If the predicate is true for all the domain (universal quantification), could it also be considered true for some of the domain (existential quantification)?

Similar to a subset within a set?

Or since 'some' implies there must also be false values as well, are these considered exclusive?


forall(x)P(x) then exist(x)P(x)?


  • 2
    $\begingroup$ Can you please clarify your question? It's not very clear what you are asking. $\endgroup$ Sep 16 '12 at 20:33
  • $\begingroup$ Edited for clarity, let me know if more is needed, thanks! $\endgroup$
    – atlas1
    Sep 16 '12 at 21:04
  1. In standard first-order logic $\forall xFx$ does entail $\exists xFx$, but this is because we conventionally assume the domain of quantification is non-empty. There is an interesting history behind this convention, and it has some minor advantages. But it is easy enough to set up systems of first-order logic which allow for empty domains (e.g. as presented in Wilfrid Hodges's much used and much admired introductory book Logic), and then the entailment fails.
  2. It is notable that even with the standard convention, restricted 'all' does not entail the corresponding restricted 'some': $\forall x(Fx \to Gx)$ does not entail $\exists x (Fx \wedge Gx)$.
  3. "'Some' implies there must also be false values as well". Not so: "some" doesn't mean the same as "only some". Smelling the smoke, the teacher says "Someone has been smoking": he doesn't thereby rule out all four boys in the dorm have been smoking. (Of course, in some contexts, we do hear plain "some" as intended to convey "only some". But that contextual implicature can be cancelled -- it is always logically in order to say "Some F are G" and then add "And I don't rule out that they all are.")
  • $\begingroup$ Thank you for such a thorough response, this certainly helps! $\endgroup$
    – atlas1
    Sep 16 '12 at 22:28

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