Necessarily, the number of planets is nine. ?!

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Reply Mon 21 Dec, 2009 06:47 am
P1. (the number of planets)=9.
P2. [](9>7).
therefore,
C. []((the number of planets)>7).

If P1 is true then every property of (the number of planets)
is equivalent to the same property of 9.

x=y =df AF(Fx <-> Fy).



(the number of planets)=9, is empirically true, but,
Necessarily((the number of planets)=9) is false.

Both premisses are true and the conclusion is false.

Why does this argument fail?
 
kennethamy
 
Reply Mon 21 Dec, 2009 07:03 am
@Owen phil,
Owen;113208 wrote:
P1. (the number of planets)=9.
P2. [](9>7).
therefore,
C. []((the number of planets)>7).

If P1 is true then every property of (the number of planets)
is equivalent to the same property of 9.

x=y =df AF(Fx <-> Fy).



(the number of planets)=9, is empirically true, but,
Necessarily((the number of planets)=9) is false.

Both premisses are true and the conclusion is false.

Why does this argument fail?


Opaque context? Quine's example.
 
Owen phil
 
Reply Mon 21 Dec, 2009 07:32 am
@kennethamy,
kennethamy;113216 wrote:
Opaque context? Quine's example.


Yes, Quine gave this example in (1962).

Opaque context??? What does this mean?

Is Leibnitz's definition of identity valid or not?
 
kennethamy
 
Reply Mon 21 Dec, 2009 07:49 am
@Owen phil,
Owen;113230 wrote:
Yes, Quine gave this example in (1962).

Opaque context??? What does this mean?

Is Leibnitz's definition of identity valid or not?


Referentially opaque.

See: Opaque context - Wikipedia, the free encyclopedia.

Definitions are not valid or invalid. But yes, Leibniz's (no "t") is accepted, but opaque contexts are an exception.
 
Owen phil
 
Reply Mon 21 Dec, 2009 10:20 am
@kennethamy,
kennethamy;113236 wrote:
Referentially opaque.

See: Opaque context - Wikipedia, the free encyclopedia.

Definitions are not valid or invalid. But yes, Leibniz's (no "t") is accepted, but opaque contexts are an exception.


LL. x=y =df AF(Fx <-> Fy).

What exception is there for (All F) in this definition?

EF(Fx & ~Fy) implies that x is different from y.

I believe that 'referrential opacity' is opaque.

But, I think the identity in question is not an instance of LL.

That is, (the x:Gx)=y means Ez(Ax(x=y <-> Gx) & z=y) by the definition of descriptions.

AF(F(the x:Gx) <-> Fy) may be different.

Perhaps this difference, if such there be, is what Quine meant.
 
Emil
 
Reply Mon 21 Dec, 2009 10:35 am
@Owen phil,
Owen;113208 wrote:
P1. (the number of planets)=9.
P2. [](9>7).
therefore,
C. []((the number of planets)>7).

If P1 is true then every property of (the number of planets)
is equivalent to the same property of 9.

x=y =df AF(Fx <-> Fy).



(the number of planets)=9, is empirically true, but,
Necessarily((the number of planets)=9) is false.

Both premisses are true and the conclusion is false.

Why does this argument fail?


I don't know but the identity in P1 is not necessarily true; it is contingent. That may have something to do with it.

What about: Substitution of identicals in a modal operator is not a valid inference. It seems that without this the argument is invalid.

You can find better symbols here.

The argument with proper symbols:

[INDENT]1. n = 8
2. □(9>7)
⊢3, □(n>7)

[/INDENT]Substitution of 9 for n is invalid since 9 is under a necessity operator. If the operator is removed the new argument is valid:

[INDENT]1. n = 8
2. 9>7
⊢3, n>7[/INDENT]
 
kennethamy
 
Reply Mon 21 Dec, 2009 10:40 am
@Owen phil,
Owen;113277 wrote:
LL. x=y =df AF(Fx <-> Fy).

What exception is there for (All F) in this definition?

EF(Fx & ~Fy) implies that x is different from y.

I believe that 'referrential opacity' is opaque.

But, I think the identity in question is not an instance of LL.

That is, (the x:Gx)=y means Ez(Ax(x=y <-> Gx) & z=y) by the definition of descriptions.

AF(F(the x:Gx) <-> Fy) may be different.

Perhaps this difference, if such there be, is what Quine meant.


I am sorry, but I cannot follow your notation.

But referential opacity is illustrated in the following: *

1. A believes that the man next door is the postman.
2. The man next door is A's girlfriend's father.
3. A believes that the postman is his girlfriend's father.

As a matter of fact, 3 is false. But 1 and 2 are true.

Therefore the argument is invalid.

So LL's law seems to fail here.

The "way out" is the opacity of the referential terms in this belief context. They do not really refer to what they appear to refer to. Modal contexts are also referentially opaque. Hence, Quine's hostility to modalities.

*But there is no need for formal machinery here. English is good enough.
 
Emil
 
Reply Mon 21 Dec, 2009 10:57 am
@kennethamy,
kennethamy;113280 wrote:
I am sorry, but I cannot follow your notation.

But referential opacity is illustrated in the following: *

1. A believes that the man next door is the postman.
2. The man next door is A's girlfriend's father.
3. A believes that the postman is his girlfriend's father.

As a matter of fact, 3 is false. But 1 and 2 are true.

Therefore the argument is invalid.

So LL's law seems to fail here.

The "way out" is the opacity of the referential terms in this belief context. They do not really refer to what they appear to refer to. Modal contexts are also referentially opaque. Hence, Quine's hostility to modalities.

*But there is no need for formal machinery here. English is good enough.


But I like formal machinery! Very Happy
 
kennethamy
 
Reply Mon 21 Dec, 2009 11:05 am
@Emil,
Emil;113285 wrote:
But I like formal machinery! Very Happy


To each his own, and you have found your own. I suppose I would like it too if I were as good at it as you are. I have to struggle with it. (I used to be better). Russell once wrote something like, when he was young he was a mathematician, when he became a little older, he was a logician. When he became much older, he was a philosopher. And when he got very old, he dabbled in politics. The steady degeneration of mind.
 
Zetherin
 
Reply Mon 21 Dec, 2009 12:18 pm
@kennethamy,
kennethamy;113289 wrote:
To each his own, and you have found your own. I suppose I would like it too if I were as good at it as you are. I have to struggle with it. (I used to be better). Russell once wrote something like, when he was young he was a mathematician, when he became a little older, he was a logician. When he became much older, he was a philosopher. And when he got very old, he dabbled in politics. The steady degeneration of mind.


That's another reason why it's great that Emil documents all of his logical and philosophical endeavors on a blog. I've been taught that it's very wise to keep notes on things while young, as memory will eventually fail. Heck, memory fails me now and I'm young!
 
Emil
 
Reply Mon 21 Dec, 2009 12:21 pm
@kennethamy,
kennethamy;113289 wrote:
To each his own, and you have found your own. I suppose I would like it too if I were as good at it as you are. I have to struggle with it. (I used to be better). Russell once wrote something like, when he was young he was a mathematician, when he became a little older, he was a logician. When he became much older, he was a philosopher. And when he got very old, he dabbled in politics. The steady degeneration of mind.


Well said. (12 characters)

---------- Post added 12-21-2009 at 07:58 PM ----------

Zetherin;113294 wrote:
That's another reason why it's great that Emil documents all of his logical and philosophical endeavors on a blog. I've been taught that it's very wise to keep notes on things while young, as memory will eventually fail. Heck, memory fails me now and I'm young!


I use writing as extended memory. It's very hard to judge if a complex argument is even valid if it is not on paper.
 
Owen phil
 
Reply Wed 23 Dec, 2009 09:18 am
@kennethamy,
kennethamy;113280 wrote:
I am sorry, but I cannot follow your notation.

But referential opacity is illustrated in the following: *

1. A believes that the man next door is the postman.
2. The man next door is A's girlfriend's father.
3. A believes that the postman is his girlfriend's father.

As a matter of fact, 3 is false. But 1 and 2 are true.

Therefore the argument is invalid.

So LL's law seems to fail here.

The "way out" is the opacity of the referential terms in this belief context. They do not really refer to what they appear to refer to. Modal contexts are also referentially opaque. Hence, Quine's hostility to modalities.

*But there is no need for formal machinery here. English is good enough.


I don't think that English is good enough to sort out these problems.

((the number of planets is nine) & [](9>7)) -> []((the number of planets is greater than 7)..is false.
ie.
(Ey(Ax(x=y <-> x numbers the planets) & y=9) & [](9>7)) ->
[]((the number of planets)>7) ..is false.

but,

((the number of planets is nine) & [](9 >7) -> ((the number of planets) is necessarily greater than 7)..is true.
ie.
(Ey(Ax(x=y <-> x numbers the planets) & y=9). & [](9>7)) -> ((the number of planets) []> 7), is true.

[](9>7) <-> (9 []> 7), is true.
[]((the x:Fx) > 7) <-> ((the x:Fx) []> 7), is false.

So much for Quine's referential opacity in modal contexts.

LL, Leibniz's Law, applied to descriptions is a different problem.

(the x:Fx)=y -> AG(G(the x:Fx) <-> Gy) is not valid.
(the x:Fx)=(the x:Hx) -> AG(G(the x:Fx) <-> G(the x:Hx)) is not valid.

Note, Russell proves that (E!(the x:Fx) & Ax(Gx)) -> G(the x:Fx) is true for all G. That is if (the x:Fx) exists, then it can be expressed as a value of the individual variable.

That is to say, for Russell...

(E!(the x:Fx) & (the x:Fx)=y) -> AG(G(the x:Fx) <-> Gy), is valid.

The restriction of predicates to material predicates seems unproblematic for him.

Note: (E!(the x:Fx) & ((the x:Fx)=y)) -> AG(G(the x:Fx) <-> Gy) is also not valid.
 
kennethamy
 
Reply Wed 23 Dec, 2009 09:28 am
@Owen phil,
Owen;113769 wrote:
I don't think that English is good enough to sort out these problems.

((the number of planets is nine) & [](9>7)) -> []((the number of planets is greater than 7)..is false.
ie.
(Ey(Ax(x=y <-> x numbers the planets) & y=9) & [](9>7)) ->
[]((the number of planets)>7) ..is false.

but,

((the number of planets is nine) & [](9 >7) -> ((the number of planets) is necessarily greater than 7)..is true.
ie.
(Ey(Ax(x=y <-> x numbers the planets) & y=9). & [](9>7)) -> ((the number of planets) []> 7), is true.

[](9>7) <-> (9 []> 7), is true.
[]((the x:Fx) > 7) <-> ((the x:Fx) []> 7), is false.

So much for Quine's referential opacity in modal contexts.

LL, Leibniz's Law, applied to descriptions is a different problem.

(the x:Fx)=y -> AG(G(the x:Fx) <-> Gy) is not valid.
(the x:Fx)=(the x:Hx) -> AG(G(the x:Fx) <-> G(the x:Hx)) is not valid.

Note, Russell proves that (E!(the x:Fx) & Ax(Gx)) -> G(the x:Fx) is true for all G. That is if (the x:Fx) exists, then it can be expressed as a value of the individual variable.


It is true that the number of planets is necessarily greater than seven? Why? (I think I smell a modal fallacy).
 
Owen phil
 
Reply Sun 27 Dec, 2009 05:50 am
@kennethamy,
kennethamy;113771 wrote:
It is true that the number of planets is necessarily greater than seven? Why? (I think I smell a modal fallacy).


The number of planets is nine, implies, the number of planets is necessarily greater than seven.

Proof:

1. The number of planets is exactly nine, is true.
2. The number of planets is exactly nine, implies, the number of planets is nine.
3. The number of planets is exactly nine, implies, the number of planets is
necessarily greater than 7.
4. The number of planets is exactly nine, implies, (the number of planets is nine & the number of planets is necessarily greater than seven).
((p -> q) & (p -> r)) -> (p -> (q & r)).
5. The number of planets is exactly nine, implies, (the number of planets is nine implies, the number of planets is necessarily greater than seven).
(p -> (q & r)) -> ((q & r) -> (q -> r)) ie. p -> (q -> r).
Therefore,
6. The number of planets is nine, implies, the number of planets is necessarily greater than seven. (because 5 and 1 are true.)
Q.E.D.

1a. Ax(x=9 <-> x numbers the planet).

2a. (Ax(x=9 <-> x numbers the planets) & 9=9) -> Ey(Ax(x=y <-> x numbers the planets) & y=9).

3a. (Ax(x=9 <-> x numbers the planets) & (9 []> 7)) -> Ey(Ax(x=y <-> x numbers the planets) & (y []> 7)).

4a. Ax(x=9 <-> x numbers the planets) -> (Ey(Ax(x=y <-> x numbers the planets) & y=9) & Ey(Ax(x=y <-> x numbers the planets) & (y []> 7))).

5a. 4a. Ax(x=9 <-> x numbers the planets) -> (Ey(Ax(x=y <-> x numbers the planets) & y=9) -> Ey(Ax(x=y <-> x numbers the planets) & (y []> 7))).

Therefore,

6a. Ey(Ax(x=y <-> x numbers the planets) & y=9) -> Ey(Ax(x=y <-> x numbers the planets) & (y []> 7)). Because of 5a and 1a are true.
Q.E.D.

Note: The number of planets is necessarily equal to nine, is also true.
Ey(Ax(x=y <-> x numbers the planets) & [](y=9)), is true.

Ken, I think your crystal ball of a nose is out of order.
Do you still smell a modal fallacy?
 
kennethamy
 
Reply Sun 27 Dec, 2009 06:24 am
@Owen phil,
Owen;114610 wrote:
The number of planets is nine, implies, the number of planets is necessarily greater than seven.

Proof:

1. The number of planets is exactly nine, is true.
2. The number of planets is exactly nine, implies, the number of planets is nine.
3. The number of planets is exactly nine, implies, the number of planets is
necessarily greater than 7.
4. The number of planets is exactly nine, implies, (the number of planets is nine & the number of planets is necessarily greater than seven).
((p -> q) & (p -> r)) -> (p -> (q & r)).
5. The number of planets is exactly nine, implies, (the number of planets is nine implies, the number of planets is necessarily greater than seven).
(p -> (q & r)) -> ((q & r) -> (q -> r)) ie. p -> (q -> r).
Therefore,
6. The number of planets is nine, implies, the number of planets is necessarily greater than seven. (because 5 and 1 are true.)
Q.E.D.

1a. Ax(x=9 <-> x numbers the planet).

2a. (Ax(x=9 <-> x numbers the planets) & 9=9) -> Ey(Ax(x=y <-> x numbers the planets) & y=9).

3a. (Ax(x=9 <-> x numbers the planets) & (9 []> 7)) -> Ey(Ax(x=y <-> x numbers the planets) & (y []> 7)).

4a. Ax(x=9 <-> x numbers the planets) -> (Ey(Ax(x=y <-> x numbers the planets) & y=9) & Ey(Ax(x=y <-> x numbers the planets) & (y []> 7)).

5a. 4a. Ax(x=9 <-> x numbers the planets) -> (Ey(Ax(x=y <-> x numbers the planets) & y=9) -> Ey(Ax(x=y <-> x numbers the planets) & (y []> 7)).

Therefore,

6a. (Ey(Ax(x=y <-> x numbers the planets) & y=9) -> Ey(Ax(x=y <-> x numbers the planets) & (y []> 7)). Because of 5a and 1a are true.
Q.E.D.

Ken, I think your crystal ball of a nose is out of order.
Do you still smell a modal fallacy?


Yes I do, since the number of planets is necessarily greater than 7 is false, but the number of planets is 9 is true. And a true proposition cannot imply a false proposition. You should smell a fallacy too. Everyone should smell a fallacy.

That necessarily, if the premises of a valid argument are true, the conclusion is true, is not the same thing, nor does it imply, that if the premises of a valid argument are true, the the conclusion is necessarily true. To imagine so is to commit a modal fallacy.
 
Owen phil
 
Reply Sun 27 Dec, 2009 07:20 am
@kennethamy,
kennethamy;114613 wrote:
Yes I do, since the number of planets is necessarily greater than 7 is false, but the number of planets is 9 is true. And a true proposition cannot imply a false proposition. You should smell a fallacy too. Everyone should smell a fallacy.

That necessarily, if the premises of a valid argument are true, the conclusion is true, is not the same thing, nor does it imply, that if the premises of a valid argument are true, the the conclusion is necessarily true. To imagine so is to commit a modal fallacy.


What? How did you decide that "the number of planets is necessarily greater than 7 is false" ...evidently your smeller stinks.

9=9 -> [](9=9) is a theorem. All truths of numbers are necessarily true.
9>7 -> [](9>7), is tautologous.

Where is the modal fallacy??

Note: The number of planets is necessarily equal to nine, is also true.
Ey(Ax(x=y <-> x numbers the planets) & [](y=9)), is true.
 
Emil
 
Reply Sun 27 Dec, 2009 07:22 am
@Owen phil,
Owen;114621 wrote:
What? How did you decide that "the number of planets is necessarily greater than 7 is false" ...evidently your smeller stinks.

9=9 -> [](9=9) is a theorem. All truths of numbers are necessarily true.
9>7 -> [](9>7), is tautologous.


Yes.

Quote:
Note: The number of planets is necessarily equal to nine, is also true.


No. The identity is not necessary but contingent.
 
Owen phil
 
Reply Sun 27 Dec, 2009 07:33 am
@Emil,
Emil;114622 wrote:
Yes.



No. The identity is not necessary but contingent.


The number of planets is equal to nine, is not necessary but contingent, is correct.

Necessarily, the number of planets is equal to the number nine..is false.

The number number of planets is necessarily equal to the number nine..is also contingently true.

Emil,
"What about: Substitution of identicals in a modal operator is not a valid inference. It seems that without this the argument is invalid."

x=9 -> [](x=9) for all numbers x.
We may indeed substitute 6+3 or 5+4 etc. for 9.

x=y -> [](x=y) is true for all numbers x and y, but,
If x or y are contingent descriptions, then it is false.

(the number of planets)=9, is true.
[]((the number of planets)=9),is false.
(the number of planets) []= 9, is true.

(the number of planets) > 7, is true.
[]((the number of planets)>7), is false.
(the number of planets) []> 7). is true.
 
kennethamy
 
Reply Sun 27 Dec, 2009 08:15 am
@Owen phil,
Owen;114621 wrote:
What? How did you decide that "the number of planets is necessarily greater than 7 is false" ...evidently your smeller stinks.

9=9 -> [](9=9) is a theorem. All truths of numbers are necessarily true.
9>7 -> [](9>7), is tautologous.

Where is the modal fallacy??

Note: The number of planets is necessarily equal to nine, is also true.
Ey(Ax(x=y <-> x numbers the planets) & [](y=9)), is true.


You mean that it was impossible that that we should have discovered that there were only 5 planets? Or 12 planets? 9 was the magic number? Hmmm.
 
Emil
 
Reply Sun 27 Dec, 2009 08:16 am
@Owen phil,
Owen;114623 wrote:
The number of planets is equal to nine, is not necessary but contingent, is correct.

Necessarily, the number of planets is equal to the number nine..is false.

The number number of planets is necessarily equal to the number nine..is also contingently true.


The last is false.
 
 

 
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