course mth 151 don't really understand this section Ӌzg}͈assignment #017
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11:38:53 `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 --> STATEMENT WOULD BE false if r was false. p q r t t t t f t f t t f f t ( if r is true) confidence assessment: 0
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11:39:00 [ (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 --> ok self critique assessment: 1
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11:40:59 `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 --> answer is invalid because of the R column now having a ""false"" in it confidence assessment: 0
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11:47:17 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 --> ok self critique assessment: 0
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12:00:28 `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 second statement is false if the truth values are TTF., because R is false , if the truth vale for R was true, then the statement would be true. confidence assessment: 0
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12:00:46 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 --> ok self critique assessment: 0
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12:06:39 `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 --> for TFF the staetement is false FTF the statement is false FFF the staement is true confidence assessment: 0
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12:06:57 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 --> OK self critique assessment: 0
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12:10:30 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 --> by adding ""therefore R"" to the equation,it makes everything true. self critique assessment: o
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12:12:05 `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 --> not sure confidence assessment: 0
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12:14:35 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 --> ok self critique assessment: 0
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12:26:30 `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 = if it snows Q = roads slippery R = safer to drive on by plugging in the representations for (R,Q,P) the statement: [ (p ->q) ^ (q ->r)^ p] ->r will be true. confidence assessment: 0
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12:28:50 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 --> ok self critique assessment: 1
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12:39:29 `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 = it doesn't rain q = picnic ~p->q = if it doesn't rain, there is a picnic ~q ->p = there is no picnic, therefore it rained. ~q = no picnic p = it rained (~p ->q ) (~q ->p ) confidence assessment: 0
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12:41:53 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 --> forgot truth table self critique assessment: 1
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