#$&* course mth151 11/25 around 7:40 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: [ (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. You could make a table, which would be useful in understanding the above explanation. STUDENT COMMENT: I still don't quite grasp this. Is this the same thing as the table? INSTRUCTOR RESPONSE: On any question where you don't understand the given solution, you should break the given explanation up into phrases and tell me what you do and do not understand about each. For example, on this problem you might break the explanation up as follows: [ (p -> q) ^ (q -> r) ^ p] -> r must be true if r is true (Do you understand what this is saying? Do you understand why it must be so? Exactly what do you understand and what do you not understand about this statement?) the statement can only be false if [ (p -> q) ^ (q -> r) ^ p] is true while r is false. (Do you understand what this is saying? Do you understand why it must be so? Exactly what do you understand and what do you not understand about this statement?) TTT, TFT, FTT, FFT are all the truth values that have r true (Do you understand what this is saying? Do you understand why it must be so? Exactly what do you understand and what do you not understand about this statement?) the truth values TTT, TFT, FTT, FFT (i.e., all the truth values that have r true) all make the statement true. (Do you understand what this is saying? Do you understand why it must be so? Exactly what do you understand and what do you not understand about this statement?) [ (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. (Do you understand what this is saying? Do you understand why it must be so? Exactly what do you understand and what do you not understand about this statement?) Now putting it all together: [ (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. (Do you understand what this is saying? Do you understand why it must be so? Exactly what do you understand and what do you not understand about this statement?) STUDENT COMMENT: so r is the term that makes it true or false ------------------------------------------------ Self-critique Rating:ent: 3 INSTRUCTOR RESPONSE: The consequent r does by itself does not necessarily determine the truth of the statement. If r is true, then the statement is true. However if r is false then the statement might be true or false. If the conclusion r is false, then if the antecedent (in this case [ (p -> q) ^ (q -> r) ^ p]) is true the statement is false. However if the antecedent is false, then the statement is true, despite the fact that r is false. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: p q r p->q q->r T T F T F since the compound statement in the [] contains ^, we know that if even one of the statements within the compound statement is false then [(p -> q) ^ (q -> r) ^ p] would be false we know that q->r is false this makes the argument F->F (since [(p -> q) ^ (q -> r) ^ p] is false and r is F) since ""if"" ""then"" statements are only false if T->F, the argument is True confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. STUDENT COMMENT I’m still confused as to why one statement being false makes the entire situation true, but I think I will figure it out. INSTRUCTOR RESPONSE If the statement is A -> B, then if A is false, the statement is true. This is because the only way for A -> B to be false is for A to be true and B false. Since A isn't false, A -> B isn't false, and if a statement isn't false, then (since it must be true or false) it must be true. Now A would be the compound statement (p -> q) ^ (q -> r) ^ p. Since this is a string of ^ statements, it can only be true of all the statments p -> q, q -> r and p are true. Since p isn't true, the string of ^ statements isn't true, so it must be false. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: as i stated above, q->r is false because q is true and r is false this makes [(p -> q) ^ (q -> r) ^ p] false as well [(p -> q) ^ (q -> r) ^ p] -> r can only be false if T->F since [(p -> q) ^ (q -> r) ^ p] if false, then [(p -> q) ^ (q -> r) ^ p] -> r is true confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: p and q are both true, so p -> q and p are true. The only candidate for a false statement among the three statements is q -> r. So we evaluate q -> r for truth values TTF. Since q is T and r is F, we see that q -> r must be F. This makes [ (p -> q) ^ (q -> r) ^ p] false. Therefore [ (p -> q) ^ (q -> r) ^ p] -> r must be true, since it can only be false and if [ (p -> q) ^ (q -> r) ^ p] is true. STUDENT QUESTION Explain to me about finding truth in these sets such as TTF. I can't find it in the book nor did the lady on the video say anything about them. INSTRUCTOR RESPONSE TFF stands for the truth values of p, q and r. TFF means the p is true, while q and r are both false. In your truth table this corresponds to the fourth line, which should read: p q r p->q q->r [(p->q)^(q->r)^p] [(p->q)^(q->r)^p] [(p->q)^(q->r)^p] [(p->q)^(q->r) ^p]->r T T F T F F T Note that [(p->q)^(q->r)^p] is false for this line, because this expression is a conjunction and at least one of the statement s in the conjunction is false. This makes [(p->q)^(q->r)^p] - r true, since a false antecedent makes the conditional true. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q004. Examine the truth of the statement [ (p -> q) ^ (q -> r) ^ p] for each of the truth sets TFF, FTF and FFF. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: p q r p->q q->r T F F F T F T F T F F F F T T this part of the table tells us everything we need to know. for TFF, p->q if false which makes [ (p -> q) ^ (q -> r) ^ p] false for FTF, p is false which makes [ (p -> q) ^ (q -> r) ^ p] flase for FFF, p is false which makes [ (p -> q) ^ (q -> r) ^ p] false confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: if r is false then that makes q->r false which makes [ (p -> q) ^ (q -> r) ^ p] false and the argument ends up being F->F which is true if r is true then it doesn't matter if [ (p -> q) ^ (q -> r) ^ p] is true or false because [ (p -> q) ^ (q -> r) ^ p] -> r can only be false if T->F confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The three given truth values, plus the TTF we examined earlier, are all the possibilities where r is false. We see that in the cases where r is false, [ (p -> q) ^ (q -> r) ^ p] is always false. This makes [ (p -> q) ^ (q -> r) ^ p] -> r true any time r is false. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q006. Explain how we have shown in the past few exercises that [ (p -> q) ^ (q -> r) ^ p] -> r must always be true. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: if r is false then [ (p -> q) ^ (q -> r) ^ p] is false and therefore [ (p -> q) ^ (q -> r) ^ p] -> r is true if r is true then [ (p -> q) ^ (q -> r) ^ p] -> r must be true confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q007. Explain how this shows that the original argument about rain, wet grass and smelling wet grass, must be valid. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: we have just proven that the statement [ (p -> q) ^ (q -> r) ^ p] -> r is always true and the original argument is symbolized by this statement therefor this argument must be vaild confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: That argument is symbolized by the statement [ (p -> q) ^ (q -> r) ^ p] -> r. The statement is always true. There is never a case where the statement is false. Therefore the argument is valid. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q008. Explain how the conclusion of the last example, that [ (p -> q) ^ (q -> r) ^ p] -> r is always a true statement, shows that the following argument is valid: 'If it snows, the roads are slippery. If the roads are slippery they'll be safer to drive on. It just snowed. Therefore the roads are safer to drive on.' Hint: First symbolize the present argument. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: it snows p roads are slippery q roads are safer to drive on r if it snows, the roads are slippery p->q if roads are slippery, then they will be safer to drive on q->r it just snowed p the roads are safer to drive on r [(p->q) ^ (q->r) ^ p] -> r this is the same statement that we have already proven cannot be false, therefore the argument is valid confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `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.' YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: it rained p there is a picnic q it doesn't rain ~p there is no picnic ~q [(~p->q) ^ ~q] -> p p q ~q ~p ~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 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We could let p stand for 'it rained', q for 'there is a picnic'. The first statement is 'If it doesn't rain there is a picnic', which is symbolized by ~p -> q. The second statement, 'There is no picnic', is symbolized by ~q. The conclusion, 'it rained', is symbolized by p. The argument therefore says IF [ (~p -> q) AND ~q ], THEN p. This is symbolized by [ (~p -> q) ^ ~q ] -> p. We set up a truth table for this argument: p q ~p ~q ~p -> q (~p -> q) ^ ~q [ (~p -> q) ^ ~q ] -> p T T F F T F T T F F T T T T F T T F T F T F F T T F F T ********************************************* Question: `q010. Symbolize the following argument: If it the sun shines, then we'll have a picnic. The sun doesn't shine. Therefore we don't have a picnic. Then set up a truth table to test the validity of the argument. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: sun shines p we will have a picnic q sun doesn't shine ~p we don't have a picnic ~q [(p->q) ^ ~p] -> ~q p q ~p ~q p->q [(p->q) ^ ~p] [(p->q) ^ ~p] -> ~q T T F F T F T T F F T F F T F T T F T T F F F T T T T T the argument is not valid. the argument does not sa that you will only have a picnic if the sun shines. you could still have a picnic even if it does not. *if the argument had been ""if the sun shines, then we'll have a picnic. we didn't have a picnic. therefore the sun did not shine"" then we could have used the same symbols but it would have looked like this...[(p->q) ^ ~q] -> ~p the true values for this would have been p q ~p ~q p->q [(p->q) ^ ~q] [(p->q) ^ ~q] -> ~p T T F F T F T T F F T F F T F T T F T F T F F T T T T T confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating:ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!