asst_17

You did well on these questions, but ran into a little confusion about [ (p -> q) ^ (q -> r) ^ p] -> r. See my notes and let me know if anything remains unclear.

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22:06:59 `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 think that this statement is a tautology which means that all arguments are true.

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22:10:09 [ (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, I missed this one but I understand your explanation. It says that for the statement to be false then r would have to be false and since r is true, then the rest of the statements must be true.

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22:12:44 `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 --> With the truth values TFF, the statement would be false because for the statement to be true, r has to be true.

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22:14:58 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 --> I thought that since one of the statements was false, then the whole thing would be false and this says that TTF makes this statement true. I'm a little confused on this one.

One of the statement in [ (p -> q) ^ (q -> r) ^ p] is false, making [ (p -> q) ^ (q -> r) ^ p] false.

This makes [ (p -> q) ^ (q -> r) ^ p] -> r true, since only a T -> F conditional is false.

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22:18:12 `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 --> With the truth values of TTF, the r is false. This would make the statement false because r would have to be true for the statement to be true.

Not so. If [ (p -> q) ^ (q -> r) ^ p] is false then [ (p -> q) ^ (q -> r) ^ p] -> r is true, no matter whether r is true or false.

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22:19:30 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 --> I think I need to go back and look at this a little closer. I'm confusing this problem with the previous one.

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22:22:36 `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 --> I think this statement would be false because with these truth values, there is no place where every statement is true.

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22:23:38 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, I got this one. I didn't give my answer in as much detail, but this is what I have on paper.

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22:24:55 `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 --> I think this is a tautology and this would mean that it is always true.

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22:27:43 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 --> Ok I looked back to my truth table to see that r is false for this statement. I understan what you have given here.

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22:30:12 `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 --> I'm really not exactly sure why this is so. I sort of understand this but not clearly enough to explain it. I will look over this further.

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22:31:21 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 --> Ok, I understand this explanation. It is saying for this statement, if r is true, then the statement is true, and if r is false, then the statement is false.

Not quite. Even if r is false, the statement is true as long as [ (p -> q) ^ (q -> r) ^ p] is false.

You can't disprove anything from a false statement, so if [ (p -> q) ^ (q -> r) ^ p] is false, [ (p -> q) ^ (q -> r) ^ p] -> r cannot be disproven and is therefore true.

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22:34:30 `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 --> This shows that if it rains (p) then the grass gets wet (q). If the grass gets wet (q), then we can smell the wet grass (r). It rained yesterday (p), therefore we were able to smell the wet grass (r). This statement shows the symbolic statement written into words.

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22:34:42 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

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22:37:56 `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 -> q) ^ (q -> r) ^ p ] -> r This shows that all the statements are true therefore the statement is valid.

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22:38:32 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, I got this one.

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22:42:18 `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 --> I think it would be: [ ( ~p -> q) ^ ~q ] -> p

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22:43:23 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 --> Ok I got this one. I wasn't quite sure that I was symbolizing correctly with the ~ negation.

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