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A final reconstruction

# A final reconstruction - t and t 1 The argument then looks...

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A final reconstruction In this version there is no confusion between instants and intervals. Rather, there is a fallacy that logic students will recognize as the “quantifier switch” fallacy. The universal quantifier, “at every instant,” ranges over instants of time; the existential quantifier, “there is a place,” ranges over locations at which the arrow might be found. The order in which these quantifiers occur makes a difference! (To find out more about the order of quantifiers, click here .) Observe what happens when their order gets illegitimately switched: 1c. If there is a place just the size of the arrow at which it is located at every instant between t 0 and t 1 , the arrow is at rest throughout the interval between t 0 and t 1 . 2c. At every instant between t 0 and t 1 , there is a place just the size of the arrow at which it is located. We will use the following abbreviations: L ( p , i ) The arrow is located at place p at instant i R The arrow is at rest throughout the interval between
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Unformatted text preview: t and t 1 The argument then looks like this: 1c. If there is a p such that for every i , L ( p , i ), then R . 5 p 2200 i L ( p , i ) → R 2c. For every i , there is a p such that: L ( p , i ). 2200 i €5 p L ( p , i ) But (2c) is not equivalent to, and does not entail, the antecedent of (1c): There is a p such that for every i , L ( p , i ) 5 p 2200 i L ( p , i ) The reason they are not equivalent is that the order of the quantifiers is different. (2c) says that the arrow always has some location or other (“at every instant i it is located at some place p ”) - and that is trivially true as long as the arrow exists! But the antecedent of (1c) says there is some location such that the arrow is always located there (“there is some place p at which it is located at every instant i ”) -and that will only be true provided the arrow does not move! So one cannot infer from (1c) and (2c) that the arrow is at rest....
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