...and?

Sure.

If one looks at previous posts by me, one would see that I also directly, explicitly and formally addressed the matter that RAA also provides:

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

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

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

And just a few posts ago, I discussed the no one is disallowed from choosing what to refute with RAA, as long as the refutation is valid. That upholds the very purpose of the logic system: It shows proofs from sets of formulas but does not dictate from which set of formulas (whether called 'axioms' or 'premises' or not called anything other than a 'set of formulas') we may derive. That was discussed with Lionino, who does understand the point, in the specific sense of axioms. People are free to choose their axioms, and to refute formulas inconsistent with those axioms. If G is a set of axioms and P (not a member of G) is inconsistent with P, then G refutes P. And if H is different set of axioms and Q (not a member of H but a member of G) is inconsistent with H, then H refutes Q. That we have this choice is a good thing, not a flaw in RAA nor in any other natural deduction rule, as it permits working from different axioms.

Also, A is not made true in that derivation. A is true or not depending on a given model. There is no requirement that any line be true, not even the conclusion, in a given model. For that matter ~A could even be a logically true formula itself. A sound logic system only requires that any interpretation in which all the members of G are true is an interpretation in which P is true.

And just a few posts ago, I discussed the no one is disallowed from choosing what to refute with RAA, as long as the refutation is valid. — TonesInDeepFreeze

The problem is that this proof of yours is invalid:

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

Once I pointed it out you edited your post to try to inject some background conditions, but we both know that this proof is invalid. There is no rule of inference that allows us to draw (4) from (1) and (2).

To be clear, I was not faulting your formulation, but rather only showing, contrary to the other poster, that it does not require an appeal to 'premise', 'assumption', 'supposition' or even 'contradiction'.

The problem is that this proof of yours was invalid:

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

I edited to provide more explanation. My edit did not alter the substance of what was there before the edit. And it was it was not "pointed out" at that time that the proof is invalid. My edit could not have been about a claim of invalidity, since there was no such claim of invalidity today.

It is valid.

Every interpretation in which "A -> (B & ~B)" is true is an interpretation in which "~A" is true. If the poster can't see that, then he can make a truth table to see it.

is no rule of inference that allows us to draw (4) from (1) and (2). — Leontiskos

[fixed quote in edit]

Still the poster doesn't know what RAA is.

The argument doesn't draw (4) from (1) and (2). The argument draws (4) from (1), as (2) is discharged.

The argument is an instance of RAA. And RAA is sound.

It is valid. — TonesInDeepFreeze

But it's not.

Every interpretation in which "A -> (B & ~B)" is true is an interpretation in which ~A is true. — TonesInDeepFreeze

All you are saying is, "ρ→¬μ," but this does not make the proof valid. What rule of inference do you think you used to draw (4)? (4) adjudicates the and-elimination.

The argument doesn't draw (4) from (1) and (2). The argument draws (4) from (1) as (2) is discharged. — TonesInDeepFreeze

Heh. Why is (2) "discharged" and not (1)?

Oh, understood.

I moved Universities between logic courses, and was first taught natural deduction and then axiomatic systems, without the difference being made clear. So I tend to mix the two styles, with somewhat idiosyncratic results.

There is no rule of inference that allows us to draw (4) from (1) and (2). — TonesInDeepFreeze

Yep. The example Leo gave is not an example of RAA. And this:

But it's not. — Leontiskos

What to conclude except that Leo does not understand validity.

Heh. Why is (2) "discharged" and not (1)? — Leontiskos

That he asks this is quite odd, since our purpose here was A -> (B & ~B) ⊢ ~A.

There is no rule of inference that allows us to draw (4) from (1) and (2).
— TonesInDeepFreeze

Yep. The example Leo gave is not an example of RAA. — Banno

Tones was quoting me, and he should have used the quote feature. If he had you would not have inadvertently agreed with me. Because as this thread shows, you would say any number of stupid things rather than do that.

Note that (4) is originally your conclusion, and we now both agree that it is invalid. My example was a quote from you, where you claimed to give a reductio.

Tones was quoting me — Leontiskos

That was obvious. A mere typo.

If he had you would not have inadvertently agreed with me. — Leontiskos

I didn't.

Note that (4) is originally your conclusion, and we now both agree that it is invalid. — Leontiskos

This is inane. (4) cannot be invalid on its own. The argument is valid in classical prop logic.

This is inane. (4) cannot be invalid on its own. — Banno

A proposition can be invalid qua conclusion, and that's precisely what I said. :roll:

The argument is valid in classical prop logic. — Banno

Again and again the simple questions go unanswered:

What rule of inference do you think you used to draw (4)? — Leontiskos

I didn't. — Banno

You quoted my words and then you said, "Yep..." You just didn't know they were my words. :roll:

Here it is again.

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

And so, A -> (B & ~B) ⊢ ~A. This is a valid argument.

You made the claim that this was RAA:

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

Which, as Tones pointed out, leaves out 3:

There is no rule of inference that allows us to draw (4) from (1) and (2). — TonesInDeepFreeze

It was this with which I was agreeing.

From a different angle, Tones says:

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

If one looks at previous posts by me, one would see that I also directly, explicitly and formally addressed the matter that RAA also provides:

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

Tones thinks that ¬(1) and ¬(2) both follow from (1, 2, 3). It goes without saying that there is no rule of inference that forces one rather than the other. I would simply say that both of these proofs are invalid. There is no rule of inference to justify (4) on either count. This all goes to the misunderstandings of reductio ad absurdum in this thread, and in particular to Tones' recent claim that there is no need to advert to a difference between an assumption/premise and a supposition.

You made the claim that this was RAA: — Banno

The conversation I am having with Tones revolves around <your argument>, which is an instance of the form of reductio that I gave.

Which, as Tones pointed out, leaves out 3: — Banno

"3" is present in the word "contradiction." :roll: You are and were nitpicking.

And again, as far as I can tell Tones was quoting me without using the quote feature, as he responded to a post where I said:

There is no rule of inference that allows us to draw (4) from (1) and (2). — Leontiskos

I would simply say that both of these proofs are invalid. — Leontiskos

And you would be wrong.

- More drool. The sort of confusion and self-contradiction you are exhibiting in this thread within a few short posts is unprecedented.

The conversation I am having with Tones revolves around <your argument>, which is an instance of the form of reductio that I gave. — Leontiskos

No, it isn't. You missed out a line.

It is valid.
— TonesInDeepFreeze

But it's not. — Leontiskos

The poster continues to indicate that he does not know what validity is in this context and that he is unwilling to read the posts to which responds. He skips that I just stated exactly why the argument is valid. If he won't look at a truth table as suggested, then there's little hope he'll understand anything here.

All you are saying is, "ρ→¬μ," but this does not make the proof valid. — Leontiskos

The poster seems to not know what validity is and that he is unwilling to read the post to which he responded. He skips that I stated exactly why the argument is valid. If he won't look at a truth table as suggested, then there's little hope he'll understand anything here.

What rule of inference do you think you used to draw (4)? (4) adjudicates the and-elimination. — Leontiskos

The poster seems to not know what RAA is and that he is unwilling to read the posts to which he responds. He asks what rule is used, when the rule used is RAA, exactly as the rule is formulated.

Again, as I said, for concision we may state RAA without conjunction elimination:

If Gu{P} |- Q and if Gu{P} |- ~Q, then G |- ~P
is equivalent with
If Gu{P} |- Q & ~Q , then G |- ~P

If Gu{~P} |- Q and if Gu{~P} |- ~Q, then G |- P
is equivalent with
If Gu{~P} |- Q & ~Q, then G |- P

So, in this case:

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

is equivalent with

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

The truth-functional logicians have no sense of the difference between these two arguments:

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... — Leontiskos

...much less 's half-baked reductio:

• ρ
• μ
• Contradiction, therefore ¬μ

The poster continues to indicate that he does not know what validity is in this context and that he is unwilling to read the posts to which responds. He skips that I just stated exactly why the argument is valid. If he won't look at a truth table as suggested, then there's little hope he'll understand anything here. — TonesInDeepFreeze

The poster continues to substitute rhetoric for argument, utterly failing to engage in rational argumentation or inferential reasoning. Why such a course is taken, one does not yet know. Diagnosis continues.

The poster seems to not know what validity is and that he is unwilling to read the post to which he responded. He skips that I stated exactly why the argument is valid. If he won't look at a truth table as suggested, then there's little hope he'll understand anything here. — TonesInDeepFreeze

The poster seems to suffer from psychological delusions and grandiosity. When faced with simple questions he retreats into himself, opting for 3rd-person rhetorical strategies and failing to engage in inferential reasoning.

/quoteWhat rule of inference do you think you used to draw (4)? (4) adjudicates the and-elimination.quote — TonesInDeepFreeze

The poster continues to evidence a significant difficulty in using fairly basic forum features, such as quotes.

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

The poster continues to assert his baseless arguments without answering the question and providing the rule of inference he purports to use in order to arrive at conclusion (4). Ongoing observation recommended. He seems to have no understanding of the difference between his truth-functional formalisms and reality, or even how to properly utilize his formalisms.

The poster is doing it again! Trying to discredit interlocutors by painting them with a brush "truth-functional", even after I had at least a few times addressed that.

For about the half-dozenth time:

I am not a "truth functionalist". I study and enjoy classical logic, and appreciate its uses. But I am interested in other logics. I do not say that classical logic is the only logic that can be studied, enjoyed and used.

But when classical logic is being discussed, especially critiqued, it is crucial to say what actually is the case with classical logic. And in bringing clarity to what classical logic actually is, one needs to explain. Providing such explanations does not make one a "truth functionalist".

The argument doesn't draw (4) from (1) and (2). The argument draws (4) from (1) as (2) is discharged.
— TonesInDeepFreeze

Heh. Why is (2) "discharged" and not (1)? — Leontiskos

https://thephilosophyforum.com/discussion/comment/922432

And "discharged" in scare quotes is silly and juvenile.

He skips that I stated exactly why the argument is valid. If he won't look at a truth table as suggested, then there's little hope he'll understand anything here. — TonesInDeepFreeze

So many of your claims have already been debunked in this thread. The truth-table approach to reductio was dispatched almost ten pages ago!

Has everyone agreed by this point that ↪Banno's truth table does not fully capture what a reductio is? (See bottom of post for truth table)

((a→(b∧¬b)) ↔ ¬a) is truth-functionally valid, but the implication in the first half of the biconditional is not the same implication that is used in a reductio ad absurdum. — Leontiskos

And in bringing clarity to what classical logic actually is, one needs to explain. — TonesInDeepFreeze

If you want to bring clarity you should explain what inference you used to draw (4). As it happens, truth tables don't adjudicate contradictions. I don't get to say:

1. P→Q
2. P
3. ~Q
4. ∴ Q {See truth table for 1, 2; avert eyes from 3 at all costs. I repeat: do not allow 3 a seat at the truth table!}

(The fact that you think this sort of thing can be adjudicated by a truth table is proof that non-truth-functionality is in your blind spot.)

https://thephilosophyforum.com/discussion/comment/922432 — TonesInDeepFreeze

https://thephilosophyforum.com/discussion/comment/922468

Your example didn't mix styles. It was fine. All I did was show how to formulate the proof without 'premise', 'assumption', 'supposition' or 'contradiction'.

A proposition can be invalid qua conclusion — Leontiskos

If you gave a definition of 'valid' in your sense, then we could evaluate your claims about it.

Meanwhile, in ordinary formal logic the common definition is the one I've stated.

Of course, we may discuss relative to different definitions. I don't at all insist that the ordinary definition is the only one that we may use. But when the discussion is about classical logic, especially a critique of classical logic, then we need to at least see what happens in classical logic with its definitions. RAA is valid in context of classical logic. But, again, if you would provide your definition, then, of course, we may find that RAA is not valid in that different context.

Also, with RAA, as with any rule, validity pertains to the relation between a set of formulas and a formula. But there is another sense 'valid' in ordinary logic too, which is that a sentence is valid if and only if it is true in all interpretations.

Again and again the simple questions go unanswered:

What rule of inference do you think you used to draw (4)?
— Leontiskos — Leontiskos

It has been answered again and again and again. The answer is:

RAA

...much less ↪Banno's half-baked reductio:
ρ
μ
Contradiction, therefore ¬μ — Leontiskos

Another inane misattribution. Nowhere have I said that, and certainly not in the post linked.

From a different angle, Tones says:

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

If one looks at previous posts by me, one would see that I also directly, explicitly and formally addressed the matter that RAA also provides:

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

Tones thinks that ¬(1) and ¬(2) both follow from (1, 2, 3). — Leontiskos

Look at the proofs exactly. They show that ~(A -> (B & ~B)) follows from (2), and ~A follows from (1).

Of course, (1) and (2) together are inconsistent, so both ~(A -> (B & ~B)) and ~A follow from (1) with (2).

Meanwhile, it's not needed to mention (3) since it comes merely by inference from (1) and (2).

From a different angle, Tones says:

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

If one looks at previous posts by me, one would see that I also directly, explicitly and formally addressed the matter that RAA also provides:

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

It goes without saying that there is no rule of inference that forces one rather than the other. — Leontiskos

Yes, in the exact sense that there is no inference rule that dictates what set G must be.

I would simply say that both of these proofs are invalid. — Leontiskos

Wrong. By the definition of 'valid' in context of classical logic, they are valid. If you have a different definition 'valid', then of course, anything goes.

There is no rule of inference to justify (4) on either count. — Leontiskos

Wrong. Both proofs provide valid inferences. The first proof validly infers ~A from (1), and the second proof validly infers ~(A -> (B & ~B)) from (2)

This all goes to the misunderstandings of reductio ad absurdum in this thread — Leontiskos

The misunderstanding in this thread is yours, as you don't even know what RAA is, despite that I've exactly formulated it for you and explained that exact formulation.

and in particular to Tones' recent claim that there is no need to advert to a difference between an assumption/premise and a supposition. — Leontiskos

It's not merely a claim. I showed you exactly, with exact examples, several times now.

So many of your claims have already been debunked in this thread. — Leontiskos

Not so. What has been clearly demonstrated is that you do not have a grasp of propositional logic.

That you do not understand validity, nor truth functionality, nor how to perform a deduction.

But perhaps you have unwittingly presented an account of the awkwardness of Aristotelian logic.

Writing "Assumption" to the right is a hangover from E. J. Lemmon.