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# Answers to testable ch2 exercises and practice test for exam 1 (30%)

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University of Toronto St. George

Philosophy

PHL246H1

Colin Howson

Winter

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Exercise Answers. Chapter Two. I will have to use O as I cannot find the upside down T in my symbols. Sorry if it looks hard to read because of this compromise. I hope it is legible enough. 1. Show that P(O) = 0 O and T are mutually exclusive. If A and ~A are mutually exclusive, then Av~A T [from above the exercise]. Let A be T and ~A be O. T v O T From rule one, we get that P(TvO)=P(T) Using rule 4, we can expand the left side to get P(T)+P(O)=P(T) Rule three says that P(T)=1 so we substitute for 1+P(O)=1 Which through simple subtractionbalancing gets us P(O)=0 QED 2. Show that for these sorts of questions, you will want to appeal to some sort of entailment occurring i. P(A) P(AvB) Any sentence A entails AvB [from first chapter]. Using the claim of monotonicity (see the text above the exercise in the chapter), we can see that since A=>AvB, P(A) P(AvB). ii. P(A^B) P(A) A^B entails A [again, from chapter one]. Using the same claim, since A^B=>A, P(A^B) P(A) iii. P(A) 1 A entails T [its from chapter one, seeing a trend]. Since A entails T, P(A) P(T). But wait! Theres more! Since rule three states that P(T)=1, then we can substitute to get P(A) 1. iv. 0 P(A) 1 O entails A [chapter one, yo]. Since is entails A, P(O) P(A). But we already gained the valuable knowledge that P(O)=0 from exercise one, so, 0 P(A). Since these are true for all sentences, and since we discovered in the above exercise that P(A) 1, we can incorporate that here but putting both halves together and saying that 0 P(A) 1. www.notesolution.com

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