Re: The king of france is ...

On Apr 19, 3:39 am, holden_o...@xxxxxxxx wrote:
On Apr 19, 12:07 am, Newberry <newberr...@xxxxxxxxx> wrote:





We note that the expression

    (x)(Fx -> Gx)        (1)

is neutral with respect to the grammatical number. Even if the set {x:
Fx} has only one element, (1) still applies. So

    (x)(Bx -> Rx)        (2)

expresses either "all the apples in my basket are red" or "the apple
in my basket is red." One would certainly agree that if there are
three apples in my basket and they are all red then the state of
affairs is expressed as (2). If there are two apples in my basket and
they are both red the state of affairs is also expressed as (2). If
there is one apple in my basket and it is red then it is certainly the
case that (2). The singular of

    "All the apples in my basket are red"        (3)

is

    "The apple in my basket is red"        (4)

Therefore (2) expresses both (3) and (4).

One could object however that since and English sentence must carry
the information about the cardinality of the subject class, neither
(3) nor (4) are equivalent to (2). In that case we would have to
express (2) as

   "The apple in my basket is red or all the apples in my basket are
red."

(4) would be expressed as

    (x)[Bx -> (Rx & (y)(By -> y=x)]        (5)

and (3) would be expressed as

   (x)[Bx -> (Rx & ~(y)(By -> y=x)]        (6)

However it is not clear how (2) could possibly become

    (Ex)[Bx & Rx & (y)(By -> y=x)]        (7)

Not so. (x)(Bx -> Rx), says all apples in my basket are red.

"(x)(Bx -> Rx)" is not expressing plural. So it says what it says even
if there is only one apple in the basket.

The apple in my basket is red, means, Ey(Ax(x=y <-> Bx) & Ry).

If there are no apples in my basket then all apples in my basket are
red, is false.

Not according to "classical" logic.

If there is one green apple in my basket then all apples in my basket
are red, is false.
etc.

If all apples are red then you are correct.
If E!(the x: x is a red apple and x is in my basket) then (x)(Bx ->
Rx), is true.- Hide quoted text -

- Show quoted text -

.

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