course Mth 151 Ÿ²êòºÂ“EÅ‘°íôí˜ôÝé¨ÉÉassignment #017
......!!!!!!!!...................................
13:37:35 `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.
......!!!!!!!!...................................
RESPONSE --> r would be true in ""since T -> T"" and it would be true in ""since F -> T"" because both statements are true. confidence assessment: 2
.................................................
......!!!!!!!!...................................
13:39:50 [ (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.
......!!!!!!!!...................................
RESPONSE --> Since the truth valuse for r is true then the truth values would be true. self critique assessment: 2
.................................................
......!!!!!!!!...................................
13:59:17 `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?
......!!!!!!!!...................................
RESPONSE --> it would be false because ""since T -> F"" is false confidence assessment: 2
.................................................
......!!!!!!!!...................................
14:12:09 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.
......!!!!!!!!...................................
RESPONSE --> I did not work through the table close enough. I came out with false for the statements but when I plugged it in the whole statement, I got it backwards some how. self critique assessment: 2
.................................................
......!!!!!!!!...................................
14:15:34 `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?
......!!!!!!!!...................................
RESPONSE --> The statement (q->r) was false. It tells us that when one of the statements is false, the truth value for the whole statement will be false. confidence assessment: 2
.................................................
......!!!!!!!!...................................
14:18:17 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.
......!!!!!!!!...................................
RESPONSE --> I got the same part of the statement being false but when I wrote the whole statement I wrote false instead of true. self critique assessment: 2
.................................................
......!!!!!!!!...................................
14:21:52 `q004. Examine the truth of the statement [ (p -> q) ^ (q -> r) ^ p] for each of the truth sets TFF, FTF and FFF.
......!!!!!!!!...................................
RESPONSE --> TFF F T F F FTF T F F F FFF T T T T the one that was all false came out true confidence assessment: 2
.................................................
......!!!!!!!!...................................
14:48:40 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.
......!!!!!!!!...................................
RESPONSE --> I got the first two correct but the last one I got that p -> q and q -> r were both true beause if p was false and q was false doesn't it make the statement true and the same for q -> r. Since both statements are true would (q -> q) ^ (q -> r) be true? making it true????? self critique assessment: 2
.................................................
......!!!!!!!!...................................
14:49:56 `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?
......!!!!!!!!...................................
RESPONSE --> Because if it was false -> true then it is true. confidence assessment: 2
.................................................
......!!!!!!!!...................................
15:03:11 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.
......!!!!!!!!...................................
RESPONSE --> self critique assessment:
.................................................
......!!!!!!!!...................................
15:04:23 `q006. Explain how we have shown in the past few exercises that [ (p -> q) ^ (q -> r) ^ p] -> r must always be true.
......!!!!!!!!...................................
RESPONSE --> If r is true then the statement is true because with it ending in true it makes it true. confidence assessment: 2
.................................................
......!!!!!!!!...................................
15:04:52 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.
......!!!!!!!!...................................
RESPONSE --> If it ends in true then the statement is true. self critique assessment: 2
.................................................
......!!!!!!!!...................................
15:05:56 `q007. Explain how this shows that the original argument about rain, wet grass and smelling wet grass, must be valid.
......!!!!!!!!...................................
RESPONSE --> If you smell the grass when it rains and it does rain then it will be true that you smell the grass. confidence assessment: 2
.................................................
......!!!!!!!!...................................
15:10:15 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.
......!!!!!!!!...................................
RESPONSE --> If it ends with true then it makes it ture. self critique assessment: 2
.................................................
......!!!!!!!!...................................
15:12:42 `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.
......!!!!!!!!...................................
RESPONSE --> ( (p -> q) would be the snow; (q -> r) would be the roads; r would be roads being safe. Since you have all three that are true then the statment has to be valid. confidence assessment: 2
.................................................
......!!!!!!!!...................................
15:13:15 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.
......!!!!!!!!...................................
RESPONSE --> It is valid because when you plug in the statement, it is true. self critique assessment: 2
.................................................
......!!!!!!!!...................................
15:16:01 `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.'
......!!!!!!!!...................................
RESPONSE --> p - rain; q - picnic (~p -> q) ^ (~q - > q) confidence assessment: 2
.................................................
......!!!!!!!!...................................
15:16:37 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
......!!!!!!!!...................................
RESPONSE --> I did not do the truth table, I didn't realize I needed to. self critique assessment: 2
.................................................