It is possible to precisely state all the conditions that apply, but in that case, the explanation becomes impenetrable. Nobody would be interested in a multidisciplinary forum. In order to keep it readable, there is no other alternative than to leave things out. — Tarskian

First, I made a mistake as I misread your disjunction for conjunction. I made edit notes for that in my posts now.

You don't know what every person is interested in. As far as posts thus far, the only person to comment on your remark is me, and I am interested in seeing the subject properly represented. Better not to post terribly poor renderings of technical matters than to mangle the subject. The fact that this is a philosophy forum doesn't entail that it is good to oversimplify to the point of foggy vagueness and/or substantive misstatement. And I don't know why you would suppose that people would care about your synopsis of Carnap if they didn't also grasp the mathematical basis. You think many (if any) people are going to read your one liner about the theorem and grasp anything about it without a clearly stated mathematical basis? Moreover, your one-liner is incorrect.

Your synopsis is poor:

It does not make clear that it pertains specifically to the mathematical theorem.

It should not say 'logic sentences' in general, since the theorem pertains to sentences in certain languages for certain theories.

You should generalize over formulas in those languages and not over properties (since there are properties not expressed by formulas).

Disjunction is inclusive, but that does not entail that it is ever the case that "P is true and Q is false" and "P is false and Q is true".

You conflate a predicate with a sentence.

(S ∧ ¬F(r(#S)) ∨ (¬S ∧ F(r(#S))

Meaning:
(S is true and F is false) or (S is false and F is true)

Meaning:
A true sentence that does not have the property, or a false sentence that has the property, or both. — Tarskian

(1) You skipped that I pointed out that:

(S is true and F is false) and (S is false and F is true)

is never the case.

(2)

"S ∧ ¬F(r(#S)" is not the same as "S & ~F".
"¬S ∧ F(r(#S)" is not the same as "~S & F".

F

does not have a truth value. What has a truth value is

F(r(#S))

Saying "F is false" is nonsense.

(3) I agree with this:

C entails that there is a sentence S such that T proves:

(S & ~F(r(#S))) v (~S & F(r(#S))).

F expresses a property. F(r(#S)) is true if and only if S has the property expressed by F.

But:

[EDIT CORRECTION: I misread the quote. The quote was a disjunction not a conjunction. Mine is not a counterexample. No true sentence is not equivalent with itself, but every false sentence is equivalent with itself.]

(1) Your quoted characterization did not have the specifications you are giving now. Your quoted characterization was a broad generalization about properties and sentences.

(2) PA doesn't say 'true' and 'false'.

(3) Inclusive 'or' allows 'both' but not regarding 'true' and 'false'.

We may have:

P is true and Q is false
or
P is false and Q is true

But we cannot have:

P is true and Q is false
and
P is false and Q is true

(4) There are properties not expressed by formulas, so the generalization should be over formulas, not properties.

(5) I did not say: "P(S) := S <-> S." I said: "with the arithmetization of syntax, both 'is a sentence' and 'is equivalent with itself' are expressible in PA." What is not expressible in PA are 'true' and 'false'.

('r' for 'the numeral for' and '#' for 'the Godel number of')

Let C be this theorem:

For certain theories T, for every formula F(x) there is a sentence S such that T |- S <-> F(r(#S)).

Let K be:

"For any property of logic sentences, there always exists a true sentence that does not have it, or a false sentence that has it, or both."

C is not correctly rendered as K.

(1) K doesn't qualify as to certain kinds of theories.

(2) C generalizes over formulas, not over properties.

(3) C doesn't say anything about 'true'.

(4) C doesn't say that for every property of sentences there is a true sentence that does not have the property. C doesn't say that for every property of sentences that there is a false sentence that does have the property.

[EDIT CORRECTION: I misread the quote. The quote was a disjunction not a conjunction. Mine is not a counterexample. No true sentence is not equivalent with itself, but every false sentence is equivalent with itself.]

Moreover:

(5) I showed a counterexample to both prongs of K.

[EDIT CORRECTION: I misread the quote. The quote was a disjunction not a conjunction. Mine is not a counterexample. No true sentence is not equivalent with itself, but every false sentence is equivalent with itself.]

You said that my counterexample is not in PA. So what? It doesn't have to be in PA, it merely needs to be a counterexample to K. And, by the way, K is not in PA, especially since PA doesn't have a predicate 'true'. And C includes PA as one of the T's, but C itself is not in PA.

(6) And with the arithmetization of syntax, both 'is a sentence' and 'is equivalent with itself' are expressible in PA. But I didn't do that, because K doesn't specify any language or kinds of theories.

/

For certain theories T, for every formula F(x) there is a sentence S such that T |- S <-> F(r(#S)).

is not remotely anything like:

For any property of logic sentences, there always exists a true sentence that does not have it, or a false sentence that has it, or both.

For any property of logic sentences, there always exists a true sentence that does not have it, or a false sentence that has it, or both. — TonesInDeepFreeze

You wrote:

"For any property of logic sentences, there always exists a true sentence that does not have it, or a false sentence that has it, or both."

That doesn't mention PA. Rather, it a universal generalization over properties and sentences.

No, I meant what I wrote, I showed you a property of sentences that every sentence has.

[EDIT CORRECTION: I misread the quote. The quote was a disjunction not a conjunction. Mine is not a counterexample. No true sentence is not equivalent with itself, but every false sentence is equivalent with itself.]

And what you wrote doesn't even make sense. # S is a number not a sentence.

ψ ↔ ¬ F(°#(ψ))

is not:

"For any property of logic sentences, there always exists a true sentence that does not have it, or a false sentence that has it, or both."

Counterexample: Let P be the property: P(S) if and only if S is equivalent with S.

[EDIT CORRECTION: I misread the quote. The quote was a disjunction not a conjunction. Mine is not a counterexample. No true sentence is not equivalent with itself, but every false sentence is equivalent with itself.]

For perspective, keep in mind that Skolem arithmetic and Presburger arithmetic are not fully analagous, since Skolem arithmetic has more detailed axioms about its operation symbol. — TonesInDeepFreeze

Qualification:

Presburger arithmetic is usually stated with a finite axiomatization. But it also can be finitely axiomatized.

On the other hand, the only axiomatization of Skolem arithmetic I can find is at Wikipedia. It seems to be a finite axiomatization (it doesn't have an induction schema), but I don't understand it because it includes exponentiation though exponentiation is not primitive. So I can't say whether that Wikipedia axiomatization makes sense.

My point is that if we compare a finite axiomatization of Presburger arithmetic with finite axiomatization of Skolem arithmetic, we may find that they are indeed "analogous" to some extent.

One point though: Godel-numbering is in the meta-theory, but we want to know why we need multiplication in the object theory. But, if I'm not mistaken, we need that it is representable in the object theory; I'd have to study the proof again.

why having both addition and multiplication entail incompleteness?
How does it entail incompleteness? — ssu

See a proof Godel-Rosser.

Is it that with both addition and multiplication you can make a diagonalization or what is the reason? — ssu

Diagonalization is available in any case. But we need multiplication for Godel numbering. We also need exponentiation, but Godel proved that exponentiation is definable.

I used to think that Carnap's theorem was the real culprit:

"For any property of logic sentences, there always exists a true sentence that does not have it, or a false sentence that has it, or both." — Tarskian

Where did Carnap write that?

the OP makes the point that there are more mathematical truths than there are symbol strings to express them. — fishfry

That depends on what things are truths.

If a truth is a true sentence, then there are exactly as many truths as there are true sentences, which is to say there are denumerably many

If a truth is "state-of-affairs", such as taken to be a relation on the universe, then, for an infinite universe, there are more "truths" then sentences.

each valid sentence in the language of the axioms is then either true or false in the model. (It could be independent, too, but we're not concerned with that here). — fishfry

(1) I would avoid the word 'valid' there, since it could be misunderstood in the more ordinary sense of 'valid' meaning 'true in every model'. What you mean is 'well-formed'. But, by definition, every sentence is well-formed, so we only need to say 'sentence'.

(2) If by 'independent' you mean 'not determined to be true, and not determined to be false', then there are no such sentences. Per a given model, a sentence is either true in the model or false in the model, and not both.

Presburger arithmetic is designed to be complete and decidable. Therefore, it cannot formalize concepts such as divisibility or primality, or, more generally, any number concept leading to multiplication of variables.

If anyone has something more to say about this and why this is so — ssu

We know it is so because having both addition and multiplication entails incompleteness, so, since Presburger arithmetic is complete, it can't define multiplication.

Ok, so what's the interesting thing with having both addition and multiplication?
— ssu

[...] Any simplification to Robinson's arithmetic will make it complete — Tarskian

Robinson arithmetic is incomplete.

Mathematical truth is always:

Axioms plus an interpretation. — fishfry

The truth of a sentence is per interpretation, not per axioms.

Some sentences are true in all models. Some sentences are true in no models. Some sentences are true in some models and not true in other models.

Axioms are sentences. Some axioms are true in all models (those are logical axioms). Some axioms are true in no models (those are logically false axioms, hence inconsistent, axioms). Some axioms are true in some models and not true in other models (those are typically mathematical axioms).

The key relationship between axioms and truth is: Every model in which the axioms are true is a model in which the theorems of the axioms are true. And every set of axioms induces the class of all and only those models in which the axioms are true.

so what's the interesting thing with having both addition and multiplication? — ssu

For starters, the system represents all the primitive and recursive functions, and is incomplete.

Skolem Arithmetic only has multiplication (no addition) and is also complete. — Tarskian

For perspective, keep in mind that Skolem arithmetic and Presburger arithmetic are not fully analagous, since Skolem arithmetic has more detailed axioms about its operation symbol.

Can you provide a very simple definition of this sort of truth in math? [...] Model theory? — jgill

Yes, there is a mathematical definition of 'true in a model'.

set theory [has] large fragments that are bi-interpretable with arithmetic — Tarskian

No, a fragment of set theory with also the negation of the axiom of infinity is bi-interpretable with PA. I pointed that out previously.

"In particular, a nonstandard model of arithmetic can have indiscernible numbers that share all the same properties."

Even though the law of identity is certainly applicable in the standard model of the natural numbers, it may fall apart in nonstandard models of arithmetic.

So, ω+7 ¬= ω+7 may be true in a nonstandard context, with ω the infinite ordinal representing the order type of the standard natural numbers. — Tarskian

The law of identity, the indiscernibilty of identicals, and the identity of indiscernibles are different. With a semantics for '=' such that '=' is interpreted as the identity relation on the universe, the first two hold.

'w+7 = w+7' is true in every model. For any term 'T', 'T = T' is true in every model.

I think your charges of "misrepresentation" are all bosh — Leontiskos

You claimed that I don't distinguish between material implication and everyday discourse. But I had explicitly said, about three times, that I do recognize that material implication does not accord with many of the everyday senses of 'implies' and even that there are other formal logics with which material implication does not accord. You lied and then lied again.

You claim that I adhere only to truth functional logic. But I had said, at least twice, that I am interested in the study of many kinds of logic. You lied and then lied again And here, you are making claims about classical logic, and so we need to at least be clear how classical logic is actually formulated, and saying how classical logic is actually formulated is not a claim that one must only view logic in terms of classical logic.

There's a lot more I want to comment on in the posts today, but I'm out of time. I hope to have time later.

In this are you saying that these two claims are not equivalent?

"If A implies B & ~B, then A implies a contradiction"
(a→(b∧¬b))→¬a — Leontiskos

One is a statement in the meta-language and the other in the object language. They are different levels of statement.

limiting yourself to truth-functional logician — Leontiskos

You lie again. You lie in the face of my having said the exact opposite. And you again make coherent discussion impossible.

I said that I study different logics; I don't limit myself to only a truth functional framework.

But we happen to be discussing your criticisms of a truth functional logic, so it is crucial to be clear what that logic is. Saying that what the logic actually is does not imply that that is the only logic that we may consider.

You repeated your tactic from earlier: Falsely painting my view about logic in general, then using that false painting to incorrectly impugn my statements about the logic we happen to be discussing.

If "A implies a contradiction" were a translation of the sentences — Leontiskos

It's not a translation of the sentences discussed. That point was recognized by @Lionios who originally claimed it to be a translation.

does he mean that it is false or that it is FALSE? — Leontiskos

Banno may speak for himself, but I don't know what difference in reference you mean by spelling 'false' without caps and with all caps.

Nothing is "reconceived" in natural deduction. It seems you don't know how a natural deduction system is formulated.

In your language we would say that it can be conceived as a particular contradiction or a non-particular contradiction (non-particular being, in my terms, "falsity incarnate," or FALSE, or ABSURD, and in Lionino's earlier phrasing, contradiction-proposition-qua-truth-value, which truth value is necessarily false as opposed to contingently false). — Leontiskos

What in the world? That is not my "language"; it's a bunch of your own verbiage.

I can't make sense of almost all of the rest of your post and similar posts. Probably, I'll focus on the parts that are most blatantly false or ill-premised about the logic you're talking about.

There are particular apples and we can generalize about them. There is no apple that is not a particular apple. But we do say things like "If x is an apple, then x has a core". That is not claiming that there is an apple that is not a particular apple, but rather we can make generalizations about apples.

Perhaps now you are beginning to see the point? — Leontiskos

Perhaps now you're beginning to see the point that a poster has no fault in posting several posts, some of them fairly long, in reply to several posts, some of them fairly long. Moreover that it is not unreasonable to post a long post in reply to a short one. And that it is a lot better not to lie about a poster by claiming he said things he did not say and indeed as he said the opposite of what he said. And that it is ridiculous to fault a posters for sometimes needing to delete posts, such as when needing to get rid of an otiose posting box left from a post that was started deferred to another post or not even posted.

/

Back to RAA: You've not shown any fault in RAA, but rather your fault in not understanding it.

what is a non-particular contradiction? — Leontiskos

'non-particular' is your word. It's up to you to say what you mean by it.

There are particular contradictions and we can generalize about them. One such generalization is that all contradictions are equivalent.

(Tones called his move a supposition whereas Banno called the same move an assumption). — Leontiskos

Again: We derive that a set of premises G implies a sentence ~P by using any of the members of G as lines, then entering P on a line, then deriving a contradiction, then inferring ~P with the line for P discharged.

The rule is the same whether we call the line entry of P an 'assumption', 'conditional assumption', 'assumption toward a contradiction', 'assumption to be discharged', 'supposition', 'conditional supposition', 'supposition toward a contradiction', 'supposition to be discharged', 'conditional premise', 'premise toward a contradiction', or 'premise to be discharged'.

Those are merely ways of referring to the line; they handles we use; the logical basis of the rule does not depend and is not affected by what handle we happen to use to describe the line entry.

The one who performs the reductio sees an opportunity to produce a contradiction and then decides to pursue it in order to achieve the inference desired (which inference is, again, a metabasis). — Leontiskos

There's no metabasis (change).

Again, to show a derivation that a set of premises G proves a sentence ~P, we may use any of the members of G as premises in the derivation, then enter P as a conditional assumption, then derive a contradiction and infer ~P with the conditional assumption P discharged.

There is no switching or "metabasis". Rather, at the very start, we state our premises and stick with them. The conditional assumption P is not one of our premises, but rather it is conditional assumption that is then discharged.

You would do well to look at a specific natural deduction system to see the exact way its rules are formulated. And also to look at a proof of the deduction theorem that is the basis of natural deduction.

Thank you for recognizing my point.

/

For any A, we have these possibilities:

(1) There is a contradiction Q such that A implies Q.

(2) There is a contradiction Q such that A does not imply Q.

(3) For all contradictions Q, A implies Q.

(4) For all contradictions Q, A does not imply Q.

If there is a contradiction that A implies, then A implies all contradictions.

If there is a contradiction that A does not imply, then there are no contradictions that A implies.

So we could state equivalences among (1), (2), (3), (4).

/

Theorems:

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

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

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

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

From (A -> (B & ~B)) we infer that A implies all contradictions.

From ~(A -> (B & ~B)) we infer that A implies no contradictions.

From (~A -> (B & ~B)) we infer that A implies no contradictions.

From ~(~A -> (B & ~B)) we infer that A implies all contradictions.

such that it can prove its own consistency. Then in this case, a proof of ¬¬a metalogically implies that ¬a isn't provable, i.e that a does not imply a contradiction. — sime

The system S could be inconsistent, in which case, if "S is consistent" is expressible in the language of S, then S proves "S is consistent" even though S is inconsistent.

I don't know what role A is suggested to have. Some formula A does not prove a contradiction, I guess. I don't know how that is supposed to bear upon the consistency of a system with other axioms. ~A and ~~A. Don't know why choosing that pair rather then A an ~A, or maybe this has to do with intuitionism.

In general, the consistency of an axiomatic system isn't provable in an absolute sense due to Godel's second incompleteness theorem — sime

What does "absolute sense" mean?

Godel-Rosser is that system of a certain kind don't prove their own consistency. That doesn't entail that there are not other systems proven to be consistent by secure means.

A reductio is not truth-functional. — Leontiskos

RAA is an inference rule.

A sentence Phi is truth functional if and only if the truth or falsity of the sentence is a function of the assignment of truth or falsity to the atomic sentences occurring in Phi.

I don't know what you mean by an inference rule being truth functional.

But an inference rule may be truth preserving, as RAA is:

RAA is among the natural deduction rules. And regarding those rules:

If the rules provide that P is provable from a set of premises G, then any model in which the sentences in G is true is a model in which P is true.

why can we read a→(b∧¬b) as "a implies a contradiction" — Lionino

I would not accept that reading, for the reasons I mentioned several posts ago.

Most briefly: Yes, if P implies Q & ~Q, then P implies a contradiction. But that is a remark about "P -> (Q & ~Q)" and not a translation of it.

I am simply misunderstanding what "→(B(x)∧¬B(x)" means, it can't be just "any contradiction", as Tones has pointed. — Lionino

What I say is that "P implies a contradiction" is not a translation of "P -> (Q & ~Q).

One must think about the difference between a reductio ad absurdum and a direct proof — Leontiskos

To prove a negation, we must have a rule to use to do that. And any alternative (that adheres to soundness and provides for the completeness of the calculus) to modus tollens or RAA would be equivalent with them in the sense that the resulting systems would provide the same inferences as each other.

in your reductio you do not treat the contradiction as false — Leontiskos

The inference rules don't opine as to falsity. Rather, syntactically, when a contradiction results from a conditional premise, then the rules allows ending the proof with the negation of the conditional premise and discharging the conditional premise. (Same for subproofs within a proof.)

Truth and falsehood are handled in the semantics. And we show that the syntax and semantics are in accord as:

For any set of sentences G and any sentence P:

P is provable from G if and only if there is no model in which all the members of G are true and P is false.

Intuitively, a 'model' represents a possible set of circumstances.

In other words, if P is provable from G, there is no set of circumstances in which all the sentences in G are true but P is false.

And we prove that rigorously.

Modus tollens is an inference rule or axiom, depending on the system. That's syntactical.

The notion of falsity is semantical.

Syntax and semantics work in synch in classical logic. But to use modus tollens in and of itself does not require mentioning falsity.

To add to my remarks about the number of posts and length of posts. A reply to a short post or to a small portion of a post can be long because of the number, extent and fundamental nature of the errors or even one error in the short post. And beyond that, anyone should be welcome to expatiate as much as they like. At least for me, that's at the heart of an open forum: People get to share their ideas, beliefs, knowledge, and disagreements without being limited by arbitrary restrictions. Again, I think about how petty, hypocritical, and close minded the poster is when he posts so very much yet begrudges someone else from answering with posts in a row, especially given such obvious conditions as catching up after being away for even a few days or posting in hours when others are not.