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Unformatted text preview: R I K SENGUPTA Homework # 8 P rofessor Delia Fara 12.14 The purported proof is clearly wrong. To see why this is so, note that even though our choice of c was arbitrary, our choice of d depended directly on c. This is because d had to be an object whose sole property had to do with c, the property of not being c. Thus for every c, the d we choose is based on that particular c, and we can add a subscript c after d to clarify that d c actually depends on c. So, on the basis of this argument, i t is clear that we cannot make the statement "since c was arbitrary, for all x, d x", because such a statement assumes that d is a fixed constant, while in reality we just showed that it depends on the c we choose. That is the step where the purported proof was wrong. As an illustration, consider a world with two objects, a and b. Choose c = a, then our only choice for d is b. Again, choose c = b, then our only choice for d is a. So even in this simple case, we cannot choose one particular object d such that i t is different from all other objects. This completes the explanation. 12.22 The sentence is logically t rue. To see why this is so, restate the sentence after replacing the quantifier, thus: Ex (Cube(x) \ / (Ay Cube(y))) So, this says that there exists some x such that either it is not a cube, or every element of the world is a cube. This is clearly t rue. There are only two possibilities, either a world has only cubes, or it has at least one element that is not a cube. If i t has only cubes, then we choose any of them as x. The statement holds, because all the elements of the world are cubes, so the second part of the disjunction is true. If the world has at least one element that is not a cube, then we choose this element as x. Then it clearly exists, and it is not a cube, so that the first part of the disjunction holds true. So in either case, the statement is true. ...
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This note was uploaded on 05/03/2010 for the course PHI 201 at Princeton.