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
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