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My question stems from the two different logical forms for the following sentence under Russell's analysis. One in which the word 'not' has a wide scope and one where it has a narrow scope.

The sunken city of legend does not exist.

With wide scope:

There does not exist a unique thing that is a sunken city of legend.
not ([the x: x is a sunken city of legend] x exists)

With narrow scope:

The x such that x is a sunken city of legend, is such that x does not exist.
[the x: x is a sunken city of legend] not (x exists)

I know how to express the wide scope version into predicate logic, but not the narrow scope interpretation. Can someone please help me with this?

Also, under the narrow scope interpretation the sentence would be false, so is it okay to reject this interpretation in favour of the wide scope version?

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It depends on how you want to capture 'exist'. If you take it to be a quantifier, then there aren't two scope readings of the sentence 'The sunken city of legend does not exist', because 'not' negates a quantifier and not a predicate, and there's just nothing else to negate. So the only reading, according to Russell's theory, would be: there's no unique thing which is a sunken city of legend. That is:

~∃x((Sx & ∀y(Sy → x = y))

If, on the other hand, you interpret 'exist' as a predicate, then the wide scope and narrow scope readings are formulated as usual. Wide: there's no unique thing which is a sunken city of legend and exists. Narrow: there's a unique thing which is a sunken city of legend and doesn't exist. So:

~∃x((Sx & ∀y(Sy → x = y) & Ex)

∃x((Sx & ∀y(Sy → x = y) & ~Ex)

This would make sense only if you quantify over more than just things that 'exist' (e.g., you quantify over possible things as well). But if you interpret 'exist' as a quantifier, these two readings don't make a lot of sense.

  • Thanks, why does it only make sense when quantifying over more then just existent things? Would it not be fine to do so and just have ∃x((Sx & ∀y(Sy → x = y) & ~Ex) as false? – Anon Apr 24 at 20:39
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    That would be fine and the formulation would be false as you say. What I meant by not making sense: if you quantify only over what exists then "there are Fs that don't exist" is vacuously false; it doesn't matter what F is. It's not meaningless, just not something that would make sense to say. – Eliran Apr 24 at 20:51
  • Just as a side note, aren't ~∃x((Sx & ∀y(Sy → x = y)) and ~∃x((Sx & ∀y(Sy → x = y) & Ex) logically equivalent, seeing as, “x exists” is usually understood as equivalent to the generalisation Ey(y = x)? – Anon Apr 25 at 13:30
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    @Anon If 'Ex' is equivalent to ∃y(y=x), then yes, the two formulations are equivalent. – Eliran Apr 25 at 15:58
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Wikipedia describes Bertrand Russell's analyses of definite description using the phrase "the Emperor of Kentucky is bald".

Thus, a definite description (of the general form 'the F is G') becomes the following existentially quantified phrase in classic symbolic logic (where 'x' and 'y' are variables and 'F' and 'G' are predicates – in the example above, F would be "is an emperor of Kentucky", and G would be "is gray"):

∃x([Fx ∧ ∀y(Fy → x = y)] ∧ Gx)

Informally, this reads as follows: something exists with the property F, there is only one such thing, and this unique thing also has the property G.

If one replaces F with "is a sunken city" and G with "legendary" this may work for the "wide" perspective ("The sunken city of legend does not exist"). The existential quantifier would go outside the expression. This statement would then evaluate to false since either "Fx" or "Gx" would evaluate to false.


Wikipedia contributors. (2018, December 3). Theory of descriptions. In Wikipedia, The Free Encyclopedia. Retrieved 14:34, April 24, 2019, from https://en.wikipedia.org/w/index.php?title=Theory_of_descriptions&oldid=871839381

  • Thanks, it is supposed to be 'sunken city of legend' throughout which I have just edited to clarify. For the wide scope, I have the logic as ~∃x(Sx & ∀y(Sy→x=y)). I thought the reason the narrow scope was wrong was because it quantifies over things that do not exist, but I do not know how to express it in logic – Anon Apr 24 at 14:59
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    @Anon I edited my answer to remove the last paragraph based on your comment. However, there is still a difference. In the wide version there is "unique" which is not in the narrow version. The "unique" is in your symbolization: ∀y(Sy→x=y). I don't think the wide or narrow versions would be symbolized differently. – Frank Hubeny Apr 24 at 15:04

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