true for a constant ‘a’ (∃-inst)o If P(a) is true →∃xP(x)is true →∃xP(x) ∨∃xQ(x) is trueo If Q(a) is true →∃xQ(x)is true →∃xP(x) ∨∃xQ(x) is trueoo if ∃x(P(x) ∨Q(x)) is falseo→ there is no constant such that P(a) ∨Q(a) is trueo→ there is no constant such that P(a) is true or there is no constant such that Q(a) is trueo→∃xP(x) is false and ∃xQ(x) is falseo→∃xP(x) ∨∃xQ(x) is falseQ9)∙Use rules of inference to show that if ∀x(P(x) →(Q(x) ∧S(x))) and ∀x(P(x) ∧R(x)) are true, then ∀x(R(x) ∧S(x)) is true.Step Reason1.∀x(P(x) → (Q(x) ∧ S(x))) Premise2.∀x(P(x) ∧ R(x)) Premise3.P(a) ∧ R(a) for arbitrary a UI from 24.P(a) for arbitrary a Simplification from 35.R(a) for arbitrary a Simplification from 36.P(a) → (Q(a) ∧ S(a))) for arbitrary a UI from 17.(Q(a) ∧ S(a)) for arbitrary a MP from 4 and 68.S(a) for arbitrary a Simplification from 79.R(a) ∧ S(a) for arbitrary a Conjunction from 5 and 810.∀x(R(x) ∧ S(x)) UG from 9∙Use rules of inference to show that if ∀x(P(x) ∨Q(x)), ∀x(￢Q(x) ∨S(x)), ∀x(R(x)→￢S(x)), and ∃x￢P(x) are true, then ∃x￢R(x) is true.Step Reason1.∀x(P(x) ∨ Q(x)) Premise2.∀x(￢Q(x) ∨ S(x)) Premise3.∀x(R(x) →￢S(x)) Premise4.∃x￢P(x) Premise5.￢P(a) for some a EI from 46.P(a) ∨ Q(a) UI from 17.Q(a) for some a Disjunctive Syllogism from 5 and 68.￢Q(a) ∨ S(a) UI from 29.S(a) for some a Disjunctive Syllogism from 7 and 810.R(a) →￢S(a) UI from 311.￢R(a) for some a Modes Tollens from 9 and 1012.∃x￢R(x) Existential Generalization (EG) from 11Q10)∙Use a direct proof to show that every odd integer is the difference of two squares.Direct Proofo Let n be an odd integer such that n=2k+1 where k is an integer (definition of odd integers) o Let squares of two integers k and (k+1) such as k2 and (k+1)2= k2+2k+1o(k+1)2 - k2 = 2k+1o So, n is the difference of two squares.o QED∙Show that if n is an integer and n3+ 5 is odd, then n is even using a proof by contraposition.Proof by Contrapositiono Assume n is an odd integer (negation of even is odd)o So, n=2k+1 where k is an integer (definition of odd integers)o n3 + 5 = (2k+1)3 + 5 = 8k3+12k2 +6k+1+5 = 2(4k3+6k2 +3k+3)o So, n3 + 5 = 2m where m is an integer such that m = (4k3+6k2 +3k+3)o Thus, n3 + 5 is an even integer (negation of “n3 + 5 is odd”)o Then, n is even (not odd)o QED.∙Prove that if n is an integer and 3n + 2 is even, then n is even using a proof by contradiction.Proof by Contradictiono Assume that n is an odd integer (negation of even is odd)o So, n=2k+1 where k is an integer (definition of odd integers)o3n+2 = 3(2k+1)+2 = 6k+5 = 2(3k+2)+1o So, 3n+2 = 2m+1 where m is an integer such that m=3k+2o Thus, 3n+2 is odd (contradiction with our assumption “3n + 2 is even”)o So, n is eveno QED.。