
16 Aug
2012
16 Aug
'12
11:07 p.m.
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. -- Live well, ~wren