course Mth 151 Sorry, this a day late. I had a friend whose baby(7 mths old) died, and I been tied up with that.
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15:23:12 query 3.5.6 all dogs love to bury bones. Archie doesn't. Therefore Archie isn't a dog .
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RESPONSE --> Valid
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15:23:48 ** You would put all dogs in a circle, and this circle would be inside another circle consisting of all things that love to bury bones. Archie is outside this bone-burying circle and since this circle contains all dogs Archie isn't a dog. This makes the argument valid. The x for Archie has to go outside the outer circle, so it has to be outside the inner circle. Thus the x can't be in the inner circle, and Archie therefore can't be a dog. The conclusion can't be contradicted. COMMON ERROR WITH INSTRUCTOR RESPONSE: I put 'all dogs like to bury bones' in one circle and 'archie likes to bury bones' in another. INSTRUCTOR RESPONSE: You don't want to use a single circle to represent a compound statement. 'All dogs like to bury bones' and 'Archie likes to bury bones' are compound statements. SIMILAR ERROR: in one circle ,I put all dogs love to bury bones, inthe other circle I put Archie, so I knew that Archie wasn't a dog, so the statement is valid . INSTRUCTOR COMMENT: See previous comment.
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RESPONSE --> ok, I had my circle drawn out the same way.
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15:25:45 query 3.5.20 all chickens have a beak. All hens are chickens. Therefore all hens have beaks.
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RESPONSE --> Valid The outside circle, for beaks The inner circle for chickens Hens, would be in the inner circle, therefore have beaks
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15:26:23 ** You need to show the relationship between chickens and things with breaks. You would make a circle for everything with a beak and another circle for chickens. Since all chickens have beaks the chicken circle has to be inside the 'beaked' circle. Then you have hens. They are all chickens so the hen circle is inside the chicken circle. Since the chicken circle is already inside the beaked circle the hen circle (inside the chicken circle) is also inside the beaked circle, and you conclude that all hens have beaks. COMMON ERROR WITH INSTRUCTOR COMMENT: In the outer circle, I put chickens with beaks. Inside that circle, I made another circle for hens are chickens. INSTRUCTOR COMMENT: 'hens are chickens' is a statement, not a thing. The circles have to be defined by things. SIMILAR ERROR WITH COMMENT: Two circles: large circle of hens are chickens and a smalled circle within of hens have beaks. Valid INSTRUCTOR COMMENT: You don't put propositions into circle (e.g., 'hens are chickens' isn't a circle). You put sets of things into circles (e.g., a circle for hens and a circle for chickens, with the hens circle inside the chickens circle). **
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RESPONSE --> ok
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15:28:12 When the diagram is drawn according to the premises, is it or is it not possible for the diagram to be drawn so that it contradicts the conclusion? If it is possible describe how.
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RESPONSE --> Yes, because you can draw the diagram, to prove that some statements are false, according to its conclusion.
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15:29:07 ** The circle for hens must be inside the circle for chickens, which is inside the circle for beaked creatures. Therefore the circle for hens must be inside the circle for beaked creatures. No other way to draw it consistent with the conditions. **
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RESPONSE --> ok, I understand, I didn't know you were referring to the previous question!
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15:36:51 3.5 27 all drivers contribute. All contributors make life a little worse. Some people in the suburbs make life a little worse. Therefore some people who contribute live in the suburbs.
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RESPONSE --> invalid, because the people who live in a suburb, can be in or out of one of the circles, we don't know, so its invalid.
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15:37:12 ** BRIEF SOLUTION: Drivers circle inside contributors circle. Contributors circle inside make-life-worse circle. Suburbs circle overlaps make-worse circle but the degree of overlap is not specified, and the circle doesn't necessarily extend all the way into the contributors circle. So the picture can be drawn to contradict the conclusion without contradicting the given premises. WITH A LITTLE MORE DETAIL: You would have circles for drivers, contributors, people who make life worse and people in the suburbs. All drivers contribute so the drivers circle would be inside the contributors circle. All contributors make life worse so their circle would be inside the 'people who make life worse' circle. The people-in-the-suburbs circle has to be inside the make-life-worse circle. The make-life-worse circle contains the other two circles, and there is a place in that circle for the people-in-suburbs circle such that the people-in-suburbs circle doesn't intersect any of the other circles (put it in the part of the make-life-worse circle that doesn't contain either of the other two circles). So it's possible to put all people in the suburbs outside of the 'contribute' circle. COMMON SENSE: In commonsense terms, which you should relate to the picture, nothing is said that forces all people in the suburbs into the drivers circle or the contributors circle, since the statement is that SOME people in the suburbs make life worse. There might be a little old lady in the suburbs who doesn't drive and does nothing but make life better--this is not contradicted by the 'some people in suburbs make things worse' statement. Thus the argument is invalid. COMMON ERROR WITH COMMENTARY: all drivers contribute goes in one circle. All contributors make life a little worse goes in another. Some people in the suburbs make life a little worse goes in a third. INSTRUCTOR COMMENT Again classes of things, not relationships, go into the circles. Everything you listed here is a relationship, not a class of objects. **
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RESPONSE --> ok
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le٬j assignment #017 űpZ Liberal Arts Mathematics I 03-13-2006
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15:38:51 Query 3.6.6 if he didn't have to set up he would be ecstatic. He's ecstatic. Therefore he doesn't have to set up. Is the argument and valid or invalid and why?
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RESPONSE --> invalid, fallacy by fallacy of converse.
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15:38:59 ** This argument is an instance of the 'fallacy of the converse'. In commonsense terms we can say that there could be many reasons why he might be ecstatic--it doesn't necessarily follow that it's because he doesn't have to set up. A Venn diagram can be drawn with 'no setup' inside 'ecstatic'. An x inside 'ecstatic' but outside 'no setup' fulfills the premises but contradicts the conclusions. Also [ (p -> q) ^ q ] -> p if false for the p=F, q=F case. **
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RESPONSE --> ok
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15:43:53 Query 3.6.12 she ecomms it or uses credit. She doesn't use credit. Therefore she orders it ecommsb Is the argument and valid or invalid and why?
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RESPONSE --> invalid?
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15:44:36 ** The argument can be symbolized as p V q ~q therefore p This type of argument is called a disjunctive syllogism. **
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RESPONSE --> ok
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15:47:17 Query 3.6.20 evaluate using the truth table: ~p -> q, p, therefore -q
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RESPONSE --> invalid, by the problem 3.6 # 20, but my problem reads p -> ~q ~p ~q
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15:48:01 ** We need to evaluate {(p--> ~q) ^ ~p} --> ~q, which is a compound statement representing the argument. p q ~p ~q (p--> ~q) {(p--> ~q) ^ ~p} {(p--> ~q) ^ ~p} --> ~q then truth table is p q ~p ~q (p--> ~q) {(p--> ~q) ^ ~p} {(p--> ~q) ^ ~p} --> ~q T T F F F F T T F F T T F T F T T F T T F F F T T T T T Note that any time p is true (p->~q)^~p) is false so the final conditional (p->~q)^p) -> ~q is true, and if q is false then ~q is true so the final conditional is true. The F in the third row makes the argument invalid. To be valid an argument must be true in all possible instances. } Another version of this problem has ~p -> q and p as premises, and ~q as the consequent. The headings for this version of the problem are: p q ~p ~q ~p -> q (~p -> q) ^ p [ (~p -> q) ^ p ] -> ~q. Truth values: T T F F T T F T F F T T T T F T T F F F T F F T T T F T The argument is not true by the final truth value in the first line. To be true the statement [ (~p -> q) ^ p ] -> ~q must be true for any set of truth values. **
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RESPONSE --> ok
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16:01:39 3.6.24 evaluate using the truth table: (p ^ r) -> (r U q), and q ^ p), therefore r U p
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RESPONSE --> This is a long truth table, my rows may be a different order: TTT T T T T T T T TTF T F F F T T T TFT F F T T T T T TFF F F T F F T T FTT F T T F F T T FTF F F T F F T T FFT F F T F F F T FFF F F T F F F T = VALID
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16:01:50 ** The headings can be set up as follows: p q r p^r rUq (p^r)->(rUq) {((r ^ p ) --> (rU q)) ^ (q^p)} {((r ^ p ) --> (rU q)) ^ (q^p)} --> (rUp) This permits each column to be evaluated, once the columns for p, q and r are filled in by standard means, by looking at exactly two of the preceding columns. Here's the complete truth table. pqr r^p q^p rUp rUq (r^p)->(rUq) [(r^p)->(rUq)]^(q^p) {[(r^p) -> (rUq)] ^ q^p} -> rUp ttt t t t t t t t ttf f t t t t t t tft t f t t t f t tff f f t f t f t ftt f f t t t f t ftf f f f t t f t fft f f t t t f t fff f f f f t f t All T's in the last column show that the argument is valid. COMMON BAD IDEA: p, q, r, (r ^ p), (rUq), (q^p), (rUp), {[(r^p)->(rUq)] ^ (q^p)}->(rUp) You're much better off to include columns for [(r^p)->(rUq)] and {[(r^p)->(rUq)] ^ (q^p)} before you get to {[(r^p)->(rUq)] ^ (q^p)}->(rUp). If you have to look at more than two previous columns to evaluate the one you're working on you are much more likely to make a mistake, and in any case it takes much longer to evaluate. **
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RESPONSE --> OK
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16:15:08 3.6.30: Christina sings or Ricky isn't an idol. If Ricky isn't an idol then Britney doesn't win. Britney wins. Therefore Christina doesn't sing.
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RESPONSE --> IF I set the problem up right?, then it is invalid
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16:16:30 ** Solution using deductive reasoning: If r stands for RM is a teen idol c stands for CA sings b stands for BS wins then the statements are c U ~r ~r -> ~b b therefore ~c. The contrapositive of ~r -> ~b is b -> r. So we have b -> r b therefore r. We now have c U ~r r therefore c by disjunctive syllogism. That is, Britney wins so Rich is an idol. Christina sings or Ricky isn't an idol. So Christina sings. The argument concludes ~c, the Christina doesn't sing. So the argument is invalid Solution using truth tables: If we let p stand for Christina sings, r for Ricky Martin is a teen idol and w for Britney Spears wins AMA award then we have p V ~r ~r->~w w Therefore ~p The argument is the statement [(pV~r)^(~r->~w)^w]-~p We can evaluate this statement using the headings: p r w ~r ~w ~p (pV ~r) (~r->~w) [(pV~r)^(~r->~w)^w] [(pV~r)^(~r->~w)^w]-~p. We get p r w ~r ~w ~p (pV ~r) (~r->~w) [(pV~r)^(~r->~w)^w] [(pV~r)^(~r->~w)^w]-~p T T T F F F T T T F T T F F T F T T F T T F T T F F T F F T T F F T T F T T F T F T T F F T F T F T F T F F T T F F F T F F T T F T T T T T F F F T T T T T F T. The argument is not valid, being false in the case of the first row. **
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RESPONSE --> OK, I had the problem layed out wrong
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16:19:00 Previous version 3.6.30 determine validity: all men are mortal. Socrates is a man. Therefore Socrates is mortal
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RESPONSE --> valid
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16:19:12 ** This can be reasoned out by the transitive property of the conditional. If p stands for 'a man', q for 'mortal', r for 'Socrates' you have r -> p p -> q therefore r -> q which is valid by the transitive property of the conditional. A truth-table argument would evaluate [ (r -> p) ^ (p -> q) ] -> (r -> q). The final column would come out with all T's, proving the validity of the argument. **
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RESPONSE --> ok
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yS{x}QѱdNӍ㩴 Student Name: assignment #016
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16:22:09 `q001. There are 6 questions in this set. Is the following argument valid? 'If it rains, the grass will get wet. If the grass gets wet, we'll be able to smell the wet grass. It rained yesterday. Therefore yesterday we were able to smell the wet grass.'
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RESPONSE --> valid
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16:22:29 This argument certainly seems valid. We say what will happen if rains, and what will happen is that happens. Then we say that it rains, so the whole chain of happenings, rained then wet grass then smell, should follow.
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RESPONSE --> ok
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16:24:23 `q002. Is the following argument valid: 'If it snows, the roads will be slippery. If the roads are slippery they'll be safer to drive on. Yesterday it snowed. Therefore yesterday the roads were safer to drive on.'
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RESPONSE --> Even though its actually not true, the argument is VALID!
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16:24:38 The validity of an argument has nothing to do with whether the statements in that argument are true or not. All we are allowed to do is assume that the statements are indeed true, and see if the conclusions of the argument therefore hold. In this case, we might well question the statement 'if the roads are slippery they'll be safer to drive on', which certainly seems untrue. However that has nothing to do with the validity of the argument itself. We can later choose to reject the conclusion because it is based on a faulty assumption, but we cannot say that the argument is invalid because of a faulty assumption. This argument tells us that something will happen if it snows, and then tells us what we can conclude from that. It then tells us that it snows, and everything follows logically along a transitive chain, starting from from the first thing.
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RESPONSE --> ok
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16:25:29 `q003. Is the following argument valid: 'Today it will rain or it will snow. Today it didn't rain. Therefore today it snowed.'
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RESPONSE --> valid
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16:25:41 If we accept the fact that it will do one thing or another, then at least one of those things must happen. If it is known that if one of those things fails to happen, then, the other must. Therefore this argument is valid.
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RESPONSE --> ok
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16:26:29 `q004. Is the following argument valid: 'If it doesn't rain we'll have a picnic. We don't have a picnic. Therefore it rained.'
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RESPONSE --> valid
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16:26:41 In this argument where told the something must happen as a result of a certain condition. That thing is not happen, so the condition cannot have been satisfied. The condition was that it doesn't rain; since this condition cannot have been satisfied that it must have rained. The argument is valid.
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RESPONSE --> ok
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16:28:50 `q005. We can symbolize the following argument: 'If it rains, the grass gets wet. If the grass gets wet, we'll be able to smell the wet grass. It rained yesterday. Therefore yesterday we were able to smell the wet grass.' Let p stand for 'It rains', q for 'the grass gets wet' and r for 'we can smell the wet grass'. Then the first sentence forms a compound statement which we symbolize as p -> q. Symbolize the remaining statements in the argument.
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RESPONSE --> p -> q p therfore r
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16:29:27 The argument gives three conditions, 'If it rains, the grass gets wet. If the grass gets wet, we'll be able to smell the wet grass. It rained yesterday.', which are symbolized p -> q, q -> r and p. It says that under these three conditions, the statement r, 'we can smell the wet grass', must be true. Therefore the argument can be symbolized by the complex statement [ (p -> q) ^ (q -> r) ^ p] -> r.
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RESPONSE --> ok
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16:32:27 `q006. The preceding argument was symbolized as [ (p -> q) ^ (q -> r) ^ p] -> r. Determine whether this statement is true for p, q, r truth values F F T.
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RESPONSE --> this row for FFT: FFT T T T F T, TRUE WITH THIS LINE OF THE TRUTH TABLE.
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16:32:40 For these truth values p -> q is true since p is false (recall that the only way p -> q can be false is for p to be true and q to be false), q -> r is false since q is false, and p itself is false, therefore [ (p -> q) ^ (q -> r) ^ p] is false. This makes [ (p -> q) ^ (q -> r) ^ p] -> r true, since the statement can only be false if [ (p -> q) ^ (q -> r) ^ p] is true while r is false.
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RESPONSE --> OK
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튟`wW Student Name: assignment #017
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16:34:15 `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 --> ?
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16:34:53 [ (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 wasn't clear on the question
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16:40:46 `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 --> TTF makes the argument true TTF T F F F T
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16:40:57 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 --> OK
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16:45:35 `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 --> FALSE= (q->r),(p->q ^ q->r), [(p->q)^(q->r) ^ p]
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16:45:45 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 --> ok
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16:46:51 `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 --> They are all false
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16:46:57 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
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16:48:09 `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 --> in a conditional statement, f -> f = always equals TRUE
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16:48:14 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
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16:49:48 `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 --> With the truth table TTT T T T T T TTF T F F F T TFT F T F F T TFF F T F F T FTT T T T F T FTF T F F F T FFT T T T F T FFF T T T F T
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16:50:03 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
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16:53:12 `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 --> you have an actecedent, that's true and a consequent that's true, so the conclusion is true
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16:53:22 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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16:54:55 `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
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16:55:05 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
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16:56:16 `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 --> (~p -> q) ^ ~q -> p
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16:56:50 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, and I had the truth table too
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