It was claimed that: (a, b) ≤ (c, d) ⇔ (a < c) ∨ (a = c ∧ b ≤ d) defines a well-ordering on N x N. Show that this is actually the case.
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It was claimed that:
(a, b) ≤ (c, d) ⇔ (a < c) ∨ (a = c ∧ b ≤ d) defines a well-ordering on N x N. Show that this is actually the case.
Step by step
Solved in 3 steps
- Match the following sentence to the best suitable answer: - A. B. C. D. for the linear congruence ax=1(mod m), x is the inverse of a, if__________ - A. B. C. D. What is -4 mod 9 ? - A. B. C. D. The solution exists for a congruence ax=b(mod m) such that GCD(a,m)=1 and - A. B. C. D. (107+22)mod 10 is equivalent to :_________ A. 5 B. c divides b C. GCD(a,m)=1 D. 9 mod 10Let R=ABCDEGHK and F= {ABK→C, A→DG, B→K, K→ADH, H→GE} . Is it in BCNF? Prove your answer.If sets A={10,20,30,40} B ={30,40,50,60} C= {50,60,70} U = {10,20,30,40,50,60,70}What is the cardinality of A? DO NOT FORGET TO ENCLOSE YOUR ANSWERS IN A BRACKET { } IF IT IS A SET.
- Using the code in the picture (Phyton 3): Find the Recurrence relation for foo(a, b) when b > 0 (Follow the format) T(n) = ___ T ( ___ / ___ ) + O ( ___ ) What is the worst-case time complexity of foo(a, b)? What is the worst-case auxiliary space complexity of foo(a,b)?Given set S and T. Both S and T are not TM-decidable but both sets are TM-recognizable. Is the set ~T TM-recognizable? Explain. is the set (S intersect T) TM-recognizable ExplainCan please help me with this problem and can you explain step by step leading up to the solution. Thank you question that i NEED HELP WITH ; Show that NP is closed under union and concatenation.
- For f (a, b) = (a | b) | b(a) Simplify f (a, b).(b) Find DNF for f (a, b).(c) Is f (a, b) satisfiable?Use the Time-Hierarchy Theorem to prove that P!= EXP.Given the following sets: A = {7, 5} B = {1, 2, 3, 4, 5} C = {1, 7} Find C x B x A Question: what is the cardinality of the resulting cartesian product?
- Prove that equivalences of sets are transitive, for example if X ~ Y and Y ~ Z then X ~ Z.) Show that ∀xP(x) ∧ ∃xQ(x) is logically equivalent to ∀x∃y(P(x) ∧ P(y)) The quantifiers have the same non empty domain. I know that to prove a proposition is logically equivalent to another one, I have to show that ∀xP(x) ∧ ∃xQ(x) ↔ ∀x∃y(P(x) ∧ P(y)) Which means I have to prove that (∀xP(x) ∧ ∃xQ(x)) → ∀x∃y(P(x) ∧ P(y)) ∧ ∀x∃y(P(x) ∧ P(y)) → (∀xP(x) ∧ ∃xQ(x)) I don't know the answer, so I saw the textbook answer. It says (1) Suppose that ∀xP(x) ∧ ∃xQ(x) is true. Then P(x) is true for all x and there is an element y for which Q(y) is true. I get this part. Because P(x) ∧ Q(x) is true for all x and there is a y for which Q(y) is true, ∀x∃y(P(x) ∧ P(y)) is true. Emm... I think ∀x∃y(P(x) ∧ P(y)) is true because ∀x only affects P(x) and ∃y only affects P(y) since their alphabets are different. So, it has the exact same meaning as ∀xP(x) ∧ ∃yQ(y). And since the domains are the same, ∀xP(x) ∧ ∃yQ(y) is actually equal to ∀xP(x) ∧ ∃xQ(x). But the textbook states that "P(x) ∧ Q(x) is…The room temperature x in Fahrenheit (F) is converted to y in Celsius (C) through the function y = f(x) = 5(x-32)/9. Let a fuzzy set B1 (in Fahrenheit) be defined by B1 = 0.15/76 + 0.42/78 + 0.78/80 + 1.0/82 + 1.0/84 What is the induced fuzzy set of B1 in terms of the extension principle? B2 = ?