the more purely formal a system is, the less this discontinuity of reductio ad absurdum is able to be recognized. — Leontiskos

The more formal the system, the more precisely we can determine exactly what is and what is not permitted by it.

The formalization doesn't overstep the ordinary, informal argument form of reductio ad absurdum. Rather, the formalization requires that the form be adhered to strictly.

Meantime, look up Euclid and Pythagoras to see that the argument form does go back to the ancients. Moreover, take a look at the quotes from the articles about the argument form.

By the way, I looked at Mates's 'Elementary Logic' (a great book) where his system uses non-intuitionistic MT (If G |- ~A -> ~B and G |- B, then G |- A), which is actually the more common primitive form for classical systems anyway.

So, RAA must be derivable from non-intuitionistic MT without even invoking LNC. Though, of course, non-intuitonistic MT does entail LEM, while my proof does not.

Mates lists LNC as a theorem in his system with non-intuitionistic MT. So, with that theorem, we would continue with the proof I gave. A full proof would have to show the details in deriving LNC.

Laws of deduction are not usually derived from one another. — Banno

In a primarily inferential system like classical propositional logic some can be derived and some cannot (i.e. some rules of inference are technically superfluous). For example, if one has access to the disjunctive syllogism then they also have effective access to the classical modus tollens (over the material implication). So there will be unique and irreducible inference-axioms in any inferential system, but my claim is that RAA is uniquely unique.

The text that we used for philosophical logic was the second edition of Harry Gensler's Introduction to Logic, which is very wide ranging. The chapter that precedes, "Chapter 6: Basic Propositional Logic," is a chapter on induction. In the last two pages of that chapter he points out the irreducibility of modus ponens and brings in Aristotle (link). Note that the beginning of the first sentence is, "Some suggest that we approach justification in inductive logic the same. . ."

Now justifying the inference-axiom of modus ponens is not overly problematic (although it does require induction and/or intellection), but justifying the inference-axiom of RAA is rather more problematic. This difficulty has shown up often in this thread in the way that folks wish to continue treating RAA glibly even when more fundamental questions are being asked. This goes to my point that it is easier to reject a RAA than a MP. To accept the premises of a MP is to already have implicitly accepted the conclusion. Not so for an RAA, as Tones' claim demonstrates.

"A does not imply a contradiction" is not a true statement about "(A→¬(B and ¬B))". — TonesInDeepFreeze

I know. I don't edit all posts that I regret. But I don't agree now with what I wrote on that post.

First, this is not a derivation of RAA. It is a putative modus tollens that looks a little bit like an RAA. As I said, there are analogical similarities. — Leontiskos

What you say here is blatantly erroneous.

There is nothing "putative" about the use of MP, and the resulting schema is an instance of RAA.

..there is a measure of discontinuity between RAA and the other inferences of classical propositional logic, such that there is no straightforward derivation of RAA from these other rules of inference.* — Leontiskos

Such derivations have been presented here by several folk, including the one from the IEP given above.

You say that I have made a number of well-documented errors in this thread. — Leontiskos

And then repeated them, despite their being shown to be wrong.

* What I have said more recently is that the more purely formal a system is, the less this discontinuity of reductio ad absurdum is able to be recognized. — Leontiskos

...for someone who has failed to engage correctly with formal logic this may indeed be so.

So there will be unique and irreducible inference-axioms in any inferential system, but my claim is that RAA is uniquely unique. — Leontiskos

In axiomatic systems RAA is derived.
Go to https://www.maths.tcd.ie/~odunlain/u11602/online_notes/pdf_propos.pdf and you will find an axiomatic system. Look at proof 10.5. It is a proof of (~A→A)⊢A in an axiomatic system. Sadly, again, you are simply wrong, RAA is not "irreducible". Whatever an "inference-axiom" is - did you mean "theorem"?

So what can be redeemed here? Perhaps this might help:

...in a dialogical context (which is my primary context) a MP cannot be rebuffed, but a reductio can.
— Leontiskos
I'll invite you to set out an example. It might be helpful. — Banno

There is nothing "putative" about the use of MP — Banno

You've pulled a 180! Earlier you literally rejected my characterization of the argument as a modus tollens and said:

Modus Tollens tells us that "Given ψ→ω, together with ~ω, we can infer ~ψ". In the first example you do not have ~ω. It might as well be a Reductio, although even there it is incomplete. — Banno

After you decried that you wanted nothing to do with the modus tollens, I replied:

What is at stake is meaning, not notation. To draw the modus tollens without ¬(B∧¬B) requires us to mean FALSE. You say that you are not using a modus tollens in the first argument. Fair enough: then you don't necessarily mean FALSE. — Leontiskos

Again, this has all been addressed at some length in these earlier posts, including the problem for the modus tollens where (B∧¬B) and ¬(B∧¬B) are equally capable of furnishing the modus tollens with its second premise. At this point I am bored of revisiting old material as you continue to contradict yourself.

Such derivations have been presented here by several folk, including the one from the IEP given above. — Banno

IEP proves my point, "Whitehead and Russell in Principia Mathematica characterize the principle of “reductio ad absurdum” as tantamount to the formula (~p →p) →p of propositional logic. But this view is idiosyncratic. Elsewhere the principle is almost universally viewed as a mode of argumentation rather than a specific thesis of propositional logic."

Look at proof 10.5. It is a proof of (~A→A)⊢A in an axiomatic system. — Banno

Which is of course not RAA, unless you want to follow Whitehead and Russell in their idiosyncratic view. You're just Googling at random to try to support a claim you made, a claim which you said was elementary.

My bad, I shouldn't have uncritically adopted your nomenclature. Laws of deduction are not usually derived from one another. But deriving equivalent schema to MT and RAA are exercises in basic logic. Here's one using MT:

ρ→(φ^~φ) (premise)
~(φ^~φ) (law of non contradiction)
:. ~ρ (modus tollens)
— flannel jesus

And the conclusion is ρ→(φ^~φ)⊢~p, one of the variants of RAA. — Banno

See the response I gave:

This is perhaps my favorite proof for the modus tollens thus far. The question is whether that second step justifies the modus tollens. Does the "law of non contradiction" in step two allow us to think of the contradiction as a simple kind of falsity, which requires no truth-assignment? And if so, does that thing (whatever it is), allow us to draw the modus tollens? These are the questions I have been asking for 12 pages.

See my posts <here> and <here> for some of the curious differences between (φ^~φ) and ¬(φ^~φ). — Leontiskos

Now you are just confabulating. That habit you have of attributing stories you made up to others.

- Deflection yet again.

:roll:

- So many of your posts evidence a strong desire to avoid serious philosophical engagement at all costs, but then every so often you make a real contribution and keep people guessing. Eventually, though, one stops playing the onerous guessing game.

I dunno. I tried to give you an opportunity to regroup and perhaps present a case that might make sense. Instead you doubled down. Trouble is, so many of your posts evidence a strong desire to avoid serious philosophical engagement at all costs, but then every so often you make a real contribution and keep people guessing.

But that's a silly game anyone can play.

Eventually, though, one stops playing the onerous guessing game.

Here's the thing: nearly every one of your posts in this thread contains factual errors.

Here's the thing: nearly every one of your posts in this thread contains factual errors. — Banno

Assertions, assertions, and more assertions. And when asked to provide substantiation, you fly like a little bird, even when one goes ahead and does the setup work for you:

You say that I have made a number of well-documented errors in this thread. This is assertion and hot air which can in no way be substantiated, but there is a way for you to show that my corollary is mistaken. . . — Leontiskos

(NB: Given the way that common speech differs from material implication, in common speech the two speakers would generally be contradicting one another.) — Leontiskos

I don't think that's quite right, depending on what you meant by "generally".

Let's continue to ignore the OP's use of "implies" and consider what's going on when someone says this sort of thing.

(1) If Smith wins the election, there will be a recession.

That claims some kind of link between one event and another. To contradict this claim, you deny that the link holds:

(2) If Smith wins the election, there may or may not be a recession.

That's the straightforward denial of (1).

But now consider

(3) If Smith wins the election, there will not be a recession.

That proposes another link, and I would suggest that in everyday reasoning the truth of (3) requires the falsity of (1), even though P→~Q does not entail ~(P→Q), which indeed does seem to be a problem for material implication.

After all, if P requires that ~Q, it can hardly require that Q.

I think people do recognize the difference even in everyday reasoning, and would accept that (2) is the simple contradiction of (1), and that (3), while also denying (1) a fortiori, is a much stronger claim.

It's clear with quantifiers too:

(4) All Englishmen are honorable.
(5) Not all Englishmen are honorable -- some are, some aren't.
(6) No Englishmen are honorable.

If no Englishmen are honorable, then it stands to reason that not all of them are, but that's a much stronger claim than simply denying that being English entails being honorable.

(Apologies if this was already covered.)

I understand that you'd think that B∧¬B should be able to be replaced by any proposition P, but that is not the case.

Example:
(A∧(B∧¬B))↔(B∧¬B) is valid
But (A∧C)↔C is invalid. — Lionino

To bring this full circle, consider your post which started us off on this long trek:

((p→q)∧(p→¬q)) and (p→(q∧¬q)) are the same formula — Lionino

Or, "((A→B)∧(A→¬B)) and (A→(B∧¬B)) are the same formula."

Now it is surely true that < ((A→B)∧(A→C)) and (A→(B∧C)) are the same formula >.

But are these the same sort of equivalence?

1. ((A→B)∧(A→¬B))↔(A→(B∧¬B))
2. ((A→B)∧(A→C))↔(A→(B∧C))

I want to say that (B∧¬B) and (B∧C) are not meta-logically equivalent, and because of this the truth tables are misleading.

-

Or going back to this:

I understand that you'd think that B∧¬B should be able to be replaced by any proposition P, but that is not the case.

Example:
(A∧(B∧¬B))↔(B∧¬B) is valid
But (A∧C)↔C is invalid. — Lionino

Logically speaking, (A∧C)↔C is invalid for any (non-A) substitution of C. That's just what it means for a formula to be invalid. Yet when we substitute (B∧¬B) for C it magically becomes valid. There is a kind of meta-contradiction here insofar as (A∧C)↔C is both invalid and not invalid. It is invalid according to the truth table, and it is not invalid given the fact that we can substitute some C which makes the formula valid.

This is what I was warning about in my first posts on metabasis eis allo genos:

[Using (b∧¬b) within formulas] is a bit like putting ethanol fuel in your gasoline engine and hoping that it still runs. — Leontiskos

(Note that to say, "It works fine, just look at the truth table!," is exactly like saying, "The ethanol fuel works fine, just look at my gas gauge!")

I don't think that's quite right, depending on what you meant by "generally". — Srap Tasmaner

Thanks for bringing it back to the beginning, which is what I was also trying to do in my last post.

That proposes another link, and I would suggest that in everyday reasoning the truth of (3) requires the falsity of (1), even though P→~Q does not entail ~(P→Q), which indeed does seem to be a problem for material implication. — Srap Tasmaner

This is what we were just talking about on page 8 of the related thread, "What can we say about logical formulas/propositions?"

I think people do recognize the difference even in everyday reasoning, and would accept that (2) is the simple contradiction of (1), and that (3), while also denying (1) a fortiori, is a much stronger claim. — Srap Tasmaner

Agreed, but I stand by my original claim that (1) contradicts (3).

If no Englishmen are honorable, then it stands to reason that not all of them are, but that's a much stronger claim than simply denying that being English entails being honorable. — Srap Tasmaner

Yes, this is the square of opposition. "Some S are P" and "No S are P" are contradictories, whereas "All S are P" and "No S are P" are "contraries." But "contraries," when used in this sense, are also contradictories.

But your point is well made. The two parties would be speaking "contraries" and not merely contradicting.

(For some reason I did not receive a notification of your post.)

Introducing some of the insights from 's <thread>, I would say that the creators of classical propositional logic intended to create a system where conditionals are conditional:

A conditional, by its very name, signifies that which is not necessary.
[It is instead hypothetical] — Leontiskos

But when we place a contradiction in the consequent of a conditional it is no longer conditional (e.g. (A→(B∧¬B)). So if it is a meta-principle of classical propositional logic that all conditionals are conditional, then allowing the contradiction has upended this meta-principle.

Thus when draws ~A, he has implicitly accepted that the conditional is no longer a conditional. The objection to his move is to say that anything which undermines the meta-principle that all conditionals are conditional should not be allowed into propositional logic. Ergo: we should not accept and affirm formulae which are known to contain contradictions. This objection is of course not open to the truth-functionalist who has no recourse to meta-objections (despite the fact that these internal contradictions destroy the formal nature of validity).

I made a similar point earlier:

The most basic objection is that an argument with two conditional premises should not be able to draw a simple or singular conclusion (because there is no simple claim among the premises). — Leontiskos

In other words, this is an invalid form: <If A then B; If C then D; Therefore, E>. When something of that form claims to be valid we have a meta-contradiction, where either the supposedly invalid form is not invalid, or else the inference in question is not permissible. More precisely, we have a metabasis eis allo genos, which is at best quasi-permissible.

(Note that these sorts of exceptions created by (b∧¬b) are popping up in many places. For example, "(A∧C)↔C is invalid for any (non-A) substitution of C," and yet (b∧¬b) creates an exception.)

One could of course be very analytical and simply say that we must choose between, say, the principle that all conditionals are conditional, and the permissibility of introducing standing contradictions into formulae. Still, I think it is fairly obvious that classical propositional logic is built for the former rather than the latter. There are special moves built-in to the system—in this case RAA—that provide a way to navigate such meta-difficulties, but unlike standard inferences RAA is "a mode of argumentation rather than a specific thesis of propositional logic" (IEP). RAA pertains to the boundary of the system, not the interior.

[Using (b∧¬b) within formulas] is a bit like putting ethanol fuel in your gasoline engine and hoping that it still runs. — Leontiskos

In my opinion the creators of classical propositional logic never intended for the system to accommodate standing contradictions. ...Or in the language I used earlier, "internal contradictions," or contradictions contained within the "interior logical flow of argumentation."

as Tones' claim demonstrates. — Leontiskos

Leontiskos uses my name for a link to his own post. That deserves a link to my reply to his post:

https://thephilosophyforum.com/discussion/comment/922505

serious philosophical engagement — Leontiskos

Leontiskos is to serious philosophical engagement as salmonella is to a seriously good dinner.

Assertions, assertions, and more assertions. — Leontiskos

What rule is used? What rule is used? What rule is used?
Questions, questions, questions.

RAA. RAA. RAA.
Answers, answers, answers.

Prove RAA from MT. Prove RAA from MT. Prove RAA from MT.
Challenges, challenges, challenges.

Proof of RAA from MT and LNC. Proof of RAA from MT and LNC. Proof of RAA from MT and LNC.
Challenge met, challenge met, challenge met.

I want to say that (B∧¬B) and (B∧C) are not meta-logically equivalent, and because of this the truth tables are misleading. — Leontiskos

What are 'B' and 'C'?

If 'B' and 'C' are atomic sentences, then
(B∧¬B)
and
(B∧C)
are not equivalent.

Buy if 'B' and 'C' meta-variables ranging over sentences, then we note that it is not the case that for all sentences B and C, we have that (B∧¬B) and (B∧C) are equivalent. But that doesn't entail that there is no sentence C such that (B∧¬B) and (B∧C) are equivalent. For example, we may instantiate both 'B' and 'C' to a sentence P (where 'P' is not a variable but is an atomic sentence), and (P & ~P) and (P & ~P) are not equivalent and indeed the same. Or, we could instantiate 'B' to P and 'C' to Q & ~Q (where 'Q' is not not a variable but rather an atomic sentence), and (P & ~P) is equivalent with (P & (Q & ~Q)).

Truth tables are not "misleading" in this regard.

Considering in context of much of everyday discourse:

(1) "If Smith wins the election, then there will be a recession."
(P -> Q in ordinary symbolic logic)

That is understood as asserting a relation (perhaps necessity, causality or other relevance) between the antecedent and consequent.

But it does agree with the material conditional in one sense: The second row of the truth table for P -> Q (Smith wins but there will not be a recession) is not the case. That is, if "If Smith wins the election, then there will be a recession" is true, then it is not the case that both "Smith wins the election" is true and "there will be a recession" is false.

(2) "It is not the case that if Smith wins the election then there will be a recession."
(~(P -> Q) in ordinary symbolic logic)

That is understood as denying that there is a relation (perhaps necessity, causality or some other relevance) between the antecedent and consequent.

But it seems not to suit the material conditional at all: No row of the truth table for ~(P -> Q) (equivalent with P & ~Q) pertains, since "It is not the case that there is a (necessary, causal, or other relevance) relation between Smith winning the election and there being a recession" would not be understood as being equivalent to "Smith will win the election and there won't be a recession".

(3) "If Smith wins the election, there will not be a recession."
(P -> ~Q in ordinary symbolic logic)

That is understood as asserting a relation (perhaps necessity, causality or other relevance) between the antecedent and consequent.

But it does agree with the material conditional in one sense: The first row of the truth table for P -> ~Q (Smith wins and there will be a recession) is not the case. That is, if "If Smith wins the election, then there will not be a recession" is true, then it is not the case that both "Smith wins the election" is true and "there will be a recession" is true.

Tabulating:

(1) implies not (2).
(1) implies not (3)

(2) implies not (1)
does (2) bear upon (3)?

(3) implies not (1)
does (3) bear upon (2)?

So:

(1) and (2) contradict each other.

(1) and (3) contradict each other.

(2) and (3) don't bear upon each other?

in everyday reasoning the truth of (3) requires the falsity of (1), even though P→~Q does not entail ~(P→Q), which indeed does seem to be a problem for material implication. — Srap Tasmaner

It's a problem for material implication if material implication were required to accord with much of everyday reasoning.

all, if P requires that ~Q, it can hardly require that Q. — Srap Tasmaner

That's true. But I don't see its connection with the analysis.

people do recognize the difference even in everyday reasoning, and would accept that (2) is the simple contradiction of (1) — Srap Tasmaner

Okay.

and that (3), while also denying (1) a fortiori, is a much stronger claim. — Srap Tasmaner

(3) denies (1). But what a fortiori are you referring to? In what sense is (3) stronger than (2)? Does 'stronger than' include that (3) implies (2)? Does it?

But when we place a contradiction in the consequent of a conditional it is no longer conditional (e.g. (A→(B∧¬B)). So if it is a meta-principle of classical propositional logic that all conditionals are conditional, then allowing the contradiction has upended this meta-principle. — Leontiskos

What does "conditionals are conditional" mean?

I don't know what the poster has in mind, but, of course, with A -> (B & ~B), the consequent is not contingent. That doesn't "upend" any principle of classical logic.

RAA pertains to the boundary of the system, not the interior. — Leontiskos

How refreshing to find a topological analysis of RAA and natural deduction. Advanced stuff indeed.

[...] what it means for a formula to be invalid. — Leontiskos

A sentence S is invalid if and only if there is an interpretation in which the sentence is false.

(A∧C)↔C is invalid for any (non-A) substitution of C. That's just what it means for a formula to be invalid. Yet when we substitute (B∧¬B) for C it magically becomes valid. — Leontiskos

Not "magically".

If 'A' and 'C' are meta-variables ranging over sentences, then there are instances of

(A & C) <-> C that are invalid and instances of it that are valid.

But if 'A' and 'C' are atomic sentences, then, of course
(A & C) <-> C
is invalid.

There's no more "magic" to that than:
x = y
is true for some values of x and y and not for other values of x and y.

/

Bottom line: We must not conflate two kinds of usage of letters:

(1) Letters as meta-variables ranging over sentences.

(2) Letters that are themselves atomic sentences.

/

Instance after instance after instance in which Leontiskos thinks he's pointed out some problem with symbolic logic, it turns out the real problem is that he's ignorant of the basics of the subject.

(A→B)∧(A→C) and A→(B∧C) are the not the same formula.

They are equivalent formulas, but not the same formula.

Though we are some 10 pages after the discussion took place, on that whole matter of reading ¬(A→(B&¬B)) as "A does not imply a contradiction", if we read it instead as "A does not imply false", the fact that it entails A becomes a bit more comfortable. On the other hand, in English, or most European languages, nobody ever says "X implies false/true", that comes off as gibberish. The reason must be because the word 'implies' has the sense of (meta)physical causation, while logical implication is not (meta)physical causation; the latter starts with the antecedent being true, the former may have a false antecedent.

I think the matter of bringing logical propositions into English and vice-versa is still quite meaningful. After all, in this thread:

I was reading various riddles and puzzles on the internet. One of the formulations of the riddles state: formulate it [the riddle] with first-order logic. — javi2541997

One of the goals of the puzzle was to state the riddle in logic. Gödel's ontological proof has renditions of logical statements into plain English. The logical argument has been verified as valid, so we know that the plain English argument is valid too.

But then, I now think that «not A without B» is the best way to translate A→B, not «A implies B» or «if A, B», whilst «A without B» for ¬(A→B); which somewhat agrees with the white bullet-points of this slideshow I found:

On the other hand, in English, or most European languages, nobody ever says "X implies false/true", that comes off as gibberish. — Lionino

If I am right then it is very likely gibberish in logic as well. It is at least clear that no one knows what it is supposed to really mean.

I think the matter of bringing logical propositions into English and vice-versa is still quite meaningful. — Lionino

I'm still not sure, but I would say, along the lines of my two previous posts, that an understanding of (1) does not necessarily provide any understanding of (2), whether that understanding has to do with logic or English:

1. A→C
2. A→(B∧¬B)

To think that (1) must provide an understanding of (2) is to think that (B∧¬B) is always substitutable for P, which it is not. And again, I think this throws the original derivation of A→(B∧¬B) into doubt in the first place.

NB: The riddle of the metabasis is the riddle of when we can take (B∧¬B) as P and when we cannot. If we don't know the answer to this question then none of the reasoning in this thread which treats (B∧¬B) as P is secure.

We can define a sentential constant 'f' (read as 'falsum'):

s be the first sentential constant:

f <-> (s & ~s)

That is not "gibberish".

Moreover, in some systems 'f' is primitive and '~' is defined by:

~P <-> (P -> f)

/

Again, if 'A' and 'C' are variables ranging over sentences, then

A -> (B & ~B) comes from A -> C by instantiation of 'B & ~B' for 'C'. There is nothing problematic with that.