- #71
Owen Holden
- 92
- 0
AKG said:Owen
We can prove that [itex]\forall \, G\, (\neg (G(1/0))[/itex]. So, why does this not hold for [itex]G(x) \Leftrightarrow \neg (x = x)[/itex]? If it does hold, then doesn't this prove [itex]\neg (\neg (1/0 = 1/0))[/itex], i.e. [itex]1/0 = 1/0[/itex]?
No it does not prove ~~(1/0 =1/0), because it does not hold.
D1. 1/z =df (the x: x*z=1 & ~(z=0))
1/0 =df (the x: x*0=1 & ~(0=0)).
(1/0) is defined by the description (the x: x*0=1 & ~(0=0)).
Both (x*0=1) and ~(0=0) are contradictory, because Ax(x*0=0) and (0=0) and ~(0=1) are theorems.
Therefore, (1/0), (the x: x*0=1 & ~(0=0)), does not exist.
(1/0) is not undefined here.
As I said in the original post; we can drop the condition ~(z=0) if we assume that Ax(x*0=0).
"If we can assert that Ax(x*0=0), then D1. is simplified.
D1a. 1/z =df (the x: 1=x*z), and the proof still works."
Therefore, (1/0), (the x: x*0=1), does not exist.
D2. G(the x:Fx) =df Ey(Ax(x=y <-> Fx) & Gy)
That is, 1/0 = 1/0 <-> Ey(Ax(x=y <-> x*0=1) & y=y)
Which was proven false! i.e. 1/0 = 1/0 is false.
~(1/0 = 1/0) <-> ~Ey(Ax(x=y <-> x*0=1) & y=y).
which is obviously true. Because, Ey(Ax(x=y <-> x*0=1) & y=y) is false.
i.e. ~(1/0 =1/0) is true.
Note: It is incorrect to say, ~(1/0 =1/0) <-> Ey(Ax(x=y <-> x*0=1) & ~(y=y)). (which seems to be what you want to do).
This is an abuse of description theory, and I think Strawson wants to do the same thing.
AKG said:[Also, given your definition D1, wouldn't you say that it doesn't make sense for z=0, since we get:
[tex]1/0 \equiv x : (x \times 0 = 1 \wedge \neg (0 = 0))[/tex]
Rather, we should say: 1/0 = (the x: contradiction).
And it is a theorem of descriptions that (the x: contradiction) does not exist!
Therefore, (1/0) does not exist, even though it is defined.
and of course, no such x can satisfy [itex]\neg (0 = 0)[/itex]. Don't you agree that we simply cannot substitute z=0 into the definition of 1/z? If so, we certainly can't do it in your proof given in your first post.
Of course it makes sense to substitute 0 for z, because z includes 0.
D1. 1/z =df (the x: x*z=1 & ~(z=0)), and, D1a. 1/z =df (the x: 1=x*z), are both definitions which have sense for all values of z, including 0.
Az(1/z = (the x: x*z=1)) is true, by the definition D1a.