"If there exists nothing, there exists something" means those are true simultaneously. — BlueBanana

No. It's an implication, not a conjunction.

So your model would look like this: ~p & ~p -> p(t>0). Correct? Well, this would be inconsistent too, because you introduce a time frame for the whole formula (~p need to be at t=0 to make sense) and so you say in ~p that nothing exists, but at the same time you say that time exists which leads to a contradiction as well. So in my and your model creatio ex nihilo would be logically impossible, just that the contradictions would lay on different spots.

How would your model of a creation out of nothing look like if you don't accept mine?

¬p→p would mean "if there exists nothing, there exists something", not that something is created out of nothing. — BlueBanana

I disagree, because I interpret the implication arrow (->) as "then", a consequence in a formal, non-physical way. So ~p -> p means "if there exists nothing, then there exists something" and isn't that pretty much what we imagine if we talk about a creation out of nothing? Remember: A creation out of nothing has to be non-physical since otherwise there were already elementary physics present which would be something already and therefore no creation out of nothing!

@bluebanana: Let's clearfy things.

1. Let p = "There exists at least one thing", so ~p = "There exists no thing at all".
2. Let creatio ex nihilo = ~p & ~p -> p.
3. By logic it follows that 2. is false (and therefore impossible).

Yes, usually we'd need Predicate Logic here, but why, it's so simple, we can use Propositional Logic instead, no different results. So where is your problem?

@Srap: We agree. The set of premises ~p and ~p -> p is inconsistent, but I ask you: isn't this set of premises modeling what we think should be going on if something is created from nothing?

@Sophicat: ~p -> p is not a contradiction! And IMO I can interpret the implication "->" as a kind of "follows from". But even if I do not: If I have ~p and ~p -> p then it is impossible that ~p is true and p is true and isn't that what the creatio ex nihilo - in it's very essence - is all about, a priori to all the physics? It's impossible that ~p and p are true at once, but that is what it would take to create something from nothing...at one moment both would need to exist at once, because otherwise it would be just a case of something from something else..

A lot is not so good here, but anything is better than nothing.

1. Citations should be in grey body to distinguish them better from the answer.
2. We should get a set of basic logical and mathematical symbols, and some more smilies.
3. We should be able to mark words ot sentences and color them differently.

All that would help serious argumentation.

Not that this trivial exercise reveals anything interesting, of course. If p stands for "something exists" and then ~p stands for "nothing exists," all that he shows is that, if nothing exists, then it is not also the case that something exists. Duh. — SophistiCat

If p stands for "something exists", ~p stand for "nothing exists" and ~p -> p for "something follows from nothing" then I can prove that ~p and ~p -> p is a contradiction and therefore 1) can't be true and 2) can't be true necessarily. But since this very logical conjunction is the model for what we call "creatio ex nihilo" we can conclude its impossibility as well and that means its negation is true and its negation goes: nihil creatio ex nihilo, nothing can be created from nothing.

The proof is bulletsafe, the only thing one could criticize is that ~p and ~p -> p is not a model of what we call "creatio ex nihilo" by giving arguments, but I think my model is the very radical model of a creatio ex nihilo. All other models are not as radical. But maybe I oversee something....

@madfool: My claim is that we can prove from logic that nothing must come from nothing. Yes, I use propositional logic in an odd way for my variables refer to objects insteads of propositions. But why not take that short-cut? A variable p would stand for any object, ~p would stand for no object and ~p -> p would be the claim that something can follow from nothing, but we can prove that ~p and ~p -> p leads to a contradiction, so it must be false that something can come from nothing. I think this proof is pretty much safe, the only question is if my model with ~p & ~p -> p is somehow misleading, but I like it and see no obvious no-no's (which doesn't mean much though).

@srap: I don't understand what you mean. The creatio ex nihilo is modeled by me as "~p & (~p -> p)" and that is logically impossible which means nothing can be created out of nothing. So there is no countermodel, all models are just false, like in any contradiction.

Thx, nagase, also for the link.

Hm...let's switch gears:

1. ~p
2. ~p -> p
3. p | mp
4. ~p |premise
5. p & ~p, so 1. or 2. must be false.

I think we agree on that proof, do we? So it's just about if 1. & 2. models what we call creatio ex nihilo. That is my question! Why are 1. & 2. not modelling it properly? SophistiCat said causuality is not a special case of inference, but then she/he would need to give me one example of casuality that we cannot express as an inference...I don't think she/he can. Anything else?

@Sophisticat: I understand "nothing can come from nothing" as: it is false that something can follow from nothing. I see causality as a special case of inference, so if I can show that such an inference is wrong then it holds even more for causality.

Again:

1. Let p stand for all objects (in the form of sentences so that propositional logic is applicable).
2. Then ~p is pretty much what we would call nothingness.
3. ~p -> p is pretty much what we would call something concluded/followed from nothing.
4. ~p & ~p -> p leads to a contradiction, so it's false that something (p) can come from nothing (~p).

@nagase: What about this? Is this a correct, but rough, sketch of the 2nd theorem?

Second Incompleteness Theorem

Because of the First Incompleteness Theorem we know that if S is consistent then G is unprovable in S. Since "G is unprovable in S" is our function G (see above) we can re-formulate that statement as: If S is consistent then G. Now, we "just" formulate this statement in S and we know it's provable in S. Now, we assume we could also prove in S that S is consistent. Then by mp G would follow (and thus be proven) in S which is impossible due to the First Incompleteness Theorem. Because of this contradiction our assumption must have been false.

So would you say my modified rough sketch of the 2nd theorem is correct? Or is it still too far off?

Ok, I added your remark about "multiplication" for the First Theorem and heavily modified the Second Theorem, let me know what you think about it, nagase.

First Incompleteness Theorem

We assume a consistent formal S(ystem) where we can formulate the following function: G (G is unprovable in S). Now there are two possible cases:

a) G is provable in S, but then G says it's unprovable, contradiction,
b) ~G is provable in S, but ~G says that G is provable, which would mean both, G and ~G, are provable, contradiction.

Neither G nor ~G are provable in S. S is incomplete.

p.s. Goedel's real accomplishment was to formulate the function G in a system that just contains propositional and predicate logic and the natural numbers with addition and multiplication.

.Second Incompleteness Theorem

Because of the First Incompleteness Theorem we know that if S is consistent then G is unprovable in S. Since "G is unprovable in S" is our function G (see above) we can shortcut: If S is consistent then G. Now, we just formulate this statement in S and we know it's provable in S. Now, we assume we could also prove in S that S is consistent. Then by mp G would follow (and thus be proven) in S which is impossible due to the First Incompleteness Theorem. Because of this contradiction our assumption must have been false.

Here is what I originally wanted to proof:

If I have my perception p1 of the world then I cannot have a second perception p2 of the world that is different of p1.

My simplified proof was this:

1. premise: Pippen
2. premise: p1
3. premise: p2
4. premise: ~(p1 <-> p2)
5. p1 <-> p2 | biconditional introdcution which leads to a contradiction
6. Since 1.-3. seems alright, 4. must be false, so it follows p1 <-> p2 and that means that p1 and p2 are equal, so I cannot have two different perceptions.

It seems to me now that I'd need a stronger logic to prove my point, right? Modal logic with identity? How would one prove the above that since it seems so trivial. This is important to me since I am a solipsist that believes that nothing exists without me perceiving it, but of course I have to make sure I just have my one and only perception and not multiple ones.

Here's another look at the previous argument:

1. premise: Pippen has a left hand (LH).
2. premise: Pippen has a right hand (RH).
3. premise: It is not the case that iff LH then RH, ~(LH <-> RH).
4. LH <-> RH, contradiction to 3.
5. Since 1.-2. seem true, 3. must be false, and so it follows: LH <-> RH, but that's absurd because it says that my left hand exists only if my right hand exists and vice versa which is obviously wrong. Here nagase can't come with modality. How would u answer this one?

You wrote that the 2nd Incompleteness Theorem finds a proof inside S to show that the statement "S is consistent" (ConS) is equivalent to G and so it would follow that ConS is not provable. Ok, but how it is shown that ConS is equivalent to G?

I think my beginning is fine, the 2nd theorem starts with: Let's assume S to be consistent. But then how would you continue if you had to explain the proof in simple words to someone else?

Here's another question that kind of belong to this topic: Is equality ("=") and biconditionality ("<->") the same and you could interchange both symbols?

Nobody?

@nagase: Ah, I see. So when I have a premise P1 and a premise P2 I can always introduce P1 <-> P2, just as I could introduce P1 & P2, in the calculus of natural deduction, yes?

Ok, let me give a more complete example:

1. premise: Pippen has a left hand (LH).
2. premise: Pippen has a right hand (RH).
3. premise: It is not the case that iff LH then RH, ~(LH <-> RH).
4. LH <-> RH, contradiction to 4.
5. Since 1.-2. seem true, 3. must be false, and so it follows: LH <-> RH, but that's absurd because it basically says that my left hand can only exist with my right hand and vice versa which is obviously wrong.

So this is a valid proof in natural deduction? (Remember that I also shortcut several things here, so don't be too picky and formal) Do you see why I'm puzzled?

x*x = -1 has no solution in IR. IC doesn't change that. It doesn't solve the problem (solution for x*x=-1 in IR), it just solves a different problem (x*x=-1 for any possible number). Therefore I am thinking: Is the liar sentence in the layer logic still the "original" liar sentence or something different?

Of course that doesn't take away anything, it just points out that those ideas mostly don't solve the original "real" problem.

I agree with you, and told you so in a german forum already. :) The problem is that with this kind of logic the language becomes not closed anymore and that is unsatisfying for many philosophers. For instance what would be the status of your Liar sentence in layer logic? You couldn't tell, just for a particular layer which by definition just describes a tiny fraction of the Liar sentence. So the Lair sentence as a whole would be as "paradox" as in classical logic, jumping between true and false. Your logic merely "visualizes" this back-and-forth-jumping.

@Nagase: What you do in 3. is using the AND-introduction. My question is if I could instead introduce an implication "A -> ~A". I doubt that. I doubt that you can just with two premises P1 and P2 follow P1 -> P2 and vice versa.

And I am confident that I am right, because look at my argument above. It is consistent, I just assume the natural numbers 1,2 and 3, but with implication-introduction I can inter a contradiction which seems absurd, because 1, 2 and 3 are obviously not inconsistent.

I think layer logic is pretty much a meta-linguistic-system where you just don't allow sentences (language) to mix up with their truth values (meta-language). You solve the Paradoxes by basically making them disappear. E.g. in Trestone's layer logic the Liar sentence can't be formulated the way it is meant, because originally it wants to say: This statement is (fully) false, but in Trestones' layer logic the Liars sentence becomes: This statement is (partially) false (depending on the layer). I'd claim that is not the Lair sentence anymore and thus a different problem.

@srap: Yes, we talking about natural deduction, take this example:

1. premise: 1
2. premise: 2
3. premise: 3
4. 1 <-> 1
5. 1 <-> 2
6. 1 <-> 3

Would the deduction of 4.-6. correct and if not (what I think) then why?

How would you explain the Second Therorem based on my version with S, G(G is unprovable in S) and my explanation of the Frist Theorem? Maybe that is the best way to show me what you mean, because I will see the difference in my and your version.

But this can't be true since it leads to contradictions. Just an example:

1. premise: 1
2. premise: 2
3. premise: 3
...
5. 1 <-> 1
6. 1 <-> 2 .
7. 1 <-> 3 .

Here basically 1 equals 2 and 3 which is false. Somehow it can't be possible to introduce a biconditional with only the premises and this must be true even for implication. But I just don't know what the right rule says to deduce an imlication A -> B. What are the precise conditions of such an introduction "p -> q"?