Quiz #4 ANSWER KEY with comments Name _________________________________PHIL 110, Spring 2014 (12 points)Hamilton2/19/2014Part 1: In the blank next to each of the following sentences, put the one capital letter (from the accompanying diagram) that labels the innermost logical region to which the sentence belongs. (4 points per blank)A: logical possibility1. SameRow(a,b) ___ A ____Two objects a and b can be in the same row, but theydon’t have to be. This is just a possibility.2. SameRow(a,a) ___ B ____An object a must be in the same row as itself, so this one isa logical necessity. However, it’s not a tautology becausethe meaning of the predicate SameRow is crucial to makingthe sentence logically necessary. And if you blinded yourself to thatmeaning by replacing the atomic sentence with a capital letter like P, you’d just have P, and without knowing what P means, you wouldn’t be able to tell if it had to always be true in every world.3. SameRow(a,b) ᴠ SameSize(a,b) ___ A ____This is only a logical possibility because it could be false in some worlds, namely, any worlds in which objects a and b are neither the same size nor in the same row.4. SameRow(a,a) ᴧ SameRow(a,a) ___ B ____This one is a logical truth for the same reason as #2 above. Conjoining another SameRow(a,a) is simply redundant; it doesn’t really add or subtract anything from the necessity of the sentence. And the sentence is not a tautology because if you blind yourself to the meaning of the atomic sentences you get simply P ᴧ P, which is no more necessarily true than is simply P. Without knowing what P means, you can’t tell if it must be true in all worlds or not.5. SameRow(a,a) ᴧ SameSize(a,a) ___ B ____This one is logically necessary for reasons similar to #4 above. An object a is necessarily the same size as itself and in the same row as itself, so both halves of the conjunction must always be true, making the entire conjunction always true. It’s not a tautology, though, for when you blind yourself to the meaning of the atomic sentences you get something like P ᴧ Q, and you can’t know if that sentence would necessarily be true without knowing what P and Q individually mean.6. SameRow(a,b) ᴠ SameRow(b,a) ___ A ____Don’t let this one fool you. For the disjunction to be true, at least one of the disjuncts (halves) must be true … so the question is whether you can think of any situation in which both disjuncts would B: logical necessityC: tautologyever be false. Of course you can: any world in which objects a and b are not in the same row. In such a world, the sentence is false, which shows that this sentence is no more than a logical possibility.7. ¬(Cube(a) ᴧ ¬Cube(a)) ___ C ____This sentence can never be false. Think about it: Inside the outer parentheses there is an illogical statement that makes no sense, namely, that object a is both a cube and not a cube at the same time. That’s crazy. But then this crazy statement is negated, which is to say that it is false. And, yes, a crazy contradictory statement like what’s inside the parentheses will be false in every world, which means that the negation of that always-false statement will always be true. So this sentence is a logical necessity. But it is more than that, it is also a tautology. Put your blinders on and you get ¬(P ᴧ ¬P) which, if you think about it, is clearly just as logically necessary as was the original version. It doesn’t matter what P means; the affirmation and the negation of P cannot both be true at the same time, so the negation of that contradictory statement will always be true.BONUS QUESTION (4 points):8. ¬(SameSize(a,b) ᴧ ¬SameSize(b,a)) ___ B ____This one is very similar to #7 above, and it is logically necessary for the same reason (i.e., the negation of a contradictory statement will always be true). However, this one is not a tautology for a simple reason: notice that the order of the arguments of the predicate differs between the two atomic sentences (i.e., [a,b] versus [b,a]), and this is enough to make these two distinct atomic sentences (even though, of course, they have the same meaning). So, when you blind yourself to the meanings of the atomic sentences, you get ¬(P ᴧ ¬Q) which doesn’t have the same logical necessity as the original.Part 2: For each of the following sentences of FOL, circle the letter next to the English translation that best captures the meaning of the FOL sentence. (4 points each)9. ¬(RightOf(d,e) ᴧ RightOf(e,d))a. d is not right of e, and e is not right of d.b. It’s not the case both that d is right of e and that e is right of d.c. It’s not the case that d is right of e and it’s also not the case that e is right of d.Choices (a) and (c) here say essentially the same thing, and both get the translation wrong because both suggest that each of the atomic sentences RightOf(d,e) and RightOf(e,d) are negated individually. However, in reality it is only the conjunction of the two atomic sentences that is negated (i.e., they can’t both be true), leaving open the possibility that just one or the other of them may be true.10. (Large(d) ᴠ Med(d)) ᴧ Tet(d)a. d is a large or medium tetrahedron.b. Either d is large or it’s a medium tetrahedronc. d is both large and medium, or it’s a tetrahedron.Notice that the main connective in 10 is the conjunction, which means that d must be a tetrahedron for the sentence to be true, and also that this tetrahedron must be either large or medium (so that the first conjunct of the sentence will also be true—which it must be if the entire conjunction is true). Only answer choice (a) gives this meaning.11. ¬(RightOf(e, g) ∧ LeftOf(e, f)) ∧ ¬(RightOf(d, g) ∧ LeftOf(d, f))a. Either e or d is not right of g and not left of f.b. Neither e nor d is right of g and left of f.c. e is not right of g, and e is not left of f, and d is not right of g, and d is not left of f.Sentence 11 is a conjunction of two negated phrases, meaning that both of the negated phrases must be true. For that to be the case, then e cannot be both right of g and left off, and likewise d cannot be both right of g and left of f. Choice (b) is the only one that captures this meaning.Part 3: Fill in ALL missing truth values in ALL of the empty cells in the truth table below. This includes ALL values in the reference columns (in the standard order that we discussed in class) and ALL requiredvalues missing
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