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Old 07-19-2003, 06:06 AM   #1
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Default A logical proof of atheism

If a purported term is described or defined by a contradictory predication, then, that term does not exist.

1. ~E!(ix: Gx & ~Gx)
It is not the case that, that which has the property G and does not have the property G exists.

-----------------------------------------
Proof:

1a. E!(ix:Fx) <-> Ey(x=y <->x Fx),
See: *14.02 Principia Mathematica, Russell and Whitehead.

1b. E!(ix: Gx & ~Gx) <-> Ey(x=y <->x. Gx & ~Gx)
1c. E!(ix: Gx & ~Gx) <-> Ey(x=y <->x (contradiction))
1d. E!(ix: Gx & ~Gx) <-> Ey(Ax~(x=y))
1e. E!(ix: Gx & ~Gx) <-> Ey(~Ex(x=y))
1f. E!(ix: Gx & ~Gx) <-> Ey(contradiction).
1g. E!(ix: Gx & ~Gx) <-> (contradiction).
1h. ~(E!(ix: Gx & ~Gx)).
1i. ~<>(E!(ix: Gx & ~Gx))

i.e. EF(~<>(E!(ix: Fx))), is a theorem.
There is some property for which it is impossible that, the x such that x has that property, exists.
------------------------------------------

If God is described or defined by a contradictory predication, then, that God does not exist.

Therefore, it is possible that God does not exist ..can be asserted.


2. God (defined by some description) does not exist, is possible.

And, if we include in the definition/description of God,
D1. Its possible existence implies its necessary existence. (from Leibnitz, I think), then:

3. God exists is possible, implies, God exists is necessary. By D1.

4. God exists is not necessary, implies, God exists is not possible.

By, 3 and (p -> q) <-> (~q -> ~p)

4. God does not exist is possible, then, God exists is not possible.

By: 3 and, It is not necessary that p is true, if and only if, It is
possible that not-p is true.
i.e. By, ~(Necessary p) <-> (Possible ~p).

But, 2. God does not exist, is possible.

Therefore,

5. God exists, is not possible.

By: 2, 4, and (p & (p -> q)) -> q.


That is to say, the following argument is valid.

1. God does not exist, is possible.
and,
2. God exists is possible, implies, God exists is necessary.
therefore,
3. God exists, is not possible.

Theists, it seems, do not need to show that 'God exists, is true', rather, what they need to show is that 'God exists, is possibly true'.

What do you think?

Witt
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Old 07-19-2003, 06:20 AM   #2
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I've apparently forgotten my logic class...why does the possible existence of something imply its necessary existence and the possible non existence of an entity implies it can't exist?
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Old 07-19-2003, 06:44 AM   #3
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Hi strubenuff,

"I've apparently forgotten my logic class...why does the possible existence of something imply its necessary existence and the possible non existence of an entity implies it can't exist?"

It is a part of the description of 'God' that its possible existence entails its necessary existence. It only applies to God-like entities.

"..and the possible non existence of an entity implies it can't exist?"
This consequence also applies to God(s), for which, it's possibility ensures its necessity, and it does not apply to all entities.

Witt
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Old 07-19-2003, 08:11 AM   #4
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Maybe you've shown problems with Leibnitz's(?) reasoning, but I don't think you've conclusively proven that it is impossible for God to exist.
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Old 07-19-2003, 01:00 PM   #5
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Witt writes:

-----------------------------------------
If a purported term is described or defined by a contradictory predication, then, that term does not exist.
-----------------------------------------

and..

-----------------------------------------
Therefore, it is possible that God does not exist ..can be asserted.
-----------------------------------------

Your handling of quotation (or the lack thereof) suggests that your argument does not adequately distinguish the use of a term from its mention. Surely the theist can grant pretty much everything you've said, and still contend that both God exists, and that the statement "God exists" is possibly false, actually false, or even necessarily false.

As the saying goes: "It all depends what you mean..."

Regards,

Bilbo.
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Old 07-21-2003, 12:11 AM   #6
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Default Re: A logical proof of atheism

There apears to me to be a serious problem with your proof. First of all, in no way have you "defined or described God by something contradictory to God." All you did was say there is a probability that God does not exist. However, had you placed this statement after your other statements, it would be clear that it is simply a logical impossibility. Either God does or does not exist, logically. Logic does not deal with probability. Even if it does make logical sense, it cannot be proven, so the whole thing is moot.

Secondly, technically statement 4. (the first 4. and the third actual statement) is still true if both "God exists is not necessary" and "God exists is not possible" are false... which basically is what a theist would argue.

Also, I think your proof just "feels" wrong... let me state teach statement again but more mathematically (and I hope still expressing the same thing).

1. P(God does not exists) > 0 [this one causes problems]
2. P(God exists)>0 --> P(God exists) = 1
3. P(God exists) not equal 1 --> P(God exists)=0
4. P(God does not exist)>0 --> P(God exists) = 0
5. P(God exists) = 0

So basically, it all works assuming 1. is correct, WHICH YOU CANNOT!
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Old 07-23-2003, 10:36 AM   #7
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Default Re: A logical proof of atheism

Quote:
Originally posted by Witt
If a purported term is described or defined by a contradictory predication, then, that term does not exist.

Isn't this just an unnecessarily complicated way of stating the law of contradiction?

And what is a "purported" term? Is that different from an "actual" term? What about an "ostensible" term?

1. ~E!(ix: Gx & ~Gx)
It is not the case that, that which has the property G and does not have the property G exists.

-----------------------------------------
Proof:

1a. E!(ix:Fx) <-> Ey(x=y <->x Fx),
See: *14.02 Principia Mathematica, Russell and Whitehead.

1b. E!(ix: Gx & ~Gx) <-> Ey(x=y <->x. Gx & ~Gx)
1c. E!(ix: Gx & ~Gx) <-> Ey(x=y <->x (contradiction))
1d. E!(ix: Gx & ~Gx) <-> Ey(Ax~(x=y))
1e. E!(ix: Gx & ~Gx) <-> Ey(~Ex(x=y))
1f. E!(ix: Gx & ~Gx) <-> Ey(contradiction).
1g. E!(ix: Gx & ~Gx) <-> (contradiction).
1h. ~(E!(ix: Gx & ~Gx)).
1i. ~<>(E!(ix: Gx & ~Gx))

i.e. EF(~<>(E!(ix: Fx))), is a theorem.
There is some property for which it is impossible that, the x such that x has that property, exists.
------------------------------------------
Well the problem with all this is that it assumes that logic is a valid means of "proving" things.
What's your basis for such an assumption (without assuming logic in your answer, please)?
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Old 07-23-2003, 10:45 AM   #8
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theophilus,
Quote:
Well the problem with all this is that it assumes that logic is a valid means of "proving" things.
What's your basis for such an assumption (without assuming logic in your answer, please)?
How can one write anything without assuming logic? You could only do so by intentionally constructing meaningless sentences.
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Old 07-23-2003, 12:08 PM   #9
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Quote:
Originally posted by spacer1
theophilus,

How can one write anything without assuming logic? You could only do so by intentionally constructing meaningless sentences.
Unfortunately, that doesn't solve the problem.

This is supposed to be a "philosophical" thread; I was posing an epistemological question.
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Old 07-24-2003, 08:54 AM   #10
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There is no 'proof' witout logic.

theophilus:
Isn't this just an unnecessarily complicated way of stating the law of contradiction?

Most definitly not.

theophilus: And what is a "purported" term? Is that different from an "actual" term? What about an "ostensible" term

Yes, in the case of non-existence, the purported term is not!

The existent present king of France, does not exist.

Witt
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