#$&* course Mth 151 017. Evaluating Arguments
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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 #$&* ********************************************* Question: `q002. At this point we know that the truth values TTT, TFT, FTT, FFT all make the argument [ (p -> q) ^ (q -> r) ^ p] -> r true. What about the truth values TTF? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: True, I wrote it out in a truth table confidence rating #$&* 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I didn’t think of that I just wrote it out ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q003. The preceding statement said that for the TTF case [ (p -> q) ^ (q -> r) ^ p] was false but did not provide an explanation of this statement. Which of the statements is false for the truth values TTF, and what does this tell us about the truth of the statement [ (p -> q) ^ (q -> r) ^ p] -> r? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: That its false as the r is false and the q is true it is false in an if then sarino confidence rating #$&* 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 #$&* ********************************************* Question: `q004. Examine the truth of the statement [ (p -> q) ^ (q -> r) ^ p] for each of the truth sets TFF, FTF and FFF. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: TFF false cause first parentheses false p->q FTF False cause second ( ) false FFF false because p is false two parentheses are true but not whole statement confidence rating #$&* 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): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q005. We have seen that for TFF, FTF and FFF the statement [ (p -> q) ^ (q -> r) ^ p] is false. How does this help us establish that [ (p -> q) ^ (q -> r) ^ p] -> r is always true? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If both first term in this case inside the brackets is always false and the r is also false this always makes an if then situation true confidence rating #$&* 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): Got it ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q006. Explain how we have shown in the past few exercises that [ (p -> q) ^ (q -> r) ^ p] -> r must always be true. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Either way the inside is dependant on the truth value of r so it will make it mastch the r outside the brackets making it true confidence rating #$&* 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 "