# Quantifier 'for some but not all'

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?

• I propose using a tophat in honor of Abraham Lincoln. – Brian Rushton Jan 9 '14 at 3:34
• 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. – Alraxite Jan 9 '14 at 19:24
• @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. – 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 $(\triangle x : P(x)) \vee (\triangle x : \lnot P(x))$

• 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. – hot_queen Jan 9 '14 at 3:49
• Oops yes you are right! – dani_s Jan 9 '14 at 3:51