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Science Forum Index » Logic Forum » The king of france is ...
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| Newberry |
 Posted: Tue Apr 15, 2008 5:11 pm |
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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? |
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| Rupert |
| Posted: Tue Apr 15, 2008 6:40 pm |
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Guest
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On Apr 15, 9:18 pm, William Elliot <ma...@hevanet.remove.com> wrote:
Quote: On Tue, 15 Apr 2008, Newberry 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?
Is (1) an assumption? Hm, (2) follows directly from (1).
No, not correct.
I think Newberry is assuming (x)~Kx, (there does not exist a king of
France), in which case it seems to me that (1) and (2) both follow. |
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| Newberry |
| Posted: Tue Apr 15, 2008 6:45 pm |
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Guest
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On Apr 15, 9:40 pm, Rupert <rupertmccal...@yahoo.com> wrote:
Quote: On Apr 15, 9:18 pm, William Elliot <ma...@hevanet.remove.com> wrote:
On Tue, 15 Apr 2008, Newberry 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?
Is (1) an assumption? Hm, (2) follows directly from (1).
No, not correct.
I think Newberry is assuming (x)~Kx, (there does not exist a king of
France), in which case it seems to me that (1) and (2) both follow.
How do you translate (2) in English? |
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| Rupert |
| Posted: Tue Apr 15, 2008 7:06 pm |
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Guest
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On Apr 15, 9:45 pm, Newberry <newberr...@gmail.com> wrote:
Quote: On Apr 15, 9:40 pm, Rupert <rupertmccal...@yahoo.com> wrote:
On Apr 15, 9:18 pm, William Elliot <ma...@hevanet.remove.com> wrote:
On Tue, 15 Apr 2008, Newberry 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?
Is (1) an assumption? Hm, (2) follows directly from (1).
No, not correct.
I think Newberry is assuming (x)~Kx, (there does not exist a king of
France), in which case it seems to me that (1) and (2) both follow.
How do you translate (2) in English?
I have an idea it's not quite the sentence you're after. We could
translate it like this:
For all x, if x is the king of France, and there is some y such that
either y is not the king of France, or y=x, then Wx.
I think maybe you wanted this sentece:
(x)((Kx & (Ay)(Ky -> y=x)) -> Wx)
which can be translated as
For all x, is x is the unique king of France, then Wx. |
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| William Elliot |
| Posted: Tue Apr 15, 2008 11:18 pm |
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Guest
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On Tue, 15 Apr 2008, Newberry wrote:
Quote: 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?
Is (1) an assumption? Hm, (2) follows directly from (1).
No, not correct. |
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| Guest |
| Posted: Wed Apr 16, 2008 1:42 am |
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On Apr 15, 11:11 pm, Newberry <newberr...@gmail.com> wrote:
Quote: 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?
Yes.
Because there is no present king of France, (Kx & ~ExKx) <-> Kx, for
all x.
(1) (x)((Kx & ~ExKx) -> Wx), is true for all W's.
(2) (x)[((Kx & ~ExKx) & (Ey)(Ky -> y=x)) -> Wx], is also true for all
W's.
Because Kx -> ExKx, (Kx & ExKx) <-> Kx.
That is, (1) becomes (x)((Kx & ExKx & ~ExKx) -> Wx.
But (ExKx & ~ExKx) is a contradiction.
therefore,
(1) (x)(contradiction -> Wx).
(1a) contradiction -> (x)Wx.
(1b) Tautology v (x)Wx.
(1c) Tautology.
ie. (1) (x)(Kx -> Wx) is tautologous for all W's
The same argument applies to (2). |
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| William Elliot |
| Posted: Wed Apr 16, 2008 2:08 am |
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Guest
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On Tue, 15 Apr 2008, William Elliot wrote:
Quote: On Tue, 15 Apr 2008, Newberry 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?
Is (1) an assumption? Hm, (2) follows directly from (1).
No, not correct.
How to express (2) in English? As
Kx -> (Ey)(Ky -> y=x)
(Kx & (Ey)(Ky -> y=x)) <-> Kx
and
(1) <-> (2).
Thus to express (2) in English is the same as expressing (1).
(1) can be expressed as
"all French kings are women".
which of course is what you mean by Wx. |
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| Newberry |
| Posted: Wed Apr 16, 2008 3:22 am |
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Guest
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On Apr 15, 10:06 pm, Rupert <rupertmccal...@yahoo.com> wrote:
Quote: On Apr 15, 9:45 pm, Newberry <newberr...@gmail.com> wrote:
On Apr 15, 9:40 pm, Rupert <rupertmccal...@yahoo.com> wrote:
On Apr 15, 9:18 pm, William Elliot <ma...@hevanet.remove.com> wrote:
On Tue, 15 Apr 2008, Newberry 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?
Is (1) an assumption? Hm, (2) follows directly from (1).
No, not correct.
I think Newberry is assuming (x)~Kx, (there does not exist a king of
France), in which case it seems to me that (1) and (2) both follow.
How do you translate (2) in English?
I have an idea it's not quite the sentence you're after. We could
translate it like this:
For all x, if x is the king of France, and there is some y such that
either y is not the king of France, or y=x, then Wx.
I think maybe you wanted this sentece:
(x)((Kx & (Ay)(Ky -> y=x)) -> Wx)
which can be translated as
For all x, is x is the unique king of France, then Wx
It sounds a little bit like "the king of France is wise", does it not? |
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| Newberry |
| Posted: Wed Apr 16, 2008 3:25 am |
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Guest
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On Apr 16, 12:08 am, William Elliot <ma...@hevanet.remove.com> wrote:
Quote: On Tue, 15 Apr 2008, William Elliot wrote:
On Tue, 15 Apr 2008, Newberry 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?
Is (1) an assumption? Hm, (2) follows directly from (1).
No, not correct.
How to express (2) in English? As
Kx -> (Ey)(Ky -> y=x)
(Kx & (Ey)(Ky -> y=x)) <-> Kx
and
(1) <-> (2).
Thus to express (2) in English is the same as expressing (1).
(1) can be expressed as
"all French kings are women".
which of course is what you mean by Wx.- Hide quoted text -
How about
(x)[(Kx -> Wx) & (Ey)(Ky -> y=x)]
? |
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| Guest |
| Posted: Wed Apr 16, 2008 6:15 am |
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On Apr 16, 10:14 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
Quote: holden_o...@yahoo.ca writes:
On Apr 15, 11:11 pm, Newberry <newberr...@gmail.com> 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?
Yes.
Because there is no present king of France, (Kx & ~ExKx) <-> Kx, for
all x.
(1) (x)((Kx & ~ExKx) -> Wx), is true for all W's.
(2) (x)[((Kx & ~ExKx) & (Ey)(Ky -> y=x)) -> Wx], is also true for all
W's.
Because Kx -> ExKx, (Kx & ExKx) <-> Kx.
That is, (1) becomes (x)((Kx & ExKx & ~ExKx) -> Wx.
But (ExKx & ~ExKx) is a contradiction.
therefore,
(1) (x)(contradiction -> Wx).
(1a) contradiction -> (x)Wx.
(1b) Tautology v (x)Wx.
(1c) Tautology.
ie. (1) (x)(Kx -> Wx) is tautologous for all W's
You seem to be using the word "tautologous" in an odd way.
--
"At the Microsoft-sponsored cocktail reception in the Galaxy Ballroom
that evening, Robert Dees urges us 'to network on behalf of the people
of Iraq,'"
-- Naomi Klein reports on Microsoft's efforts to further democracy.- Hide quoted text -
- Show quoted text -
Exactly what is it that you believe is odd? |
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| Chris Menzel |
| Posted: Wed Apr 16, 2008 9:02 am |
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Guest
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On Wed, 16 Apr 2008 12:30:25 -0400, Jesse F. Hughes
<jesse@phiwumbda.org> said:
Quote: holden_owen@yahoo.ca writes:
ie. (1) (x)(Kx -> Wx) is tautologous for all W's
You seem to be using the word "tautologous" in an odd way.
Exactly what is it that you believe is odd?
A tautology is usually taken to be a kind of necessary truth.
More exactly still, a tautology in mathematical logic is usually taken
to be a logical truth of propositional logic -- that is, a proposition
that is true simply in virtue of the meanings of its constituent boolean
operators. |
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| Jesse F. Hughes |
| Posted: Wed Apr 16, 2008 9:06 am |
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Guest
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Newberry <newberryxy@gmail.com> writes:
Quote: On Apr 15, 10:06 pm, Rupert <rupertmccal...@yahoo.com> wrote:
On Apr 15, 9:45 pm, Newberry <newberr...@gmail.com> wrote:
On Apr 15, 9:40 pm, Rupert <rupertmccal...@yahoo.com> wrote:
On Apr 15, 9:18 pm, William Elliot <ma...@hevanet.remove.com> wrote:
On Tue, 15 Apr 2008, Newberry 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?
Is (1) an assumption? Hm, (2) follows directly from (1).
No, not correct.
I think Newberry is assuming (x)~Kx, (there does not exist a king of
France), in which case it seems to me that (1) and (2) both follow.
How do you translate (2) in English?
I have an idea it's not quite the sentence you're after. We could
translate it like this:
For all x, if x is the king of France, and there is some y such that
either y is not the king of France, or y=x, then Wx.
I think maybe you wanted this sentece:
(x)((Kx & (Ay)(Ky -> y=x)) -> Wx)
which can be translated as
For all x, is x is the unique king of France, then Wx
It sounds a little bit like "the king of France is wise", does it
not?
No, not much.
As I'm sure you know, "The King of France is wise" implicitly asserts
that there *is* a king of France, whereas your sentence (and Rupert's
sentence, which is surely what you meant) assert no such thing.
This is a pretty silly attempt. You have to know that your attempted
formalization falls pretty far from the mark. After all, anyone
puzzling over statements involving the current French monarch should
surely be familiar with the usual translations and be able to see at a
glance how your translation misses the mark.
--
"And God Himself won't help you if this goes bad as despite your
beliefs I can assure you that angry people will against all law if
necessary tear their rage out of your hides if it goes badly."
-- James S. Harris, on the dangers of criticizing his mathematics |
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| Jesse F. Hughes |
| Posted: Wed Apr 16, 2008 9:10 am |
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Guest
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Newberry <newberryxy@gmail.com> writes:
Quote: How about
(x)[(Kx -> Wx) & (Ey)(Ky -> y=x)]
?
How about it? It doesn't say "The king of France is wise." It says
something like, "For all x, if x is the king of France, then x is wise
and every king of France is equal to x."
As a consequence of this statement, if there is a king of France, then
there is only one thing in the universe! After all, suppose that
there are two things, x and z and that y is the king of France. Then
the above odd sentence implies that y = x and y = z.
You're really grasping here.
The above sentence is clearly *not* equivalent to:
(Ex)(Kx & Wx & (Ay)(Ky -> x=y)),
which is the formalization of "The king of France is wise."
--
Jesse F. Hughes
"If the world weren't rather strange, by now I should at least be with
some research group talking about my number theory research."
-- James S. Harris learns the world is a funny place |
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| Jesse F. Hughes |
| Posted: Wed Apr 16, 2008 9:14 am |
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Guest
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holden_owen@yahoo.ca writes:
Quote: On Apr 15, 11:11 pm, Newberry <newberr...@gmail.com> 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?
Yes.
Because there is no present king of France, (Kx & ~ExKx) <-> Kx, for
all x.
(1) (x)((Kx & ~ExKx) -> Wx), is true for all W's.
(2) (x)[((Kx & ~ExKx) & (Ey)(Ky -> y=x)) -> Wx], is also true for all
W's.
Because Kx -> ExKx, (Kx & ExKx) <-> Kx.
That is, (1) becomes (x)((Kx & ExKx & ~ExKx) -> Wx.
But (ExKx & ~ExKx) is a contradiction.
therefore,
(1) (x)(contradiction -> Wx).
(1a) contradiction -> (x)Wx.
(1b) Tautology v (x)Wx.
(1c) Tautology.
ie. (1) (x)(Kx -> Wx) is tautologous for all W's
You seem to be using the word "tautologous" in an odd way.
--
"At the Microsoft-sponsored cocktail reception in the Galaxy Ballroom
that evening, Robert Dees urges us 'to network on behalf of the people
of Iraq,'"
-- Naomi Klein reports on Microsoft's efforts to further democracy. |
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| Jesse F. Hughes |
| Posted: Wed Apr 16, 2008 11:30 am |
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Guest
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holden_owen@yahoo.ca writes:
Quote: On Apr 16, 10:14 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
holden_o...@yahoo.ca writes:
On Apr 15, 11:11 pm, Newberry <newberr...@gmail.com> 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?
Yes.
Because there is no present king of France, (Kx & ~ExKx) <-> Kx, for
all x.
(1) (x)((Kx & ~ExKx) -> Wx), is true for all W's.
(2) (x)[((Kx & ~ExKx) & (Ey)(Ky -> y=x)) -> Wx], is also true for all
W's.
Because Kx -> ExKx, (Kx & ExKx) <-> Kx.
That is, (1) becomes (x)((Kx & ExKx & ~ExKx) -> Wx.
But (ExKx & ~ExKx) is a contradiction.
therefore,
(1) (x)(contradiction -> Wx).
(1a) contradiction -> (x)Wx.
(1b) Tautology v (x)Wx.
(1c) Tautology.
ie. (1) (x)(Kx -> Wx) is tautologous for all W's
You seem to be using the word "tautologous" in an odd way.
Exactly what is it that you believe is odd?
A tautology is usually taken to be a kind of necessary truth. Since
the fact that there is no king of France is *not* a necessary truth,
the fact that
(x)(Kx -> Wx)
is true is also not a necessary truth. It is a contingent truth (true
because, as it happens, ~(Ex)(Kx) is also true) and hence it is not a
tautology.
--
"Kim liked the math I did for her and gave me quite a few
groceries... likely so many groceries that they would have cost Kim
about what she pays for two whole packages of cigarettes. Few people
have ever rewarded me for my work as much as Kim did." -- Usenet nut |
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