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.

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.

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?

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.

But I like formal machinery!

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.

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!

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.

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.

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.

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.

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

Yes.



No. The identity is not necessary but contingent.

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.

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.