Ontological arguments

[Newton_meter]

@Libertarian

The assertion ~<>G is a denial of a positive ontological proposition. How can there even be a denial of a positive ontological proposition? In other words, how can the statement “The greatest possible being cannot possibly exist” be defended? Clearly, that proposition is absurd. The more subtle assertion, <>~G, is equally absurd. “It is possible that the greatest possible being does not exist.” The only reasonable modal assertion with respect to G as a premise is <>G. “It is possible that the greatest possible being exists.”

You might have thought that I agreed with you, but I didn’t. You want the proposition G to be “a greatest possible being exists”. How to define “greatest”? Your greatest is a little tricky, because we can’t look at just a single possible world and identify the greatest being in that world.

Instead, the greatest being in a world has to be something like the one that exists in the most other possible worlds. But even then, you can’t insist that G -> G in every possible world. You need “greatest possible” to be “exists in all possible worlds”.

So G = “there is a being that exists in all possible worlds”.

Now, you set out to prove the truth of G. And you want to use, as justification for one of your assumptions, G.

~<>G (= ~G) is an assertion that in all possible worlds it is not the case that there is a being that exists in all possible worlds. That seems perfectly reasonable to me. I can find nothing contradictory (symbolically) with it. If you perceive a contradiction, then you have not rendered the argument into symbols correctly. We are not obligated to accept the truth of G until you prove it.

<>~G is an assertion that in some possible world it is not the case that there is a being that exists in all possible worlds. It seems like a weaker form of ~G, but they turn out to be equivalent given your other assumption.

<>G V ~<>G seems to be the agnostic (or possibly “weak atheist”) position. Certainly one of the disjuncts is true and one false, but the agnostic does not claim to know which.

<>G & <>~G is your (I think) characteristic of the agnostic position as one that is contradictory. But I think you mistake the possible in “it’s possible that there is a God and possible that there isn’t” as metaphysical possibility, when it is really just an admission that the agnostic makes no assumption about the dijunction G V ~G.

That is acceptable if, as in this case, you have postulated that if it is true in all possible worlds, then it is true in the actual world. Of course, there is nothing about the nature of the conjecture that makes any positive ontological proposition about it reasonable. In fact, it does not make an ontological commitment of any kind.

I believe that the truth value of Goldbach’s conjecture, like the truth value of “2 + 2 = 5” is a logical necessity. Just as “2 + 2 = 5” is not merely contingently false, but necessarily false; I believe that Goldbach’s conjecture is either necessarily true or necessarily false. We just don’t know which.

Thus, I can say:

G = every even number greater than two is the sum of two primes

(G -> G): in all possible worlds, if every even number greater than two is the sum of two primes then in all possible worlds every even number greater than two is the sum of two primes

<>G: in some possible world every even number greater than two is the sum of two primes

Therefore G: every even number greater than two is the sum of two primes

Of course, you can spot the bug in my proof. Nobody would (or should) accept an assertion that Goldbach’s conjecture is true in some possible world as one the assumptions in a proof of Goldbach’s conjecture. It’s the same bug as in:

G = there is a being that exists in all possible worlds

(G -> G): in all possible worlds, if there is a being that exists in all possible worlds then in all possible worlds there is a being that exists in all possible worlds

<>G: in some possible world there is a being that exists in all possible worlds

Therefore G: there is a being that exists in all possible worlds

And as I have pointed out for as many years, the accessiblity relation of the MOP is Euclidean, not merely symmetric. You are NOT free to invalidate a proof by disregarding one of its premises. G -> G is a premise of the proof. Ignoring the accessibility relation is enthymemic.

A Euclidean relation is one that satisfies: for all x, y, z. xRy & xRz -> yRz

Posit two worlds, 0 and 1. R = {(0,0),(0,1),(1,0),(1,1)}. R is Euclidean, here’s a simple proof: choose an arbitrary x, y, and z each in {0,1} such that xRy and xRz. Then yRz is true, because R relates all y and z each in {0,1}.

  World 0    World 1
A   F          T
B   F          F

Now, we have a countermodel to (A -> B) -> (~B -> ~A). A -> B is true in 0 (by virtual of A being false). The premise ~B is true in world 0 (since ~B is true in all worlds x such that 0Rx). However, the conclusion ~A is not true, because ~A is not true in all worlds x such that 0Rx.

That is rather like attacking an arithmetic proof by pointing out that 1 is not a number, it’s a numeral. I think you ought to know that when Eris refers to G as a proposition, he means “G”. G is indeed a being. “G” is a statement.

I was not aware that erislover refered to G as a proposition at all, which is what I was pointing out. He referred to it as a being. In the logic that I know, truth values are assigned to propositions, not beings.

[erislover]

@Newton meter

He uses a rule like this (A -> B) -> ~B -> ~A, which is not valid. To see why…

You can derive the modal modus tollens.

(A -> B) & ~B -> ~A
I trust this isn’t arguable. We may apply necessitation to this.

(A -> B) & ~B ->~A

We know from the modus ponens that when
(p -> q & p) -> q

We know from (M) that
p -> p

so then
( A -> B ) & ~B -> ~A

QED

Maybe you don’t like that. Well, let’s look at your argument again:

consider that A -> B and ~B are true in this world. Now, there are two kinds of possible worlds, those (like this one) where A -> B is true, and those where A -> B is false (nothing compels A -> B to be true at every possible world).

Those where (A -> B) is possible. Actuality demands necessary possibility… all worlds must account for the possibility of ( A -> B ). So what you’re suggesting is we consider the conjunction of (1) and (2) below…

(1) <>( A -> B) [must be true!!]
(2) ~B [assumption]
(3) ( <>A -> <>B ) [distribution]
(4) (~~A -> ~~B) [definition of <>]
(5) ~A V ~~A [excluded middle]
(6) ~A V ~~B [from 4, 5]
(7) ~A [from 2, 6]
QED

Think about it this way. In our world, we want to consider an inventor of eyeglasses B and eyeglasses A. If there are eyeglasses, then there is an inventor of eyeglasses. A->B. You’re suggesting that in some possible world, even if there are no eyeglass inventors in any of them (~B), there is still the possibility that there is a pair of eyeglasses. Does this really make sense to you?

I was not aware that erislover refered to G as a proposition at all, which is what I was pointing out. He referred to it as a being. In the logic that I know, truth values are assigned to propositions, not beings.

There is the proposition and the sense of the proposition. It is tiresome to be perfectly rigorous all the time when English speakers presumably do not need it. When the proposition G obtains, that being actually exists. I can’t believe this is a point of contention. But yes, it was shorthand to suggest that the being is true. Of course I did mean that the proposition “this being actually exists” is true. As a programmer, do you always make sure that when people are talking about character arrays they don’t use the word “character” and instead refer to a byte? How do you deal with the fact that signed short is really an integer, even if it isn’t an unsigned int? Come on.

[zwaldd]

@Libertarian

For example, by declaring that the argument must be flawed, you are denying that it may be valid. Why does your own criticism not apply to your own argument?

Because I am denying that the argument may be valid. If you presume G=necessary existence before establishing G, both <> G and <>~G become undefined.

[MaxTheVool]

@Libertarian

Point well taken, Max. But wouldn’t you agree that it is untenable to hold both a respect for my views and an opinion that I am intentionally equivocating? (That’s what he said in his third point.) How can anyone intellectually respect a man who has not made a mistake, but rather has intentionally set out to deceive, manipulate, and well, basically troll?

I don’t think it’s a black and white issue. I don’t think you “intentionally set out to deceive”, and I certainly don’t think you’re a troll. On the other hand, if someone posted a GD thread with a incendiary title like “Resolved: All non-atheists are deluded simpletons”, I would fully expect you to jump in with both ontological barrels blazing, without first stopping to discuss the differences between God, g, <>G, and so forth. Not that there’s any particular reason why you should, in a situation like that…

In the general literature, the MOP is controversial, yes. But not because of the definition. Not because of some perceived circularity. Not because philosophers are suspicious of the symbology, the formulation, or the intent of Plantinga et al. The controversy is over whether <>G is a more reasonable premise than <>~G.

Is there no discussion in the philosophical literature of the meta-questions that I find interesting, ie, “suppose the ontological proof were found to be utterly and completely convincing and without flaws… what would be the implications towards faith and belief in the real world?”.

It is unreasonable to the extreme to call for people simply to shut up about it because of inevitable recalcitrance and belligerence. Don’t you agree?

I absolutely agree… I was responding to your statement that you believe the MOP should convince people not to think you’re an idiot for believing in God, which is a statement I still find a bit puzzling and pointless, in that:
(a) as far as I can tell, you yourself do not believe in God because of the MOP
(b) in fact, the God you believe in, spiritually, is a superset of the MOP god, so I don’t see much of a connection between your beliefs and the MOP at all
and
© it’s entirely possible (and quite common, I would hope) for even hard core atheists to respect the intellect, integrity and spirituality of theists, and vice versa, proof or no proof
I also find it interesting that you think that a reasonable person could find the proof persuasive, sound and correct, but still not believe in God. Which I suppose is the difference between Philosophy and something like math. When I see people who (as happens in GQ from time to time) remain convinced that .999999… is not the same as 1, I have to admit that I lose respect for them. You asked a reasonable question, but a Proof-with-a-capital-P was offered that demonstrated that you were incorrect. Accept it.
Related anecdote: Several years after I graduated from college, I saw one of my friends from high school, who in the intervening years had become a protestant minister (I think a Methodist, although I may be mistaken). We were talking, and I asked her a question phrased something like “I hope you don’t find this to be a condescending question, but are there really any meaningful differences between Methodists, Presbyterians, Lutherans, and so forth”. She responded that when she heard me ask a question beginning with “I hope you don’t find this to be a condescending question…” she was sure that it was going to end up with “… but how can any smart college-educated person believe in God?”.

[Newton_meter]

@erislover

You can derive the modal modus tollens.

(A -> B) & ~B -> ~A
I trust this isn’t arguable. We may apply necessitation to this.

(A -> B) & ~B ->~A

Ooops. This is not valid:

(A -> B) -> (A -> B)

I believe you were thinking of

(A -> B) -> (A -> B)

Consider a countermodel, again we only need two worlds

   0   1
A  T   T
B  T   F

In world 0, (A → B) is true and A is true. B is not true in world 0. Thus, A → B is false in world 0.

QED

Maybe you don’t like that.

Only because it’s wrong.

Well, let’s look at your argument again:Those where (A → B) is possible. Actuality demands necessary possibility… all worlds must account for the possibility of ( A → B ). So what you’re suggesting is we consider the conjunction of (1) and (2) below…

(1) <>( A → B) [must be true!!]
(2) ~B [assumption]
(3) ( <>A → <>B ) [distribution]

Ooops, wrong again. This is not valid:

<>(A → B) → (<>A → <>B)

<>(A → B) means that A → B is true in some possible world. <>A means that A is true in some possible world. Nothing constrains the world that witnesses the truth of <>(A → B) and the one that witnesses the truth of <>A to be the same.

   0   1
A  F   T
B  F   F

In world 0, A -> B is true, and thus <>(A -> B) is true (with world 0 as a witness). Also, in world 0 <>A is true, with world 1 as the witness. However, in world 0 (and in world 1), <>B is not true.

Think about it this way. In our world, we want to consider an inventor of eyeglasses B and eyeglasses A. If there are eyeglasses, then there is an inventor of eyeglasses. A->B. You’re suggesting that in some possible world, even if there are no eyeglass inventors in any of them (~B), there is still the possibility that there is a pair of eyeglasses. Does this really make sense to you?

You’re trying to reason about the binary logical function “implies” by thinking of eyeglass makers? Come on, that’s a freshman mistake. A -> B has the same truth table as ~A V B. That’s it.

Let’s look at your example. We have satisfied ourselves that in this world, if there are eyeglasses then there is an inventor of eyeglasses. A -> B. We have also satisfied ourselves that there is no inventor of eyeglasses in any possible world ~B. One consequence of this is that there are certainly no eyeglasses in this world, ~A, so let me take mine off before proceeding.

Clearly, these eyeglasses things are something that don’t occur in our world, so we should really use a different word for them lest we get confused. Now, what we haven’t shown that there are no eyeglasses (whatever those are) in any possible world, because there could be possible worlds where A -> B does not hold. That is, where the existence of eyeglasses (whatever those are) does not imply an inventor, perhaps they are the result of some natural process.

Of course, there is a fix if you want A -> B in every world, you can require (A -> B). And that’s all I’ve been saying.

There is the proposition and the sense of the proposition. It is tiresome to be perfectly rigorous all the time when English speakers presumably do not need it.

What I am suggesting is that the only reason that you can even claim that <>G with a straight face is because you are deliberately not being precise about what you mean by G. <>G does not translate to “God is possible”, it translates to “in some possible world there is a being that exists in all possible worlds”. And you assume this in your proof that “there is a being that exists in all possible worlds”.

You may think I was merely being specious with my “proof” of Goldbach’s conjecture, but from where I sit the situation is exactly the same.

As a programmer, do you always make sure that when people are talking about character arrays they don’t use the word “character” and instead refer to a byte? How do you deal with the fact that signed short is really an integer, even if it isn’t an unsigned int? Come on.

Only when there might be confusion. But if you want to draw an analogy, I suggest that a better one is the distinction between an expression in the syntax of a language, and a runtime value. Or between a statement in the syntax of the language and a state transition in the operational semantics. Yes, I am unabashedly rigorous about such distinctions.

[erislover]

@Newton meter

Ooops. This is not valid:

(A -> B) -> (A -> B)

Good thing I didn’t take that step, then!

Ooops, wrong again. This is not valid:

<>(A -> B) -> (<>A -> <>B)

Negation and necessity are distributive.
(1) <> (A -> B)
(2) ~~(A -> B)
(3) ~(~A ->~B)
etc.

Unless you suppose that ~(A → B) → (~A → B)? Or is it … → (A → ~B)?

Nothing constrains the world that witnesses the truth of <>(A -> B) and the one that witnesses the truth of <>A to be the same.

Show me why it is forbidden by the axioms and definitions of modal logic. Use your insistence on rigor to bust out the symbols. Seriously.

You’re trying to reason about the binary logical function “implies” by thinking of eyeglass makers? Come on, that’s a freshman mistake.

I’m trying to do something until you can point out which step was actually flawed by the axioms and definitions. So far you’ve just shown that you can write the equation “1+1=4” and that this somehow acts as a counter-example to arithmetic.

Now, what we haven’t shown that there are no eyeglasses (whatever those are) in any possible world, because there could be possible worlds where A -> B does not hold.

These worlds, however, would still require that <>(A -> B). If ~B, they cannot achieve this. This is why ~A. Please review the second proof.

And that’s all I’ve been saying.

This is true. But it is a weaker version than is possible.

What I am suggesting is that the only reason that you can even claim that <>G with a straight face is because you are deliberately not being precise about what you mean by G. <>G does not translate to “God is possible”…

I felt I was reasonably precise already. “The greatest being that can exist.” This entails two claims, about which I will be even more precise than is necessary.
The statement “The … being that can exist” possibly obtains. <>G
The statement “The greatest being…” obtains, where ‘greatness’ indicates a relation between actuality and necessity. G → G
Conclusion: The statement “The greatest being that can exist actually exists” obtains. G

You may think I was merely being specious with my “proof” of Goldbach’s conjecture, but from where I sit the situation is exactly the same.

Which is unsurprising, and I meant to comment on it in my last post: all axioms, tautologies, and theorems are necessary, so yes, if GC is the case then GC is necessarily the case. However, I would like to see how you break down the GC into those two propositions.

Yes, I am unabashedly rigorous about such distinctions.

I don’t mind when confusion is possible, but I felt rather confident this wasn’t the case here.

[erislover]

Newton, Let’s focus on the second proof, please.

[zwaldd]

@erislover

The statement “The … being that can exist” possibly obtains. <>G

But since <>~G is ruled out a priori, the statement is “The being that must exist” possibly obtains.
“Must exist” AND “possibly obtains”? Not valid.
“Must” assumes G. Circular.

[erislover]

@zwaldd

But since <>~G is ruled out a priori, the statement is “The being that must exist” possibly obtains.

It is not a matter that <>G is ruled out a priori but that the definition of the being includes <>G. Maybe you want to discuss impossible beings, or beings that don’t have to be possible. But that’s not what this proof is about. If you’d like <>~G, the definition would be “The greatest being that doesn’t have to exist”. In this case, we might be talking about a being who can manifest paradoxes, or exist and not exist at the same time, or any number of logical impossibilities. And it certainly wouldn’t have anything to do with the being we’re talking about.

[zwaldd]

@erislover

It is not a matter that <>G is ruled out a priori but that the definition of the being includes <>G.

I didn’t say <>G is ruled out, I said <>~G is ruled out. Without <>~G, <>G is simply G. And that is both the starting premise and the conclusion. Circular.

[Anduril]

@Libertarian

It fails because of the coherent definition of g as the greatest possible being. The assertion “<>~g” reads “it is possible that the greatest possible being does not exist”. If it is the greatest possible being, then it cannot be impossible. Therefore, <>~g (or ~g) is a contradiction.

There is no contradiction. You seem to be applying a false dichotomy i.e. either G is necessary or impossible. But G could simply not exist without being impossible. (If G exists, G is necessary, If G does not exist, either G is impossible or G simply does not exist.)

[erislover]

@zwaldd

Without <>~G, <>G is simply G.

I don’t see that at all.

[Faldage]

@erislover

Good thing I didn’t take that step, then!
Negation and necessity are distributive.
(1) <> (A -> B)
(2) ~~(A -> B)
(3) ~(~A ->~B)
etc.

Unless you suppose that ~(A → B) → (~A → B)? Or is it … → (A → ~B)?

This modal logic is mighty peculiar stuff if ~(A -> B) is equivalent to (~A -> ~B). This is certainly not the case in real life. If A is ‘I clap my hands’ and B is ‘my cat jumps into my lap.’ It is clearly the case that ~(A -> B) and it is also clearly the case that ~(~A -> ~B). Of course, that’s just in real life. If modal logic only works in conditions that we cannot possibly know anything about then maybe that explains things.

[ultrafilter]

~(A -> B) <-> (A & ~B)
(~A -> ~B) <-> (A v ~B)

Take A and B true to see the difference.

zwaldd’s claim is that ~<>~G -> (<>G -> G). This is true, as shown below:

~<>~G
G (modal quantifier negation)
G (axiom, modus ponens)
<>G -> G (I forget the name of this rule but it’s valid)

However, I don’t see that ~<>~G is a premise of the argument.

[zwaldd]

@ultrafilter

However, I don’t see that ~<>~G is a premise of the argument.

Libertarian: The more subtle assertion, <>~G, is equally absurd. “It is possible that the greatest possible being does not exist.” The only reasonable modal assertion with respect to G as a premise is <>G. “It is possible that the greatest possible being exists.”

[Newton_meter]

erislover, I think you’re just making a few little mistakes that are leading you to invalid conclusions. I (essentially) do this for a living, and I still have to double and triple check technical content that I produce by hand.

First, let me explain what I mean by a countermodel. Since Lib is talking about S5, things are easy (we can ignore the accessibility relation). A model is a set of possible worlds and a valuation that assigns to each pair of a world and atomic proposition a truth value in {false, true}.

Now, we could proceed to give a semantics to the propositional connectives and the modal quantifiers, but I will skip that. The semantics defines what it means for a proposition to be true at a world in a model.

A proposition which is true at every possible world in every model is valid.

To demonstrate that a proposition is not valid, it suffices to show a world in a model where the proposition fails. That’s what I’ve been doing above.

The thing is, it’s fairly easy for me to construct these countermodels for you (I have lots of practice), but it’s not so easy for me to tell you where you went wrong in the proof. That is partly because you didn’t show all your work.

Let’s try to prove your “modal modus tollens”. We start with plain old classical modus tollens:

(A -> B) & ~B -> ~A

This is a classical tautology, so we can apply the rule of necessitation:

((A -> B) & ~B -> ~A)

The K axiom is that (P -> Q) -> (P -> Q). Using this axiom and a modus ponens (OK, even I can skip a step), we get:

((A -> B) & ~B) -> ~A

Now, provable in S5 is:

(P & Q) -> (P & Q)

And we use (P -> Q) & (Q -> R) -> (P -> R) to get

(A -> B) & ~B -> ~A

Which is what I would probably call “modal modus tollens”, and is what I suggest Libertarian use in his proof. You want to replace (A -> B) above with A -> B, using the axiom scheme:

(P -> Q) & (Q & R -> S) -> (P & R -> S)

but in order for that to be valid, you would need:

(A -> B) -> (A -> B)

Which is not valid (it would be the inverse of the K rule).

Quite honestly, the easiest way to demonstrate that this is not valid is to provide a countermodel:

   0  1
A  T  F
B  F  F

Now, in world 0, A is false so (A -> B) is true. However, (A -> B) is not true in world 0 (in fact, A -> B fails in world 0).

These worlds, however, would still require that <>(A -> B). If ~B, they cannot achieve this. This is why ~A. Please review the second proof.

You’re right about the first part, but not the second. Again, here is a countermodel:

   0  1
A  F  T
B  F  F

A -> B is true in world 0, as is ~B. However, ~A is not true in world 0 (or world 1). You are right, that <>(A -> B) is true in world 1 (world 0 witnesses its truth), but that is still consistent with ~B.

[Newton_meter]

@erislover

Newton, Let’s focus on the second proof, please.

Of course. Sorry.

You have a step where you distribute possibility over implication. That is not valid.

I’ve already shown a countermodel, here is some simple symbol manipulation:

<>(A -> B)
{defn of ->}
== <>(~A V B)
{DeMorgan’s for V}
== <>~(A & ~B)
{defn of <>}
== ~(A & ~B)
{distribution of wrt &}
== ~(A & ~B)
{DeMorgan’s for &}
== ~A V ~~B
{defn of <>}
== ~A V <>B
{defn of ->}
== A -> <>B

[erislover]

Well I am certainly corrected. Thank you for the gentle corrections, all.

What fallacy am I committing class? Affirming the consequent! sigh

Newton, here is the foundation you explain:

<>( a -> b )
~~( a -> b )
~( a & ~b ) (thanks, ultrafilter)
~( a & ~b) (I checked, this IS valid! :rolleyes:@me)
a -> ~~b
~b hypothesis
~a
<>~a but certainly not ~a.

Thanks again for not kicking my ass, though my obstinacy deserved it.

[Lamia]

@MaxTheVool

Is there no discussion in the philosophical literature of the meta-questions that I find interesting, ie, “suppose the ontological proof were found to be utterly and completely convincing and without flaws… what would be the implications towards faith and belief in the real world?”.

I mentioned Spinoza some time back, and he may have some of what you’re looking for in Part One of his Ethics, “Concerning God”. If you can manage to suffer through his unfortunate choice of writing style (holy geometric proofs!), Spinoza is pretty interesting. He presents an ontological argument, but he doesn’t stop there. He also describes what the God he has proven to exist must be like.

And what Spinoza’s God is like is…the universe. God is the greatest thing that exists because God is everything that exists. According to Spinoza and contrary to popular belief, this God does not act from a purpose or towards a goal, does not respond to prayers, and does not make value judgements or care what human beings get up to in their spare time.

So what are the implications towards faith and belief in the real world? The end of religion as it is known in the West. What we have is a God that exists, but whose existence renders religion worthless – a mere collection of misguided superstitions. (Funnily enough, as an atheist I already believed this!)

I absolutely agree… I was responding to your statement that you believe the MOP should convince people not to think you’re an idiot for believing in God, which is a statement I still find a bit puzzling and pointless, in that:
(a) as far as I can tell, you yourself do not believe in God because of the MOP
(b) in fact, the God you believe in, spiritually, is a superset of the MOP god, so I don’t see much of a connection between your beliefs and the MOP at all
and
(c) it’s entirely possible (and quite common, I would hope) for even hard core atheists to respect the intellect, integrity and spirituality of theists, and vice versa, proof or no proof
*

Personally, I have always mistrusted religious people who pull out ontological proofs because ontological proofs never seem to be the actual basis of their faith. They’re always something they discovered or throught of after they already accepted the existence of and began to worship a particular deity. If their “logical proof” isn’t what convinced them in the first place, why should it convince me? Why aren’t they admitting to their real reasons? Are their real reasons actually less compelling?

If anything, ontological proofs have confirmed my atheism. Apparently the best, most rational evidence theists can come up with is something even they don’t find particularly inspiring, and that can only prove the existence of a “God” that neither expects nor deserves faith or worship.

[olanv]

@Loopydude

It’s not even that original. Godel had a similar idea, I think. Necessity is not a valid presupposition. It forces you to accept that if God can exist somehow, He must exist anyhow. If you simply deny necessity, there’s nothing left to talk about. And there’s nothing stoping you from doing that. All you’re left with is somebody who believes in God, and someone who doesn’t, same as before. An empty exercise.

I have no problem with someone saying that something exists in some way.
Let’s say that way is as a self refuting concept.

So now you’re left with the burden of proving whether or not self refuting concepts exist. Then you’re left with the burden of proving that if you exclude the concept of “God” that it’s impossible to make the claim.

This is easier to think about when you have a statement like “otherness necessarily exists in all worlds”. There are ways to prove that, which act self referentially on the ability and/or purpose to type or believe you are typing or are interpreting or even making the claim of the statement in the first place.

Is there anything in this proof that even addresses the difference between a concept and a linguistic token, or self referrential claims?

I recall a couple of us going over this already where the idea came up a couple times to define perfection with regards to omnipotence as creating existence while not existing.

This acts self referrentially upon the meaningfulness of the proof that God exists, as God would have to not exist in order for omnipotence to parse with regards to “perfect”. We were taking these nebulous (undefined) terms like “exists”, “better” and “perfect” and using it to state that a God who creates existence who doesn’t exist is better than one that creates existence and does exist – with regards to the potence of that being, therfor God doesn’t exist… etc…