Kant didn’t know anything about this proof. He had been dead a long time. Regarding <>~G, once again, I’m waiting to hear defense of the position that the greatest possible being might not be possible.
G is not the greatest possible being. In logic, we deal in statements, not beings. You know that.
So G is the statement that the greatest possible being exists. Without knowing what you mean by greatest, how can we say one way or another?
Depends on the definitions you use. On the one hand, I can’t offer a defense, but in that case, you cannot allow for the possibility of ~G (this is what I find disingenuous). Also, this definition is also tautological, hence trivial. On the other hand, if you define existence as merely hypothetical in N(G) v N(~G), then there’s a third alternative that G simply doesn’t exist in actuality and that <>~G and <>G are both equally possible.
I agree that the move from (G -> G) to (G -> G) is not controversial. However, it is still a move that needs to be made.
(G -> G) -> <>G -> G is not a theorem of S5. If you present a proof that it is, then there is an invalid step in the proof. So, the criticism of the invocation of modal modus tollens is not a criticism of the argument, it is merely to point out how you can seemingly prove something in S5 that is not valid in S5. The criticism of the argument is that it is not valid, but that criticism comes with an easy and not very controversial fix.
Anyway, here is my real criticism of the proof. It is empty. Circular. Allow me to illustrate.
Suppose you and I sat down over a beer, and examined your proof. I asked you to present your premises one at a time. I agreed completely with your first premise, (G -> G). Now, before we even go on to the second premise, it is interesting to ask what we are committing ourselves to by agreeing to this first one.
What we are agreeing to is that G is either true in every world, or false in every world. That is, G V ~G. An informal argument should suffice (though a symbolic one can be given). Presume that there is (at least) a world where G is true. Then, G -> G is true in that world and G is true in that world. Thus, G is true in every world if G is true in one world.
On the other hand, if there is (at least) a world where ~G, then G is not necessary and we have ~G in every world. By (classical propositional) modus tollens and G -> G in every world, we have ~G in every world. Thus, G is false in every world if G is false in one world.
And this is solely a consequence of your noncontroversial first premise. In fact, I would claim that this observation is what the symbols really “mean” in your first premise. But then, if all worlds agree on the truth or falsity of G, we don’t need modal logic at all as long as we don’t try to talk about any modal propostion that involves any atomic proposition other than G. In that case, we can just erase all the boxes and diamonds on every modal proposition and use propositional logic for the proof.
Again, as a consequence of your first premise, we don’t need to speak of possible worlds at all. The first premise is essentially G -> G (and conversely, ~G -> ~G), which is utterly incontroversial and obviously tautological. You still haven’t proven that G is true, though, and I invite you to do so.
Libertarian: “Assume G…”.
Though there is no logical fallacy in the argument, I submit that there is a rhetorical fallacy: it is having the opposite effect on at least one (I imagine more) listeners.
(G -> G) -> (<>G -> G) is valid in S5. The same proof can be used for any proposition that is not logically contingent to conclude <>P -> P.
Again, off to la la land. Suppose you were one of my CS colleagues, and let’s choose an open question in the field: does P = NP. You don’t even have to know what the question is, just that it is an (important) open question.
You tell me that you believe that P = NP. This is probably (by far?) a minority position in the field, but since the question is open, it is a perfectly valid, non-contradictory position. I will grant you that this is a rational position, even if your only justification is “intuition” (in fact, that’s the only justification that anyone has for ~(P = NP) at this point anyway).
But then you go a step further. You tell me that you have a “proof” of P = NP. See, you explain, the truth of P = NP is a matter of logical necessity (I agree). Then, you say, you can prove that P = NP is true with the simple assumption that it is true in some possible world. Further, you tell me, this is the real reason that you believe P = NP, because of this proof.
Now, my respect for your position has fallen a little. Before you offered up the proof, I found your belief perfectly rational. Now, because of the reason you give for your belief I am starting to wonder about how rational that belief is.
Now back to the real world. In the real world, and in the interests of full disclosure, I share your belief in God. I find belief and disbelief, on nothing more solid than “intuition” to both be rational positions (though some forms of belief and disbelief are extremely irrational in practice). I don’t find your proof compelling, however.
Well, stated or not, one of these premises holds true a priori: <>~G or ~<>~G. Which is it?
Here is one defense.
Let’s consider the “greatest being”. What does that mean? To me, it means that there is a total order on the set of all possible beings, so that we can identify the “greatest” elements of that set. Note that there may be more than one.
Now, if the set of all possible beings is nonemtpy, then there exists a greatest possible being, and you can posit <>G = “it is possible that the greatest possible being exists”, and make the claims you are making about its denial.
However, you are making a stronger claim about these greatest possible beings. You are claiming that a being is greater than another if it exists in more worlds than the other, and that the greatest imaginable (I am deliberately avoiding “possible” here) being is one that exists in all worlds, because there is no way to get greater than that. Now, I think you are compelled to render G as
G = “the greatest imaginable possible being exists”
And nothing compels us to accept the possibility of this (and there is nothing contradictory in its denial).
More pithily, I think that you are conflating two different meanings of “possible”. Using parentheses to force the parse, there is a difference between “the (greatest possible) being” and “the greatest (possible being)”. Your (G -> G) is about the former and <>G is about the latter.
Well, in Lib’s defense, the answer is we don’t know.
Either Goldbach’s conjecture is true or it isn’t. Which is it?
Friggn’ make me look up Goldbach’s conjecture…:D. “True or not” is not the same as possibly not true or not possibly not true. So I would say Goldbach’s conjecture is possibly not true (however unlikely), pending enough info to say ‘not possibly not true’.
That said…if the answer to my question to Lib is “we don’t know”, then <>~G must hold true, my objection to which I stated above.
No, no, no.
“We don’t know” means that we accept the truth of <>~G V ~<>~G (it is just an instance of the excluded middle), but we don’t know which of the (mutually exclusive) disjuncts is true.
You want to try to render this as <>G & <>~G, however that is not generally valid. Not all things are possible, and Libertarian is deliberately formulating his proposition G so that it is either possible or impossible but not both.
Using my trick of erasing all the modal operators, my “we don’t know” is G V ~G, but we don’t know which disjunct is true. Your “we don’t know” seems to be G & ~G, which is a contradiction.
Well, I don’t actually have a Ph.D., but that should be considered a personal and not a technical failing.
OK, good enough. In that case G hasn’t been fully fleshed out as the greatest possible being, because a greater existence, ~<>~, holds true a priori for all things known to exist.
Your parenthetical splicing exactly mirrors the justification I presented earlier for the assumptions. Yet you suggest this is conflating two meanings of possible? In this breakdown, G is simply the proposition that “a being exists.” The assumptions are used to describe the logical properties of the being in question.
Which, of course, is the proof.
For kicks, I want to show you something I’ve been working on lately. Now, under S5 we can demonstrate that the actuality of a proposition implies its possibility. I’ll use Brouwer because it is quick:
A -> <>A (Brouwer)
<>A -> <> A (M applied to <>A)
A -> <>A (2, 1); ( p -> (q -> r) ) -> (p -> r)
What’s interesting is what we can derive from that.
p -> <>p (our above proof applied to p)
~(p & ~<>p) (def’n of ->)
~(p & ~~~p) (def’n of <>)
~(p & ~p) (double negation)
~(p & ~p) (distribution of with respect to &)
~~(p -> p) (def’n of ->)
<>(p -> p).
Interestingly, this means that at least one world must define something with that property.
I could not, however, demonstrate that
<>(p ->p)
which makes sense, because under S5 it can be shown (according to Stanford, though I didn’t try to demonstrate it myself) that
<>A -> A.
So if we could show that <>(p ->p), we’d show that
(p -> p)
p -> p (M applied to (p ->p)
is a theorem, which would pretty much destroy S5 by declaring all actuality necessary.
However, noting that modal status is always necessary, every world must account for the possibility of our first assumption (well, the not strict version), including the world where G obtains. The question here would be whether G obtains in the same world as (G -> G). If
(G -> G)
then this must be the case… which is the proof.
Taking Suber’s lead for the content of the proposition, this is to suggest that there is (at least) one world where perfection obtains (<>G), and that all worlds must define perfection as non-contingent (that is, (G -> G)). I would humbly suggest that this is not a particularly specious translation (into logical symbols) of perfection.
zwaldd wants us to consider that it is possible that there is a world where perfection does not obatin; that is, <>~G. This is, of course, weaker than the assertion that perfection is impossible (~G), but it is still a stronger claim than <>G. At least, to my mind it is–denial of something is always stronger, as a rule, then allowing for its possibility. So, to me, anyone who even thought to suggest <>~G must agree to <>G, and agreeing to <>G entails G (given our proof). The reason why is, I think, intuitive: if any time we suggest something is possible we may also suggest that its opposite is possible, then necessary existence is impossible–something I think even zwaldd would object to (wouldn’t you?). This means that it is not generally the case that asserting something is possible is just as reasonable as asserting it is not possible. To suggest otherwise is to necessarily deny necessity.
If, for example, we suggest that perfection is identical with necessary existence, then zwaldd is suggesting that there is a world where necessary existence doesn’t obtain–which is not the case, since necessary existence by definition always obtains. And, at a minimum, perfection entails necessary existence if we accept the definition of perfection as non-contingency. That is, at a minimum, necessary existence is sufficient for [logical] perfection.
So, summary:
The rejection of the proof, given Suber’s content, requires that we feel non-contingency is not a necesssary property of perfection. This would reject
(G -> G).
Otherwise, it requires the assertion that there is some world where logical perfection doesn’t exist. This would replace
<>G with <>~G.
If we suggest that perfection necessarily entails non-contingency, and we do not rule out perfection existing somewhere, then I’d suggest that the proof is not only valid but sound. I think it is necessarily shifting the burden of proof on those that would reject that definition of perfection by providing their own, or demonstrating the rejection of the possibility of perfection. (Of course, neither of these rejections would be symbolic proofs.)
For me, after all these debates, I’d have to suggest the proof is sound. What this means is that logical perfection necessarily exists. This is the semantic limit of the proof. Which, of course, is what I said in the first post to this thread (if necessary existence is sufficient for perfection–something I’ll have to think on and will make no commitment to at this time.)
Note, however, that a finite implimentation of this no longer requires the proof to be sound. That is, in a finite application, like the networking semantics you brought to light (I didn’t mean to imply it was your semantics, of course), <>~G is certainly reasonable: we simply never created that “world”. But what I am suggesting is that if we built a network (keeping with these semantics) which had all permutations of worlds, the proof would be sound. Is that clear? If we were not constrained by physical limitations in building every possible point on a network, then we would have the result of the proof. Since logic is an ideal case, all possible permutations are “built”, so every point on the network has G. Hence it is sound.
This, of course, brings to light the rejection of the “communication” objection which Ludovic brought up on the first page, which would suggest that the network semantics is not reasonably equivalent to [an intuitionist] S5. In light of that paper, and your presentation of the relevant points of it here Newton, I think we can dismiss that objection.
Summary of the summary (:p): so long as necessary existence is a sufficient condition for perfection, and perfection is non-contingent, the proof (to me, who will not accept <>~G given the discussion above) is sound and entails that perfection is necessary. Which, again, is trivial. But it would definitely be more interesting if necessary existence is not sufficient for perfection (it is only necessary), in which case the proof is much stronger. Again, I’ll have to think about this.
Yes, I suggest that there are two different meanings of “possible” here. The being which is greatest among all “possible beings” certainly exists in some possible world.
The (unproven) greatest imaginable being that exists in all possible worlds cannot be assumed to exist in the proof of the proposition that he exists.
I see the argument that (G -> G) is true of some proposition of the form “being X exists”. I see that <>G is true of some proposition of the form “being X exists”. There is no reason to believe that there is a being that simultaneously makes both propositions true.
If we do make that assumption, we are neglecting to consider an important case: that all beings are contingent. To construct the proof, we have to assume that not all beings are contingent, and that is exactly the conclusion of the proof.
We (or at least, I) keep forgetting that the only characteristics of Libertarian’s proposition G is that it is a proposition that is not contingent and is possible. The proof works for any proposition with these characteristics. What the proof actually shows is that if there is any proposition that is not contingent and possible, then there is a necessary proposition.
The first claim justifies us to say that (P -> P), if the premise is true then it is necessary, and this fact itself is necessary. The second allows us to claim <>P, that the premise is possible.
I don’t find this astounding. We already knew that there were necessary propositions. There are infinitely many such propositions. For instance:
P = “the successor of eight is nine”
Now, I still fail to see how choosing a particular logically necessary proposition and calling it “God” qualifies as a rational “belief in God”.
And, lest you think that it sounds like I’m agreeing, I still see no evidence that any of those propositions is of the form “being X exists in all possible worlds”, unless propositions are beings that have existence?
So, of the competing propositions ~~G and <>~G which is the denial of something and which is allowing for its possibility?
To my mind, a proposition P is “at least as strong as” Q if P |- Q, that is, Q is provable from P. P is “stronger than Q” is P |- Q and not Q |- P. In this case, under the assumption (G -> G), <>~G |- ~G and ~G |- <>~G, so neither proposition is stronger (under the assumption).
Consider some other G of which we may say (G -> G) but which we don’t know the truth of <>G, like Goldbach’s conjecture.
Which premise, “Goldbach’s conjecture is possible” or “Goldbach’s conjecture is possibly false” should be used? How does that constitute a “proof” of Goldbach’s conjecture?
I think zwaldd is making the same point that I am: there is nothing contradictory with assuming <>~G instead of <>G (or even ~G instead of <>G). There is no “default” truth value for the possibility of a proposition. And since we can easily show that <>G <-> G under the assumption (G -> G), then I think that the obvious conclusion is that the truth value of <>G is as problematic as the truth value of G.
I have read that Plantinga and Hartshorne claim that this proof demonstrates that a belief in God is as rational as disbelief. What I think it shows is something more trivial: “this proof does not give us any reason to prefer belief over disbelief or vice versa”. That is why I characterize it as empty.
Well, obviously, that depends on the content of G.
This is one analysis that I might agree with, depending on the propositions in question. Another analysis might be the number of assumptions necessary to produce Q. If, for example,
(A & B & C) |- Q
and
D |- Q
where ~( (A & B & C) |- D )
then I would suggest D is the stronger assumption. I do not, however, suggest that this is the only way to measure strength, because if we are left with the case where
A & C |- Q
and
B & C |- Q
this method does not allow us to make a comparison. This, of course, reduces to your suggestion. This is only to suggest, though, that the consequence of a proof should not necessarily affect our reading of the propositions that entail it… something I think Lib has been pushing for a while now (that is, knowing the non-contradictory result of the proof is exactly what causes us to question the premises). We cannot reject the proof just because we may prove something different with different assumptions. That would be silly.
Since Goldbach’s conjecture’s possiblity is not obviously entailed by its definition, the question doesn’t make sense. However, perfection does (to me) entail its own possibility; for example, minimally: as a logical system being consistent and complete.
Yes, this is so. What is missing, though, is the demonstration that
(G -> G) & <>G
means the same thing as
(G -> G) & <>~G.
I believe I have deconstructed the two and shown they are not demonstrating the same thing, which is why their results are different. When G is “this being exists” and the first axiom is “greatness” and the second only entails possiblity, the grouping MOP suggests is " ‘The greatest being that can exist’ is the case," while zwaldd’s suggestion is " ‘The greatest being that doesn’t have to exist’ is not the case." But this is not surprising to me.
I think you misunderstood me. I’m not suggesting <>~G (although I may have assumed it earlier). I’m saying that <>~G, ~<>~G, and now, as Newton meter has suggested, <>~G V ~<>~G would all be problematic with respect to this proof. I just needed to know which applied before I could state “here’s the problem, IMO, with this proof”.
Let me elaborate, because that sounds like that’s my definition of perfection that that it is obvious that we should consider that possible, which is not what I meant to imply. Perfection as such does not need to be possible (indeed, the Possibly-Not MOP shows us that) but that means we are only considering a kind of perfection that might not exist. Instead, we want to focus on the kind of perfection that can exist–whatever that is. This is simply the greatest thing that can possibly exist == perfection (or God, as it may be to some). The conjunction of possibility and non-contingency entails necessity. What satisfies these assumptions? The greatest possible being.
Note that I showed in S5 that the greatest provable possibility of non-contingency is (A ->A). This does leave us with the case where the place this non-contingency holds does not find A in actuality. We are not logically compelled to assume
(A -> A) & <>A
because, as I explained, if we were then all actuality would entail necessity which is, shall we say, absurd. Yet I feel that this is the only way you or zwaldd would be satisfied in accepting the two premises.
So again, we can’t accept the plausibility of a possiblyA and possibly~A without denying necessary existence: we must always find a way to ensure one prevails over the other. How we do this, though, isn’t obviously (and I’d say it necessarily isn’t) a matter of the logic itself. So we will get nowhere by simply comparing the two statements and expecting one to just fall into our lap.
I feel comparing how the definition impacts the premises is the best way to do that, and I believe I have shown why choosing <>~G over <>G is dealing with something else entirely.
Since Goldbach’s conjecture is a part of mathematics, if it was actual then it would be necessary, yes, if we extend necessitation to all of mathematics. (And why wouldn’t we?–the beauty of mathematics is its non-contengency.) But if we extend necessitation in this way, then the possibility of Goldbach’s Conjecture is itself a mathematical proof. So long as this proof is not in place, we cannot select <>G or <>~G, because we are not selecting between ontological commitments but epistemological ones (which do show indifference). Suppose instead we wanted to consider the mathematical system that has the most proofs that can be proven by Newton Meter in one sitting (and that this is how we describe the perfect mathematical system). Are we still compelled to scratch our heads at <>G or <>~G?
<>G does not compell us to be able to find a “proof” of <>G. It is only an ontological commitment. Take the axiom of choice, for example. It has been shown, to my understanding, that it or its absence does not entail a contradiction, that we are as safe in assuming it as we were in not assuming it. Some people have chosen to only consider mathematical systems without the axiom of choice, <>~C. Others work with it, <>C. This is an epistemological claim, though, that it is rational to believe in one as it is to believe in the other. As an ontological matter, though, that’s a different question.
For example, one historical debate in philosophy has been whether or not objects exist independent of minds. Given the history of philosophy, it could be said that “it is of no epistemological consquence whether or not objects exist independent of minds.” (So far as I can tell, the only time “objects exist indpendent of minds” is needed is when we are trying to show that “objects exist independent of minds”.) Epistemology doesn’t concern itself with ontology any more than validity concerns itself with soundness. Which is not to say the two never meet, of course.
We’ve constructed a valid proof of G, when those properties apply to G. You note,
If any proof gives you a reason to believe one thing over another then I wonder about how you view proofs. Does 2 exist? Well, we can create a proof of the of 2 given certain axioms and assumptions and definitions (in fact, there are several such systems that intuitively match what we think of as “2”). It is your ontological commitment to the soundness of [one of] those systems that would tell you 2 exists (however numbers exist, which is a totally different thread). The proof should not do anything for you in that respect.
This is why I suggest we must choose <>G or <>~G: because an ontological indifference denies the existence of necessity. That is, we must make an ontological commitment to
<>G V <>~G.
An epistemological indifference is simply a matter of validity, which has nothing to do with the truth of propositions, only that we do not break what is already true. For example, the scientific method could be what we consider an empirical epistemological method. Yet we conclude nothing from the scientific method per se. Instead we make ontological commitments and apply the method to make ontological conclusions. Epistemology is ontologically neutral, apart from the skepticism which denies the possibility of epistemology (I trust no one here is that much of a skeptic). Again: this is not to say the two have nothing to do with each other. What is can certainly affect what we can know, and what we say we can know will affect what we say is the case. But they are seperate in the sense that epistemological validity is not the same as ontological soundness.
Why choose <>G? Why choose the axiom of choice? Why choose the law of the excluded middle? --Because we feel it is indicative of the way things are. Perfection is possible. We can choose elements from a set of exclusive sets as we like. A proposition and its negation cannot be true at the same time.
I have a question, Newton, about the networking paper that you might be able to address.
First, I’ve shown that, given all possible nodes, at least one must have:
p -> p.
However, it is clear that without that relationship being necessary, if p doesn’t obtain there then p doesn’t obtain everywhere. But can this world do anything with a reference to a remote god (<>p)? Or is the “mobile code for… ()” statement meant to indicate that this node where
p->p
exists has to exporte that “relationship” to the location of <>p to do its thing? And without then it can’t be exported, it must stay resident on the node? Is that right?
In answer to Ultrfilter’s question above, greatness has been defined as ontological perfection — existence in every possible world. Even though you might have missed every time that that was stated, it isn’t rocket science to figure out that for a modal ontological proof, that is exactly what greatness would mean.