Re: The king of france is ...

On Apr 15, 8:11 pm, Newberry <newberr...@xxxxxxxxx> wrote:
Let Kx be x = king of France. I take it that

    (x)(Kx -> Wx)        (1)

is true. And so is

    (x)[(Kx & (Ey)(Ky -> y=x)) -> Wx]    (2)

Correct?

How about this one

(x)[(Kx -> Wx) & (y)(Ky -> x=y)]

If x is a king of France then x is wise, and x is unique?


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