#$&* course mth 151 8-7 7 017. 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 #$&*3 ********************************************* Question: `q002. At this point we know that the truth values TTT, TFT, FTT, FFT all make the argument [ (p -> q) ^ (q -> r) ^ p] -> r true. What about the truth values TTF? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If the truth values are TTF and r is F, then the entire statement is true because the “if” statement is false as well. When the if…then statement has a false “then” and a false “if”, then the entire statement is true. confidence rating #$&* 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q003. The preceding statement said that for the TTF case [ (p -> q) ^ (q -> r) ^ p] was false but did not provide an explanation of this statement. Which of the statements is false for the truth values TTF, and what does this tell us about the truth of the statement [ (p -> q) ^ (q -> r) ^ p] -> r? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (pq) is true because p and q are true. (qr) is false because q is true and r is false. (pq) ^ (qr) is false because it has “and” in the statement, since one part is true and the other false, this part is false. (pq) ^ (qr) ^ p is false as well, if one statements is false and the other true, the “and” makes the statement false. [(pq) ^ (qr) ^ p]r is true because both parts of the “if…then” statement are false, the overall value of the “if” part is false because “and” was used, meaning if there is one component that is false, it makes the statement false confidence rating #$&* 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 #$&*3 ********************************************* Question: `q004. Examine the truth of the statement [ (p -> q) ^ (q -> r) ^ p] for each of the truth sets TFF, FTF and FFF. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: TFF is false FTF is false FFF is false In a statement using “and”, if only one of the components is false, the entire statement is false. All of these combinations of values have at least one part of that is false, one false inside the statement makes the rest of the statement false. confidence rating #$&* 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 #$&*3 ********************************************* Question: `q005. We have seen that for TFF, FTF and FFF the statement [ (p -> q) ^ (q -> r) ^ p] is false. How does this help us establish that [ (p -> q) ^ (q -> r) ^ p] -> r is always true? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The fact that r is considered false, as well as the first part of the “if..then” statement, makes the entire statement true. If both parts of the statement are false, then it entire statement is true. confidence rating #$&*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): OK ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q006. Explain how we have shown in the past few exercises that [ (p -> q) ^ (q -> r) ^ p] -> r must always be true. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The statement must always be true as long as r remains false and the part within the [ ]s are false. The first part of the statement is made up of several components connected by “and”, if any part of that statement is false, the “and” makes the entire portion of the statement false. confidence rating #$&* 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 #$&*3 ********************************************* Question: `q007. Explain how this shows that the original argument about rain, wet grass and smelling wet grass, must be valid. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The original argument is valid because the components are both false. In a “if…then” statement if both parts are false then the statement is true. If the first part is true and the second part is false, then the statement would be false, making the argument invalid. confidence rating #$&* 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 #$&*3 ********************************************* Question: `q008. Explain how the conclusion of the last example, that [ (p -> q) ^ (q -> r) ^ p] -> r is always a true statement, shows that the following argument is valid: 'If it snows, the roads are slippery. If the roads are slippery they'll be safer to drive on. It just snowed. Therefore the roads are safer to drive on.' Hint: First symbolize the present argument. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: p represents it snows, q represent the roads are slippery, and r represent safe to drive on. (pq) would be true, if it snow, then the roads are slippery (qr) would be false, if roads are slippery, then they are safe to drive on. (pq) ^ (qr) would be false because “and” is used and one of the statements is false. If it snow, then the roads are slippery (true) “and” if roads are slippery, then they are safe to drive on (false) [(pq) ^ (qr) ^ p] would be false as well, no matter what the truth value of p is, the “and” makes the entire statement false because the first part was false. [(If it snow, then the roads are slippery (true)) “and” (if roads are slippery, then they are safe to drive on (false)) “and” it snows (true)] [(pq) ^ (qr) ^ p] r would be true because both parts are false. [(If it snow, then the roads are slippery (true)) “and” (if roads are slippery, then they are safe to drive on (false)) “and” it snows (true)], “then” they are safe to drive on (false). confidence rating #$&*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 #$&*3 ********************************************* Question: `q009. Symbolize the following argument and show that it is valid: 'If it doesn't rain there is a picnic. There is no picnic. Therefore it rained.' YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: p represent it rained, q represent there is a picnic [(~pq) ^ ~q] p The statement (~pq) “if it doesn’t rain (~p) there is a picnic (q)” is true If the (~pq) is true, and (~q) “there is no picnic” is connected by “and”; this part is false If [(~pq) ^ ~q] is false and (p) “it rained” is true, the entire statement is true and therefore valid. [(T) ^ (F)] T or [F]T means the statement is true and vaild confidence rating #$&* 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 "