On Thu, Aug 16, 2012 at 8:07 PM, wren ng thornton <wren@freegeek.org> wrote:

On 8/15/12 2:55 PM, Albert Y. C. Lai wrote:

On 12-08-15 03:20 AM, wren ng thornton wrote:

(forall a. P(a)) -> Q <=> exists a. (P(a) -> Q)

For example:

A. (forall p. p drinks) -> (everyone drinks)
B. exists p. ((p drinks) -> (everyone drinks))

In a recent poll, 100% of respondents think A true, 90% of them think B
paradoxical, and 40% of them have not heard of the Smullyan drinking
paradox.

:)

Though bear in mind we're discussing second-order quantification here, not first-order.

Can you expand on what you mean here? I don't see two kinds of quantification in the type language (at least, reflexively, in the context of what we're discussing). In particular, I don't see how to quantify over predicates for (or sets of, via the extensions of the predicates) types.

Is Haskell's 'forall' doing double duty?