This gives us an argument that can be set out like this

# This gives us an argument that can be set out like this - a...

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This gives us an argument that can be set out like this: a. We know that no Z-run reaches G. b. Suppose R makes all the Z -runs and no other runs. c. Then R has not reached G . [From (a). Reason: no Z -run reaches G , and R has not made any other runs.] d. But R cannot be to the left of G . [From (b). Reason: R has made all the Z runs, and if R were to the left of G , there would still be Z -points between R and G , so not all the Z -runs would have been made.] e. So R has reached G . [From (d)] Since (b) leads to a contradiction [(c) contradicts (e)], it is logically impossible for (b) to be true. Therefore, f. It is impossible for R to make all the Z -runs. Does the argument work? Thomson thinks that it does not, because it does not prove (f), but only a weaker conclusion: It is impossible for R to make all the Z -runs without making a run that is not a Z -run . So Thomson’s response to Zeno is that one can make all the Z -runs simply by making a run (e.g.,
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Unformatted text preview: a G-run) that is not a Z-run. To think this is impossible requires an equivocation on the word ‘run’. This turns out to be a weak reply. For although it shows that Zeno doesn’t get his contradiction from the assumption that R makes all the Z-runs, it concedes too much to Zeno. For it supposes that there is some description of a super-task (“making all the Z-runs and no other runs”) that does lead to a contradiction. That is, Thomson maintains both: i. “ R makes all the Z-runs and no other runs” entails “ R reaches G ” and ii. “ R makes all the Z-runs and no other runs” entails “ R does not reach G .” But as Paul Benacerraf has shown (in the article “Tasks, Super-tasks, and the Modern Eleatics,” on e-reserve) neither of these entailments holds. That is, we cannot derive a contradiction even from the assumption that R makes all the Z-runs and no others ....
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