To prove: That [ p ∨ ( ( ¬ r ) → ( ¬ s ) ) ] ∨ [ s → ( ( ¬ t ) ∨ p ) ∨ ( ( ¬ q ) → r ) ] neither a tautology nor a contradiction.

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Answer 1

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Chapter 1.1, Problem 8E

To determine

To prove: That [p∨((¬r)→(¬s))]∨[s→((¬t)∨p)∨((¬q)→r)] neither a tautology nor a contradiction.

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Use the Euclidean algorithm to find two sets of integers (a, b, c) such that 55a65b+143c: Solution = 1. By the Euclidean algorithm, we have: 143 = 2.65 + 13 and 65 = 5.13, so 13 = 143 – 2.65. - Also, 55 = 4.13+3, 13 = 4.3 + 1 and 3 = 3.1, so 1 = 13 — 4.3 = 13 — 4(55 – 4.13) = 17.13 – 4.55. Combining these, we have: 1 = 17(143 – 2.65) - 4.55 = −4.55 - 34.65 + 17.143, so we can take a = − −4, b = −34, c = 17. By carrying out the division algorithm in other ways, we obtain different solutions, such as 19.55 23.65 +7.143, so a = = 9, b -23, c = 7. = = how ? come [Note that 13.55 + 11.65 - 10.143 0, so we can obtain new solutions by adding multiples of this equation, or similar equations.]

Celina is picking a new frame for a custom piece of artwork. She has to select a frame size, material, and color. There are four different frame sizes, three different frame materials, and six different frame colors. She must chose one option only from each category. How many different possible frames could Celina pick from?

In a company with 80 employees, 60 earn $10.00 per hour and 20 earn $13.00 per hour. Is this average hourly wage considered representative?

Answer 2

Question

Chapter 1.1, Problem 8E

To determine

To prove: That [p∨((¬r)→(¬s))]∨[s→((¬t)∨p)∨((¬q)→r)] neither a tautology nor a contradiction.

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Answer 3

Question

Chapter 1.1, Problem 8E

To determine

To prove: That [p∨((¬r)→(¬s))]∨[s→((¬t)∨p)∨((¬q)→r)] neither a tautology nor a contradiction.

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Answer 4

Question

Chapter 1.1, Problem 8E

To determine

To prove: That [p∨((¬r)→(¬s))]∨[s→((¬t)∨p)∨((¬q)→r)] neither a tautology nor a contradiction.

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Answer 5

Chapter 1.1, Problem 8E

To determine

To prove: That [p∨((¬r)→(¬s))]∨[s→((¬t)∨p)∨((¬q)→r)] neither a tautology nor a contradiction.

Answer 6

Chapter 1.1, Problem 8E

To determine

To prove: That [p∨((¬r)→(¬s))]∨[s→((¬t)∨p)∨((¬q)→r)] neither a tautology nor a contradiction.

Answer 7

Chapter 1.1, Problem 8E

To determine

To prove: That [p∨((¬r)→(¬s))]∨[s→((¬t)∨p)∨((¬q)→r)] neither a tautology nor a contradiction.

Answer 8

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Answer 9

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Answer 10

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Answer 11

Ch. 1.1 - True/False Questions “” means “” Ch. 1.1 - A truth table based on four simple statements...Ch. 1.1 - True/False Questions 2. If is true, then is also...Ch. 1.1 - If p and q are both false, the truth value of...Ch. 1.1 - If pq is false, the truth value of (pq)(pq) is...Ch. 1.1 - pq andqp are logically equivalent.Ch. 1.1 - True/False Questions 7. A statement and its...Ch. 1.1 - (pq)(pq) is a tautology.Ch. 1.1 - True/False Questions 9. If B is a tautology and A...Ch. 1.1 - True/False Questions 10. If A and B are both...

To prove: That [ p ∨ ( ( ¬ r ) → ( ¬ s ) ) ] ∨ [ s → ( ( ¬ t ) ∨ p ) ∨ ( ( ¬ q ) → r ) ] neither a tautology nor a contradiction.

Solution Summary: The author explains that the compound statement which is always true irrespective of the truth values assigned to its variables is a tautology.

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To prove: That [p∨((¬r)→(¬s))]∨[s→((¬t)∨p)∨((¬q)→r)] neither a tautology nor a contradiction.

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