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all! I was wondering if I could get help translating this phrase in to first order logic.

I'm trying to say: There exists a u such that u is Russian and there exists a b such that u shot b.

Would it be:

∃u(R(u)) /\ ∃b(Shot(u, b))

Or would it be:

∃u∃b(R(u) /\ (Shot(u, b))) ?

E...
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London Jennings
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  • The first is correct. The second is equivalent but inefficient. b does not appear in R(u) so that the ∃b has no effect. By inefficient I mean that it's a little obfuscated. I'm not sure what rule says that the first form is preferred but it's the one I'd use, it just seems more clear. – user4894 Jul 12 '16 at 05:32
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    Sanity check: the first is correct if you mean ∃u ( (R(u)) /\ ∃b(Shot(u, b)) ) but is nonsensical if you mean ( ∃u(R(u)) ) /\ ∃b(Shot(u, b)) –  Jul 12 '16 at 06:28
  • @Hurkyl is right. Unfortunately it is not possible for others to correct the problem since an edit must be at least 6 characters long. Probably OP would be allowed to do it. – Colin McLarty Jul 12 '16 at 12:10
  • @Hurkyl why is it nonsensical? Is ∃u(R(u)) not a valid statement? Doesn’t it just mean “there exists a u such that the predicate R can be applied to it”? – London Jennings Jul 13 '16 at 09:27
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    Check the parentheses. ∃b(Shot(u, b)) is not a statement at all, since it has free variable u. – Colin McLarty Jul 13 '16 at 13:08
  • @ColinMcLarty Understood! Thank you for your support. – London Jennings Jul 14 '16 at 05:18

1 Answers1

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Either one is correct. For some technical purposes it is convenient to bring all the quantifiers out to initial position as in your second version. For that, see "prenex normal form" on Wikipedia. But other purposes might favor the first version and first order logic per se just finds them equivalent.

Colin McLarty
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