Which of the statements cannot possibly be true when p: n is an odd integer and q: n is an even integer. ¬p∧q p∧q ¬(p∧q) ¬p∧¬q p∧¬q
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Which of the statements cannot possibly be true when p: n is an odd integer and q: n is an even integer.
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¬p∧q
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p∧q
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¬(p∧q)
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¬p∧¬q
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p∧¬q
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- Which of the following statements are true. Do not show your explanations. (j) A cycle Cn is bipartite if and only if n is even.Recall that the biconditional p↔q stands for(p→q)∧(q→p). Construct a truth table to verify that thestatement p↔q is equivalent to the statement (p∧q)∨((¬p)∧(¬q)).Prove using the concept of interpretations and the value of a formula vI under an interpretation I that [∀x p(x) ∨ ∃x q(x)] → ∃x[p(x) ∨ q(x)] is valid.
- Q1 The periodic function sin(2x) has multiple roots between x values of -5π and 5π. If xL = -15 and xU = 15, which of the following statements is true using a bracketed method? Select one: a. All roots will be returned b. The middle root will be returned c. The chosen bracket is invalid for bracketed methods d. A single root will be returned e. The algorithm will be stuck in an infinite loop Q2 Consider x and y to represent data points (xi,yi), where i = 1, 2, 3, … n. What is the length of pafter running the following command? p = polyval(x,y) Select one: a. n b. n - 1 c. n + 1 d. Empty variable e. 1 Q3 Consider a system of linear equations in the form of AX = B, where X is the unknown vector. Which of the following can be used to solve for X? Select one: a. X = A\B b. X = B./A c. X = inv(B)*A d. X = inv(A)./B e. X = B\AAlgorithm to An iterative solution to Towers of Hanoi.in: triplet S = s0, s1, s2 representing the current game stateout: triplet R = r0, r1, r2 representing the new game statelocal: pole indices a, b, z ∈ {0, 1, 2}; disc numbers g, h ∈ [2, n]; last(Q) = Q|Q|−1, if1 ≤ |Q|, otherwise, last(Q) = +∞Write an algorithm, called Decomposition_Powers_Three, which produces thedecomposition of each integer using powers of 3, namely 1, 3, 9, 27, and 81, and the +and – operators. Each power of 3 should appear at most once in the decomposition.Examples: 1 = 1 2 = 3 – 1 3 = 3 4 = 3 + 1 7 = 9 – 3 + 1 14 = 27 – 9 – 3 – 1 43 = 81 – 27 – 9 – 3 + 1 121 = 81 + 27 + 9 + 3 + 1
- dont dont dont copy existing one strict waring answer only sure else skip1): ¬( p ∨ q ) ≡ ¬p ∧ ¬q The above law is called Group of answer choices De Morgans Law Absorption Law Complement Law Double Negation Law 2): To prove the this identity ¬p → (q → r) ≡ q → (p ∨ r) The following reasoning is valid: ¬p → (q → r)¬¬p ∨ (q → r) Conditional identity p ∨ (q → r) Double negation law p ∨ (¬q ∨ r) Conditional identity (p ∨ ¬q) ∨ r Associative law (¬q ∨p) ∨ r Commutative law ¬q ∨ (p ∨ r) Associative law q → (p ∨ r) Conditional identity Group of answer choices True False2) (L2) Prove using laws of logic that the conditional proposition (p ∧ q) → r is equivalent to (p ∧ ¬ r) →¬ q. 3) (L3) Show that the converse of a conditional proposition p: q → r is equivalent to the inverse of proposition p using a truth table. 4.1) (L4) Show whether ((p ∧ (p→q)) ↔ ¬p) is a tautology or not. Use a truth table and be specific about which row(s)/column(s) of the truth table justify your answer. 4.2) (L4) Give truth values for the propositional variables that cause the two expressions to have different truth values. For example, given p ∨ q and p ⊕ q, the correct answer would be p = q = T, because when p and q are both true, p ∨ q is true but p ⊕ q is false. Note that there may be more than one correct answer. r ∧ (p ∨ q) (r ∧ p) ∨ q
- Prove or disprove "Suppose that m and n are integers. If m > n ≥ 0, then gcd (m, n) = gcd (m − n, n)."Prove or disprove that the two propositions in each pair are equivalent. (p (q r)) , ((p q ) ( p r )) (p (q r)) , (( p q ) ( p r )) (( p q ) r ) , ( p ( q r))Assume the propositions p, q, r, and s have the following truth values:p : False (F). q : True (T). r : False (F). s : True (T). Evaluate the truth value for each of the following compound propositions. Show your work for yourevaluation.a) (p ∨ q) ∧ ¬ rb) p ∨ q ∧ ¬ rc) ¬(r ∨ s) → p ∨ qd) r ⊕ ¬s ↔ q → pe) (p ∧ ¬q) ⊕ (r → (¬s ∨ q))