Part of this is the difference between RAA in a consistent formal system and in other areas. In a consistent formal system if we reach a contradiction we know there is a false assumption. In informal, incomplete, or inconsistent systems it is not so easy. @Leontiskos tries to bring that ambiguity into the formal system, and it doesn't go.

We don't say: If Gu{P} |- Q and Gu{P} |- ~Q, then ~P |- G.
— TonesInDeepFreeze

I made a mistake. I meant to say:

"The bulk of the debate here between Banno and Leontiskos (and me interjecting sometimes) is why say G |- ~P instead of P |- ~G." — Lionino

You can use the notation however you wish, but in my formulations, G is a set of formulas not a formula, and on the left side of the turnstile is a set of formulas while on the right side of the turnstile is a formula.

The rule:

(1) If Gu{P} |- Q and Gu{P} |- ~Q, then G |- ~P.

For concision, we can formulate that as:

(2) If Gu{P} |- Q & ~Q, then G |- ~P.

So, the point you are making could be stated ('\' stands for set difference):

(3) If Gu{P} |- Q & ~Q, then there could be an R in G such that

(G\{R}) u {P} |- ~R
and
(G\{R}) u {P} is consistent.

That is a formal way of saying what, as you mention, has been correctly said in this thread for a long time:

If ~P is derived by RAA, then it might be the case that there could be a premise R, different from P, such that we derive ~R instead.

And I gave specific examples of that in exact RAA formulation. My point in that regard has always been in agreement with anyone who correctly makes that observation.

But I don't take that to be a problem with RAA.

As I mentioned, RAA, along with the rest of the natural deduction rules, provide all and only the valid inferences (that is the completeness and soundness theorems).

[From previoulsy and adding to it:] Of course, the choice of axioms or added premises is subject to whatever criteria we wish to subject them to. That we have that option is not a fault in RAA. It's not a matter of a fault in RAA but rather we have the obvious fact that we are free to choose whichever axioms or premises we want to choose. It is good that natural deduction allows us to work from whatever axioms or premises we want to choose. The purpose of the deduction system is not to dictate our starting assumptions, but rather merely to permit all and only the valid inferences from whatever assumptions we choose.

Banno — Lionino

((P & R) -> (Q & ~ Q)) |- (P -> ~R) & (R -> ~P)

That is the idea with a pair of formulas P, R; while my formulation above is way of saying the same idea but generalized with any set of formulas G.

Indeed, his idea of looking at conjunction elimination, not just RAA, in this regard is insightful.

It has happened before in the history of science where we had to reject G when finding out that Gu{P} is contradictory, because P was so evidently true. — Lionino

I would think so. And that is what this basic logic provides.

You got it wrong. I know what I meant with my posts. "We" there refers to me, I was not talking about anyone else. The specific post you quoted did not help sort out the issue, specifically the nitpick on "translation", which is why I had to make a whole new thread for that topic specifically. — Lionino

Ugh^ugh (that's ugh to the power of ugh).

I lightly jibed, aiming at myself as much as anyone; and you turn it a thing. But since your revisionist attack on me is ill-premised I respond.

(1) I haven't claimed that you don't know what you mean. (2) I haven't claimed that you used the word 'we' in any particular way.

(3) You can look at the posts, and I as I mentioned some of them in my previous reply:

You asked me for a translation of "~(A -> (B & ~B))". So I obliged your request.

You claimed that "A does not imply a contradiction" is a translation.

I explained why that is not a translation of "~(A -> (B & ~B))", a few times. As you still didn't see the point, I gave even more details, to which you replied:

"Yes, granted. I used the word "translation" wrong in basically all of my posts."

(4) The matter of the translation is not a nitpick. You posted the incorrect translation at least a few times and presumably it was important enough to you to do that. So I was right on the money to note that it is not a translation and to explain why.

In fact in my thread you corrected yourself about something midway through the discussion — Lionino

Yes, to my credit, I corrected myself as soon as I realized I erred in a formulation. But that was not related to my correct explanations as to why "A does not imply a contradiction" is not a translation of "~(A -> (B & ~B))".

(5) In my previous post, I pointed out that you falsely claimed (why?) that my reply "You're welcome for that" was in response to a post from before I entered the threads. But my reply was from 13 days ago; I entered the thread 19 days ago. More importantly, my explanations as to why "A does not imply a contradiction" is not a translation of "~(A -> (B & ~B))" were also before that reply.

if we reach a contradiction we know there is a false assumption — Banno

We know that the assumptions are inconsistent, not merely that one of them is false. But, of course, if a set is inconsistent, then for any interpretation, at least one of the members is false in that interpretation.

I lightly jibed, aiming at myself as much as anyone — TonesInDeepFreeze

I couldn't tell since you often jump into a conversation that happened several pages before the last post of the respective thread.

But since your revisionist attack on me is ill-premised I respond. — TonesInDeepFreeze

:rofl: alright however you wanna fly it

I couldn't tell since you often jump into a conversation that happened several pages before the last post of the respective thread. — Lionino

I linked exactly to your post. No matter how many intervening posts, all you had to do was click on your name under my quote of you.

But you chose to (1) Make a false claim that I was commenting on a post that was made before I came into this thread, (2) Get the conversation completely wrong by falsely saying that I had not explained why your translation was incorrect, (3) Falsely claim that my point about the translation was a nitpick. (4) Try to get a cheap advantage out the fact that I corrected myself in a formulation, even though that is unrelated to the subject of the translation.

alright however you wanna fly it — Lionino

It's not how I choose to see it. It actually is as I showed.

Ok, I am in agreement. Everything makes sense. Leontiskos was further saying that the RAA is not strictly logical because it does not tell you which side of the conjunct to rule out. I disagreed in the last post of page 21.

So, the point you are making could be stated ('\' stands for set difference): — TonesInDeepFreeze

More like the point Leontiskos is making.

You can use the notation however you wish, but in my formulations, G is a set of formulas not a formula, and on the left side of the turnstile is a set of formulas while on the right side of the turnstile is a formula. — TonesInDeepFreeze

I am aware.

We know that the assumptions are inconsistent — TonesInDeepFreeze

Yep. Good point.

This is for classical logic. I am puzzling, as a sideline, with how conjunction plays out in paraconsistent logics - Non-Adjunctive systems in which {A,B}⊭A∧B). Jaśkowski's discursive logic, by way of example.

I wonder if, and how, RAA would function in such a case. Is it that two folk in discourse could agree that A∧B imply C, yet not agree that ~C implies ~(A∧B)...

Anyway, that's a whole different minefield.

More like the point Leontiskos is making. — Lionino

I'm referring to your correct point (shared by you, Leontiskos, Banno, and me) that instead of refuting P, we could refute one of the other premises. You adapted that point to my notation. I just pointed out that your notation doesn't suit the way I was using the notation and so I rephrased your point so that it does accord with my notation.

Leontiskos was further saying that the RAA is not strictly logical because it does not tell you which side of the conjunct to rule out. I disagreed in the last post of page 21. — Lionino

That's good. I disagreed with it many pages ago, as I saw immediately that it's wrong. (Not too very bumptious of me to say. winky face emoji whatever.)

(2) Get the conversation completely wrong by falsely saying that I had not explained why your translation was incorrect — TonesInDeepFreeze

I don't think the word 'translation' has an exclusive definition in logic, and if it does, it doesn't matter, I wasn't using logical jargon, I meant 'translation' to mean exactly what it does everyday, bringing from one language to another, in that case from logic to English. Translation doesn't always mean translating literally term-by-term, which is what you did in your translations. The only reason I said I used translation wrong every time was to accept your definition and move on as to not waste time on semantics, it was tangential to my goal, which led to my thread, where in the end the only good translation provided was by a user in the 1st page.

Don't matter. I was talking about myself. If I recall correctly, I was thinking of specifically my post with the several examples compared line by line which was what finally established that that translation was not possible.

I disagreed with it many pages ago — TonesInDeepFreeze

Amazing!

Whether tangential or not, it raised an important and interesting point. The reason yours was not a correct translation is instructive. And it wasn't a matter of definitions, let alone a definition identified with me. Meanwhile, my translation was utterly obvious, therefore dull, but the point of such a translation is not to entertain but to be as pinpoint faithful to the formula as possible. If it had been a creative writing exercise then I would have tried to come up with something a lot more colorful.

If I recall correctly, I was thinking of specifically my post with the several examples compared line by line which was what finally established that that translation was not possible. — Lionino

Actually, this:

"If A implies B & ~B, then A implies a contradiction" is true, but it is a statement about the sentences, not a translation of them.
— TonesInDeepFreeze

Yes, granted. I used the word "translation" wrong in basically all of my posts. I meant "is a true statement about..." instead. — Lionino

So far, so good. There you're saying what I had spent a few posts explaining.

Now, the conclusion that I arrived at is that "A does not imply a contradiction" is not an accurate statement about ¬(A→(B and ¬B)), it would be a true statement about (A→¬(B and ¬B)) instead. — Lionino

That last clause is wrong, obviously. (Maybe you corrected it subsequently.)

When it comes to ¬(A→(B and ¬B)), as it is the same as (¬A→(B and ¬B)), "not-A implies a contradiction" is a true statement about it. — Lionino

Right.

Amazing! — Lionino

I like that word. When I was a kid, my dear friend was a San Francisco Giants fan amidst all the other kids who were Dodgers fans. He chose that path as yet another of his exercises in non-conformity. When he got to be the captain of one of the pickup teams, he named them "The A-Mays-ers". That was a wry little bit in a dull and gray time of life. Mays died recently. So deservedly a beloved man.

Then last week I was getting tech support on the phone, and the tech agent would say "Amazing!" in a kind of lilting modern way when I executed each step in the walkthrough so that I got to the right element each time. As if each click to the next element was a cause for a moment of minor celebration. I found that charming.

.

and the tech agent would say "Amazing!" in a kind of lilting modern way when I executed each step in the walkthrough so that I got to the right element each time. As if each click to the next element was a cause for a moment of minor celebration. I found that charming. — TonesInDeepFreeze

AT&T? Or perhaps Proximus? They are trained to do exactly that with customers in the US.

Neither of those, but what you say is interesting. I guess it is a thing.

I guess it is a thing. — TonesInDeepFreeze

It is. I did a gig at tech support years ago, it was part of our training to basically baby the customer and quite literally treat it like a king, saying "No" to the caller was strictly forbidden.

It's a hard job. Some companies don't mind too much agents spending whatever time is needed on a call. Other companies hector agents to get off calls as absolutely quickly as possible, thus a lot of pressure.

But the quality of tech support and customer support in general has plummeted over the last several years. Even when agents aren't pressured to wrap up the calls quickly, some of them are atrocious - not listening to you, mindlessly giving you steps that couldn't be expected to address your particular problem, and giving misinformation and lying. And resisting to the last breath transferring you to a tier 2 tech. And of course, long wait times, calls cut off, and failed promises to continue pursuit of a resolution with a followup call. And then, after you've been through weeks of runaround in multiple calls, chats and emails, they have the gall to send you an email survey asking how they did, and keep pestering you with even more emails. On the other hand, there are some companies that provide sterling support; agents that will stay on the phone and give you all kinds of extra info and tips.

By the way, your comments about dictionaries led me to discover that, unlike years ago, OED is free online. It definitely is a much richer resource than Merriam. Thanks for that. I'll be using them side by side from now on (and with my old print edition of unabridged Merriam).

I haven't read through all 22 pages of this thread, so I'll just ask this question: has the subject of Dialetheism come up? From the Stanford Encyclopedia of Philosophy article:

"A dialetheia is a sentence, A, such that both it and its negation, ¬A, are true. If falsity is assumed to be the truth of negation, a dialetheia is a sentence which is both true and false. ....Dialetheism is the view that there are dialetheias. If we define a contradiction as a couple of sentences of which one is the negation of the other, or as a conjunction of such sentences, then dialetheism amounts to the claim that there are true contradictions."

The Aristotelians here don't cotton to it.

Ok. That is much clearer than your other posts. I suppose I agree with what is conveyed. However, the RAA has formally either rho or mu as a premise, so no choice between the conjunct is needed within the RAA, the RAA is logical. — Lionino

Rho is assumed and Mu is supposed, and if someone doesn't know the difference between an assumption/premise and a supposition then they won't understand a reductio. As I have argued at many points throughout this thread, the difference between an assumption and a supposition is a meta-linguistic difference. The object language cannot account for this difference.

The modus tollens and the reductio are two different things:

• A1. μ→¬ρ
• A2. ρ
• A3. Therefore, ¬μ

• B1. ρ
• B2. Suppose: μ
• B3. Contradiction, therefore ¬μ

You can say that "the RAA is logical," but the fact remains that B3 is not as secure as A3. There are many ways to explain why, but I will stick with my original answer that an assumption and a supposition are only different at the level of meta-logic. A3 follows without any recourse to the meta-language or to meta-logic. B3 does not.

From the supposition we learn (P→¬Q), at which point P must be further asserted beyond supposition if we are to actually arrive at ¬Q: — Leontiskos

...Refactoring this idea for a proper reductio:

• C1. ρ
• C2. Suppose: μ
• C3. Contradiction, therefore (ρ xOR μ)
• C4. Therefore, ¬μ {Assumption/premise trumps supposition}

(In order for the RAA to function as a dialogical proof one's interlocutor must agree that C1 is more plausible than C2.)

B1. ρ
B2. Suppose: μ
B3. Contradiction, therefore ¬μ — Leontiskos

That's obviously not the reductio.

That's obviously not the reductio. — Banno

Given Apokrisis' demonstrations that you don't engage in good faith (as most of us already knew), I've simply put you on ignore, where you belong.

This in reply to my pointing out that what you claim is a reductio is not a reductio.

B1. ρ
B2. Suppose: μ
B3. Contradiction, therefore ¬μ — Leontiskos

It's not even a valid argument.

I do not think the bad faith is mine alone.

Understanding RAA doesn't require reference to 'premise', 'assumption', 'suppostion' or 'contradiction'.

Here is RAA in exact formulation:

If P, along with possibly other lines, shows a formula Q and a formula ~Q, then infer ~P and charge it with all the lines used to show Q and to show ~Q except the line for P.

If ~P, along with possibly other lines, shows a formula Q and a formula ~Q, then infer P and charge it with all lines used to show Q and used to show ~Q, except the line for ~P. [not intuitionistic]

Those rules are equivalent with (formulated equivalently this time with conjunction):

If Gu{P} |- Q & ~Q , then G |- ~P

If Gu{~P} |- Q & ~Q, then G |- P [not intuitionistic]

There is no mention of 'premise', 'assumption' or 'supposition' nor, for that matter, 'contradiction'. — TonesInDeepFreeze

What is "secure" supposed to mean? At least RAA is as secure as modus tollens in the sense that they are sound.

For all sets of formulas G and formulas P, if G |- P then G |= P. That is:

For all sets of formulas G and formulas P, if G |- P then for all models M, if M is a model of G then M is a model of P. That is:

For all sets of formulas G and formulas P, if P is derivable from G, then any interpretation in which all the members of G is true is an interpretation in which P is true. That is:

Classical natural deduction (which includes RAA as a primitive rule, modus tollens as a derived rule and contrapostion as a derived theorem schema) does not permit inference of a false conclusion from true premises.

No matter whether we start with RAA as primitive and derive modus tollens and contraposition, or start with modus tollens as primitive and derive RAA, or start with contraposition and derive RAA and modus tollens, we arrive at the exact same set of allowable inferences.

1. A→(B∧¬B) assumption
2. A assumption
3. B∧¬B 1,2, conditional proof
4. ~A 2, 3 reductio — Banno

1. A→(B∧¬B) assumption
2. A assumption
3. B∧¬B 1,2, conditional proof
4. ~A 2, 3 reductio — Banno

Without the word 'assumption':

1. A -> (B & ~B) {1}
2. A {2}
3. B & ~B {1, 2}
4. ~A {1}

G is {A -> (B & ~B)}
P is A
Q is B

G u {P} |- Q & ~Q, so G |- ~P.

Without the word 'assumption':

1. A -> (B & ~B) {1}
2. A {2}
3. B & ~B {1, 2}
4. ~A {1} — TonesInDeepFreeze

What if we reject (1) instead? Then A is made true, but it does not imply (B∧¬B). Your proof for ¬A depends on an arbitrary preference for rejecting (2) rather than (1). — Leontiskos