course Mth 151 â¡Oê…ÒØò‰Ýò³Ì–L‹ÏÄf–®[•ÌØó
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10:53:20 `q001. There are 9 questions in this set. Explain why [ (p -> q) ^ (q -> r) ^ p] -> r must be true for every set of truth values for which r is true.
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RESPONSE --> becuase p is true if q is true and q is true if r is true, which means p is true confidence assessment: 2
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10:56:47 [ (p -> q) ^ (q -> r) ^ p] -> r must be true if r is true, since the statement can only be false if [ (p -> q) ^ (q -> r) ^ p] is true while r is false. Therefore the truth values TTT, TFT, FTT, FFT (i.e., all the truth values that have r true) all make the statement true.
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RESPONSE --> so r is the term that makes it true or false self critique assessment: 3
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10:57:32 `q002. At this point we know that the truth values TTT, TFT, FTT, FFT all make the argument [ (p -> q) ^ (q -> r) ^ p] -> r true. What about the truth values TTF?
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RESPONSE --> T T F would make the statement false confidence assessment: 3
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11:00:14 It would be possible to evaluate every one of the statements p -> q, q -> r, etc. for their truth values, given truth values TTF. However we can shortcut the process. We see that [ (p -> q) ^ (q -> r) ^ p] is a compound statement with conjunction ^. This means that [ (p -> q) ^ (q -> r) ^ p] will be false if any one of the three compound statements p -> q, q -> r, p is false. For TTF we see that one of these statements is false, so that [ (p -> q) ^ (q -> r) ^ p] is false. This therefore makes the statement [ (p -> q) ^ (q -> r) ^ p] -> r true.
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RESPONSE --> if the first statement is false, but if r is true the statement is true self critique assessment: 3
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11:07:32 `q003. The preceding statement said that for the TTF case [ (p -> q) ^ (q -> r) ^ p] was false but did not provide an explanation of this statement. Which of the statements is false for the truth values TTF, and what does this tell us about the truth of the statement [ (p -> q) ^ (q -> r) ^ p] -> r?
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RESPONSE --> the false statement is ^p, in this case is saying that i p is false than r is true confidence assessment: 3
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11:12:46 p and q are both true, so p -> q and p are true. The only candidate for a false statement among the three statements is q -> r. So we evaluate q -> r for truth values TTF. Since q is T and r is F, we see that q -> r must be F. This makes [ (p -> q) ^ (q -> r) ^ p] false. Therefore [ (p -> q) ^ (q -> r) ^ p] -> r must be true, since it can only be false and if [ (p -> q) ^ (q -> r) ^ p] is true.
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RESPONSE --> it is q->r that are a false statement. self critique assessment: 3
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11:15:50 `q004. Examine the truth of the statement [ (p -> q) ^ (q -> r) ^ p] for each of the truth sets TFF, FTF and FFF.
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RESPONSE --> the statement is false confidence assessment: 3
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11:18:03 In the case TFF, p is true and q is false so p -> q is false, which makes [ (p -> q) ^ (q -> r) ^ p] false. In the case FTF, p is false, making [ (p -> q) ^ (q -> r) ^ p] false. In the case FFF, p is again false, making [ (p -> q) ^ (q -> r) ^ p] false.
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RESPONSE --> so all the thruths gave a false result for the statement self critique assessment: 3
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11:18:47 `q005. We have seen that for TFF, FTF and FFF the statement [ (p -> q) ^ (q -> r) ^ p] is false. How does this help us establish that [ (p -> q) ^ (q -> r) ^ p] -> r is always true?
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RESPONSE --> becuase when the statement is false, r is true confidence assessment: 3
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11:24:02 The three given truth values, plus the TTF we examined earlier, are all the possibilities where r is false. We see that in the cases where r is false, [ (p -> q) ^ (q -> r) ^ p] is always false. This makes [ (p -> q) ^ (q -> r) ^ p] -> r true any time r is false.
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RESPONSE --> the statement is true always even if the 1st part is false or if r is false self critique assessment: 3
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11:25:05 `q006. Explain how we have shown in the past few exercises that [ (p -> q) ^ (q -> r) ^ p] -> r must always be true.
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RESPONSE --> because if p is true r is true confidence assessment: 3
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11:25:52 We just finished showing that if r is false, [ (p -> q) ^ (q -> r) ^ p] is false so [ (p -> q) ^ (q -> r) ^ p] -> r is true. As seen earlier the statement must also be true whenever r is true. So it's always true.
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RESPONSE --> the statement is always true self critique assessment: 3
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11:27:51 `q007. Explain how this shows that the original argument about rain, wet grass and smelling wet grass, must be valid.
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RESPONSE --> because everytime the grass gets wet by rain we can smell the wet grass, all the actions are connected therefore is either valid or not, since the statement told us that it did rain, we could smell the grass so statement is valid. confidence assessment: 3
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11:28:44 That argument is symbolized by the statement [ (p -> q) ^ (q -> r) ^ p] -> r. The statement is always true. There is never a case where the statement is false. Therefore the argument is valid.
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RESPONSE --> because the wet smell of grass occurs when the grass gets wet by rain the statement is valid self critique assessment: 3
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11:32:53 `q008. Explain how the conclusion of the last example, that [ (p -> q) ^ (q -> r) ^ p] -> r is always a true statement, shows that the following argument is valid: 'If it snows, the roads are slippery. If the roads are slippery they'll be safer to drive on. It just snowed. Therefore the roads are safer to drive on.' Hint: First symbolize the present argument.
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RESPONSE --> p= is it snows q= the roads are slippery r= the roads are safer to drive on p->q= if is snows the roads are slippery q->r= is the roads are slippery they'll be safer to drive on ^p]->r= it just snowed therefore the roads are safer to drive on the statement is valid confidence assessment: 3
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11:33:59 This argument can be symbolized by letting p stand for 'it snows', q for 'the roads are slippery', r for 'the roads are safer to drive on'. Then 'If it snows, the roads are slippery' is symbolized by p -> q. 'If the roads are slippery they'll be safer to drive on' is symbolized by q -> r. 'It just snowed' is symbolized by p. 'The roads are safer to drive on' is symbolized by r. The argument the says that IF [ p -> q, AND q -> r, AND p ] are all true, THEN r is true. In symbolic form this is [ (p -> q) ^ (q -> r) ^ p] -> r. This is the same statement as before, which we have shown to be always true. Therefore the argument is valid.
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RESPONSE --> the statement is true so is valid self critique assessment: 3
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11:37:32 `q009. Symbolize the following argument and show that it is valid: 'If it doesn't rain there is a picnic. There is no picnic. Therefore it rained.'
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RESPONSE --> p= if it doesn't rain q= there is a picnic ~q= there is no picnic ->p [(p->q) ^ ~q]->p confidence assessment: 3
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11:40:46 We could let p stand for 'it rained', q for 'there is a picnic'. The first statement is 'If it doesn't rain there is a picnic', which is symbolized by ~p -> q. The second statement, 'There is no picnic', is symbolized by ~q. The conclusion, 'it rained', is symbolized by p. The argument therefore says IF [ (~p -> q) AND ~q ], THEN p. This is symbolized by [ (~p -> q) ^ ~q ] -> p. We set up a truth table for this argument: p q ~p ~q ~p -> q (~p -> q) ^ ~q [ (~p -> q) ^ ~q ] -> p T T F F T F T T F F T T T T F T T F T F T F F T T F F T
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RESPONSE --> if it doesn't rain should have been represented by ~p self critique assessment: 3
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±pxaÚ×á„«yƒÊ±eáÕ–éÆ«ñ´æ·éÅé™è– assignment #017 017. Evaluating Arguments Liberal Arts Mathematics I 04-12-2009 â¡Oê…ÒØò‰Ýò³Ì–L‹ÏÄf–®[•ÌØó assignment #017 017. Evaluating Arguments Liberal Arts Mathematics I 04-12-2009
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10:53:20 `q001. There are 9 questions in this set. Explain why [ (p -> q) ^ (q -> r) ^ p] -> r must be true for every set of truth values for which r is true.
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RESPONSE --> becuase p is true if q is true and q is true if r is true, which means p is true confidence assessment: 2
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10:56:47 [ (p -> q) ^ (q -> r) ^ p] -> r must be true if r is true, since the statement can only be false if [ (p -> q) ^ (q -> r) ^ p] is true while r is false. Therefore the truth values TTT, TFT, FTT, FFT (i.e., all the truth values that have r true) all make the statement true.
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RESPONSE --> so r is the term that makes it true or false self critique assessment: 3
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10:57:32 `q002. At this point we know that the truth values TTT, TFT, FTT, FFT all make the argument [ (p -> q) ^ (q -> r) ^ p] -> r true. What about the truth values TTF?
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RESPONSE --> T T F would make the statement false confidence assessment: 3
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11:00:14 It would be possible to evaluate every one of the statements p -> q, q -> r, etc. for their truth values, given truth values TTF. However we can shortcut the process. We see that [ (p -> q) ^ (q -> r) ^ p] is a compound statement with conjunction ^. This means that [ (p -> q) ^ (q -> r) ^ p] will be false if any one of the three compound statements p -> q, q -> r, p is false. For TTF we see that one of these statements is false, so that [ (p -> q) ^ (q -> r) ^ p] is false. This therefore makes the statement [ (p -> q) ^ (q -> r) ^ p] -> r true.
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RESPONSE --> if the first statement is false, but if r is true the statement is true self critique assessment: 3
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11:07:32 `q003. The preceding statement said that for the TTF case [ (p -> q) ^ (q -> r) ^ p] was false but did not provide an explanation of this statement. Which of the statements is false for the truth values TTF, and what does this tell us about the truth of the statement [ (p -> q) ^ (q -> r) ^ p] -> r?
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RESPONSE --> the false statement is ^p, in this case is saying that i p is false than r is true confidence assessment: 3
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11:12:46 p and q are both true, so p -> q and p are true. The only candidate for a false statement among the three statements is q -> r. So we evaluate q -> r for truth values TTF. Since q is T and r is F, we see that q -> r must be F. This makes [ (p -> q) ^ (q -> r) ^ p] false. Therefore [ (p -> q) ^ (q -> r) ^ p] -> r must be true, since it can only be false and if [ (p -> q) ^ (q -> r) ^ p] is true.
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RESPONSE --> it is q->r that are a false statement. self critique assessment: 3
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11:15:50 `q004. Examine the truth of the statement [ (p -> q) ^ (q -> r) ^ p] for each of the truth sets TFF, FTF and FFF.
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RESPONSE --> the statement is false confidence assessment: 3
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11:18:03 In the case TFF, p is true and q is false so p -> q is false, which makes [ (p -> q) ^ (q -> r) ^ p] false. In the case FTF, p is false, making [ (p -> q) ^ (q -> r) ^ p] false. In the case FFF, p is again false, making [ (p -> q) ^ (q -> r) ^ p] false.
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RESPONSE --> so all the thruths gave a false result for the statement self critique assessment: 3
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11:18:47 `q005. We have seen that for TFF, FTF and FFF the statement [ (p -> q) ^ (q -> r) ^ p] is false. How does this help us establish that [ (p -> q) ^ (q -> r) ^ p] -> r is always true?
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RESPONSE --> becuase when the statement is false, r is true confidence assessment: 3
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11:24:02 The three given truth values, plus the TTF we examined earlier, are all the possibilities where r is false. We see that in the cases where r is false, [ (p -> q) ^ (q -> r) ^ p] is always false. This makes [ (p -> q) ^ (q -> r) ^ p] -> r true any time r is false.
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RESPONSE --> the statement is true always even if the 1st part is false or if r is false self critique assessment: 3
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11:25:05 `q006. Explain how we have shown in the past few exercises that [ (p -> q) ^ (q -> r) ^ p] -> r must always be true.
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RESPONSE --> because if p is true r is true confidence assessment: 3
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11:25:52 We just finished showing that if r is false, [ (p -> q) ^ (q -> r) ^ p] is false so [ (p -> q) ^ (q -> r) ^ p] -> r is true. As seen earlier the statement must also be true whenever r is true. So it's always true.
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RESPONSE --> the statement is always true self critique assessment: 3
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11:27:51 `q007. Explain how this shows that the original argument about rain, wet grass and smelling wet grass, must be valid.
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RESPONSE --> because everytime the grass gets wet by rain we can smell the wet grass, all the actions are connected therefore is either valid or not, since the statement told us that it did rain, we could smell the grass so statement is valid. confidence assessment: 3
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11:28:44 That argument is symbolized by the statement [ (p -> q) ^ (q -> r) ^ p] -> r. The statement is always true. There is never a case where the statement is false. Therefore the argument is valid.
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RESPONSE --> because the wet smell of grass occurs when the grass gets wet by rain the statement is valid self critique assessment: 3
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11:32:53 `q008. Explain how the conclusion of the last example, that [ (p -> q) ^ (q -> r) ^ p] -> r is always a true statement, shows that the following argument is valid: 'If it snows, the roads are slippery. If the roads are slippery they'll be safer to drive on. It just snowed. Therefore the roads are safer to drive on.' Hint: First symbolize the present argument.
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RESPONSE --> p= is it snows q= the roads are slippery r= the roads are safer to drive on p->q= if is snows the roads are slippery q->r= is the roads are slippery they'll be safer to drive on ^p]->r= it just snowed therefore the roads are safer to drive on the statement is valid confidence assessment: 3
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11:33:59 This argument can be symbolized by letting p stand for 'it snows', q for 'the roads are slippery', r for 'the roads are safer to drive on'. Then 'If it snows, the roads are slippery' is symbolized by p -> q. 'If the roads are slippery they'll be safer to drive on' is symbolized by q -> r. 'It just snowed' is symbolized by p. 'The roads are safer to drive on' is symbolized by r. The argument the says that IF [ p -> q, AND q -> r, AND p ] are all true, THEN r is true. In symbolic form this is [ (p -> q) ^ (q -> r) ^ p] -> r. This is the same statement as before, which we have shown to be always true. Therefore the argument is valid.
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RESPONSE --> the statement is true so is valid self critique assessment: 3
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11:37:32 `q009. Symbolize the following argument and show that it is valid: 'If it doesn't rain there is a picnic. There is no picnic. Therefore it rained.'
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RESPONSE --> p= if it doesn't rain q= there is a picnic ~q= there is no picnic ->p [(p->q) ^ ~q]->p confidence assessment: 3
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11:40:46 We could let p stand for 'it rained', q for 'there is a picnic'. The first statement is 'If it doesn't rain there is a picnic', which is symbolized by ~p -> q. The second statement, 'There is no picnic', is symbolized by ~q. The conclusion, 'it rained', is symbolized by p. The argument therefore says IF [ (~p -> q) AND ~q ], THEN p. This is symbolized by [ (~p -> q) ^ ~q ] -> p. We set up a truth table for this argument: p q ~p ~q ~p -> q (~p -> q) ^ ~q [ (~p -> q) ^ ~q ] -> p T T F F T F T T F F T T T T F T T F T F T F F T T F F T
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RESPONSE --> if it doesn't rain should have been represented by ~p self critique assessment: 3
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±pxaÚ×á„«yƒÊ±eáÕ–éÆ«ñ´æ·éÅé™è– assignment #017 017. Evaluating Arguments Liberal Arts Mathematics I 04-12-2009"