#$&* course mth151 016. Translating Arguments
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Given Solution: This argument certainly seems valid. We say what will happen if rains, and what will happen is that happens. Then we say that it rains, so the whole chain of happenings, rained then wet grass then smell, should follow. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q002. Is the following argument valid: 'If it snows, the roads will be slippery. If the roads are slippery they'll be safer to drive on. Yesterday it snowed. Therefore yesterday the roads were safer to drive on.' YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Not valid confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. STUDENT COMMENT: so it does not matter that the roads are not safer when they're slippery, what matters is that the statement said they are when snows and snowed yesterday therefore the roads were safer yesterday INSTRUCTOR RESPONSE: Right. The statements don't have to be true for the argument to be valid. Of course, if the statements aren't true then even though the argument is valid the conclusion might not be true. The old saying is 'garbage in, garbage out'. If you put 'garbage' (i.e., false statements) into a logical argument, that argument can indeed result in 'garbage' (i.e., a false statement as the logical conclusion). STUDENT COMMENT: According to the statement it is true, but I might question this about driving on slippery roads. INSTRUCTOR RESPONSE: That assumption is deliberately absurd, to help make a clear distinction between correct assumptions and correct logic. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q003. Is the following argument valid: 'Today it will rain or it will snow. Today it didn't rain. Therefore today it snowed.' YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: valid confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If we accept the fact that it will do one thing or another, then at least one of those things must happen. If it is known that if one of those things fails to happen, then, the other must. Therefore this argument is valid. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. Is the following argument valid: 'If it doesn't rain we'll have a picnic. We don't have a picnic. Therefore it rained.' YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: valid confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: [ ( p- >q) ^ (q- > r) ^ p }- > r confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The argument gives three conditions, 'If it rains, the grass gets wet. If the grass gets wet, we'll be able to smell the wet grass. It rained yesterday.', which are symbolized p -> q, q -> r and p. It says that under these three conditions, the statement r, 'we can smell the wet grass', must be true. Therefore the argument can be symbolized by the complex statement [ (p -> q) ^ (q -> r) ^ p] -> r. STUDENT COMMENT: becuase the statment is valid r will be on the outside of the parenthesis INSTRUCTOR RESPONSE: It doesn't matter whether the statement is valid or not. The premises go into the parentheses or brackets, the conclusion follows the -> sign. The form of the argument is [premises] -> conclusion, where the premises inside the brackets are joined by conjunctions. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q006. The preceding argument was symbolized as [ (p -> q) ^ (q -> r) ^ p] -> r. Determine whether this statement is true for p, q, r truth values F F T. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: false confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: false confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: For these truth values p -> q is true since p is false (recall that the only way p -> q can be false is for p to be true and q to be false), q -> r is false since q is false, and p itself is false, therefore [ (p -> q) ^ (q -> r) ^ p] is false. This makes [ (p -> q) ^ (q -> r) ^ p] -> r true, since the statement can only be false if [ (p -> q) ^ (q -> r) ^ p] is true while r is false. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course mth151 016. `query 16
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Given Solution: `a** 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery 3.5.20 all chickens have a beak. All hens are chickens. Therefore all hens have beaks. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: valid confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** You need to show the relationship between chickens and things with breaks. You would make a circle for everything with a beak and another circle for chickens. Since all chickens have beaks the chicken circle has to be inside the 'beaked' circle. Then you have hens. They are all chickens so the hen circle is inside the chicken circle. Since the chicken circle is already inside the beaked circle the hen circle (inside the chicken circle) is also inside the beaked circle, and you conclude that all hens have beaks. COMMON ERROR WITH INSTRUCTOR COMMENT: In the outer circle, I put chickens with beaks. Inside that circle, I made another circle for hens are chickens. INSTRUCTOR COMMENT: 'hens are chickens' is a statement, not a thing. The circles have to be defined by things. SIMILAR ERROR WITH COMMENT: Two circles: large circle of hens are chickens and a smalled circle within of hens have beaks. Valid INSTRUCTOR COMMENT: You don't put propositions into circle (e.g., 'hens are chickens' isn't a circle). You put sets of things into circles (e.g., a circle for hens and a circle for chickens, with the hens circle inside the chickens circle). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhen the diagram is drawn according to the premises, is it or is it not possible for the diagram to be drawn so that it contradicts the conclusion? If it is possible describe how. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Not possible confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q(formerly 3.5.27). The text 'dumbed down' the problems in this section. This question is a bit challenging but is similar in nature to assigned problems: Using Venn diagrams evaluate the following argument:{}{}All drivers contribute to traffic congestion. All contributors to traffic congestion make life a little worse. Some people who live in the suburbs make life a little worse. Therefore some people who contribute to traffic congestion live in the suburbs. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: invalid confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ________________________________________ #$&*