Quantifier 'for some but not all'

Question

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?

Answer 1

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.

Answer 2

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

Answer 3

Strictly as stated this is not possible without resorting to a trick of some kind.

I'll use $Y x \mathop. \varphi(x)$ as the quantifier for some but not all, following its representation in the logical hexagon.

If your domain has only a single element in it, then the value of $Y x \mathop. \varphi(x)$ is always false. However $\varphi$ might hold or might fail for the sole element of the domain.

Here's a partial solution.

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Suppose the domain has cardinality at least $1$ and that we have a constant symbol $c$.

$$ \exists x \mathop. \varphi(x) \;\;\text{iff}\;\; \varphi(c) \lor (Y x \mathop. \varphi(x)) $$

As proof, suppose $\varphi$ holds for $c$, then $\varphi(c)$ holds and $\exists x \mathop. \varphi(x)$ holds.

Suppose $\varphi$ does not hold for $c$, but for holds for some $d \neq c$. In this case, $\exists x \mathop. \varphi(x)$ holds and $Y x . \varphi(x)$ holds.

Suppose $\varphi$ does not hold for any element in the domain. Then $\exists x \mathop. \varphi(x)$ fails and $\varphi(c)$ fails and $Y x \mathop. \varphi(x)$ also fails.