Let's consider the quantifier corresponding to the expression 'for some but not all'. Is it possible to define the universal quantifier in terms of this quantifier and sentence connectives only?

  • 5
    $\begingroup$ I propose using a tophat in honor of Abraham Lincoln. $\endgroup$ – Brian Rushton Jan 9 '14 at 3:34
  • $\begingroup$ The negation of your quantifier is, 'either for all or for none' and I doubt that it's possible to define $\forall$ in terms of this. $\endgroup$ – Alraxite Jan 9 '14 at 19:24
  • $\begingroup$ @Alraxite that's exactly why this quantifier should not be defined in a way that renders It's negation as 'either for all or for none', I think. $\endgroup$ – user76568 Jan 9 '14 at 19:47

You want to define $\forall$ and $\exists$ in term of the expression :

'for some but not all'.

1) May we assume that the translation of the expression is :

$\exists x \phi(x) \land \lnot \forall x \phi(x)$ ?

This sentence implies (if I'm right) that :

  • the universe is not empty

  • something in the universe is $\phi$

  • something is not $\phi$

  • there are at least two thing in the universe.

But the sentence is also equivalent to :

$\exists x \phi(x) \land \exists x \lnot \phi(x)$

2) If my "translation" is right and if my inferences are also right, I doubt that we can define $\exists$.

The $\exists$ quantifier implies the existence of at least one object in the universe, whlie the new "quantifier" implies the existence of at least two objects.


Let's call $\triangle$ the quantifier. Then $\exists x : P(x)$ should be equivalent to $(\triangle x : P(x)) \vee (\triangle x : \lnot P(x))$

  • 2
    $\begingroup$ One direction is false: $\Delta x ( P(x) )\vee \Delta x (\neg P(x))$ is false when $P(x)$ holds of every x but $\exists x (P(x))$ is true. $\endgroup$ – hot_queen Jan 9 '14 at 3:49
  • $\begingroup$ Oops yes you are right! $\endgroup$ – dani_s Jan 9 '14 at 3:51

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