course MTH 151 This exercise I did not get very well. I don't know why I could not understand it. Sorry for the bad job. assignment #017017. Evaluating Arguments
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17:58:06 `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 --> I'm not sure confidence assessment: 0
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18:04:06 [ (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 --> The truth values ttt,tft,ftt,fft all make the statement true self critique assessment: 2
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18:09:24 `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 --> I'm not sure confidence assessment: 0
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18:12:52 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 --> for ttf we see that one of these statements is false. self critique assessment: 2
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18:13:55 `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 --> I'm not getting this stuff confidence assessment: 0
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18:14:52 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 --> p and q are both true, so p-> q and p are true. The only candidate for a false statement amoung the statements is q-> r self critique assessment: 2
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18:17:15 `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 --> examine and do what confidence assessment: 0
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18:18:29 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 --> tff is false ftf is false fff is false