I seem to have reduced you to reciting gobbledygook. My apologies.

- Yikes. :yikes: You're doubling down on that?

I wonder if you will have trouble sleeping on such non-existent arguments?

A reductio is as much a proof in classical propositional logic as is modus tollens. — Banno

A reductio is as much a proof in classical propositional logic as is modus tollens. — Banno

I am concerned that logicians too often let the tail wag the dog. The ones I have in mind are good at manipulating symbols, but they have no way of knowing when their logic machine is working and when it is not. They take it on faith that it is always working and they outsource their thinking to it without remainder. — Leontiskos

...but they have no way of knowing when their logic machine is working and when it is not. — Leontiskos

"Machine", singular. So back to my point, that

Each of these systems sets out different ways of dealing with truth values. How the truth value of a contradiction is treated depends on which of these systems is in play. — Banno

and so

Asking, as you do, how to treat the truth value of a contradiction apart from the system that sets out how a truth value is to be dealt with makes little sense. — Banno

What I hope to have done over the last page is to show that you are mixing logics, resulting in your own confusion. You do not succeed in showing that "...something and its negation can both be false" in classical propositional logic.

You're reaching. :wink:

I have given my arguments, I have already responded to these charges.

At this point you either have an argument for "∴¬A" or you don't. Do you have one? If not, why are you still saying that ¬A is implied?

I have already responded to these charges. — Leontiskos

Maybe not as much as you think.

At this point you either have an argument for "∴¬A" or you don't. Do you have one? If not, why are you still saying that ¬A is implied? — Leontiskos

I'm not seeing a salient point here. Pretty demonstrably, you have made a series of claims that have been shown to be in error.

At this stage it is unclear what your general point concerning "metalogic" might be, beyond an "esoteric mystery".

In your conclusion you rejected assumption (2) instead of assumption (1). — Leontiskos

I think calling them both assumptions has led to your confusion. Premise 1 is more of a GIVEN than an assumption.

We start out the scenario with it GIVEN that a -> (b and ~b). With that given, we say "let's see what happens when I assume a is the case". What happens is a contradiction, so we take a step back and realise, if it's given that a -> (b and ~b), a must be false.

I don't think there is any mystery around (A→(B∧¬B)) |= ¬A, if something implies a contradiction we may say it is false. My curiosity was more around ¬(A→(B∧¬B)). We know that ¬(A→(B∧¬B))↔A is valid, (A→(B∧¬B)) entails ¬A, and ¬(A→(B∧¬B)) entails A. Tones gave a translation of the latter as:
"It is not the case that if A then B & ~B
implies
A"
I still can't make sense of it.

I still can't make sense of it. — Lionino

This is one of those funny places where symbolic logic seems to take a detour from what we mean in natural language.

This is one of those funny places where symbolic logic seems to take a detour from what we mean in natural language. — flannel jesus

It is what I ask here

Do you think it is correct to translate this as: when it is not true that A implies a contradiction, we know A is true? — Lionino

Tones replied that that is not true for all contradictions but for some interpretations. I couldn't make sense farther past it.

I can kind of explain it.

It seems as though, the right thing to say about basic classic symbolic logic is that EVERY statement is either true or false. So, you claim A, that's either true or false, period.

If A is false, then A implies anything. You can check the truth-table on implication: A -> B is always true if A is false. https://math.stackexchange.com/questions/1306280/implication-truth-table

So, no matter what B is, if A is false, A implies B - even if B ic (C and ~C).

So, if you KNOW that A doesn't imply (C and ~C), but you also know that if A was false, A has to imply (C and ~C) by the fact that anything follows from falsehood, then you must know that A must be true.

This makes sense in the universe of classic symbolic logic, where everything has explicit truth values and implication means what it means there.

I'm interested in a system of symbolic logic that doens't deviate that drastically from what we normally mean by those expressions - a system of logic where you can say "I don't think A implies (C and ~C)" without simultaneously saying "A is true".

Maybe that system of symbolic logic is just... English? Normal logical language? Idk.

What about a system of logic whereby, if the antecedent of an implication is false, rather than that making the entire statement of implication true it makes the entire statement meaningless, or undefined, or in flux, or something else like that? Something which is neither true nor false?

I don't think there is any mystery around (A→(B∧¬B)) |= ¬A, if something implies a contradiction we may say it is false. — Lionino

I think there is a mystery why we can say it is false in this case. What rule of inference in classical logic are we appealing to? my point has been that the only legitimate rule of inference that we can appeal to itself turns out to be a metabasis. This is to say that such a rule of inference will never be valid in the same way that a direct proof is valid.

I think calling them both assumptions has led to your confusion. Premise 1 is more of a GIVEN than an assumption. — flannel jesus

Some call it a "supposition," but they are fooling themselves if they think this answers my objection.

What rule of inference in classical logic are we appealing to? — Leontiskos

rule of noncontradiction, no?

What rule of inference in classical logic are we appealing to? — Leontiskos

Funnily enough, the rules of inference we're appealing to are in fact the very first ones listed on the Wikipedia page:

https://en.wikipedia.org/wiki/List_of_rules_of_inference
Reductio ad absurdum

in classical propositional logic contradictions are false. — Banno

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

I noticed that there is in fact a second problem with your reductio. You told me that in classical logic contradictions are to be treated as false, but in your reductio you do not treat the contradiction as false. You treat it as a contradiction, as an outer bound on the logic:

...Or in other words, the metabasis is usually acknowledged to be a metabasis. As an example, when we posit some claim and then show that a contradiction would follow, we treat that contradiction as an outer bound on the logical system. We do not incorporate it into the inferential structure and continue arguing. Hence the fact that it is a special kind of move when we say, “Contradiction; Reject the supposition.” In a formal sense this move aims to ferret out an inconsistency, but however it is conceived, it ends up going beyond the internal workings of the inferential system (i.e. it is a form of metabasis). — Leontiskos

You are appealing to the move of a reductio, “Contradiction; Reject supposition.” But there is no rule, “False; Reject supposition.” Therefore you are clearly not being consistent in treating the contradiction as false.

Now you could of course continue your proof and try to use a modus tollens to arrive at (A∧¬A), But not only will this result in the same exact problem, but it would result in the additional problem of utilizing FALSE in the way I pointed out <here>.

(It would seem that you are wrong in claiming that classical logic treats contradictions as false. In fact it treats them as <ABSURD>. is correct that classical logic treats whatever "implies" a contradiction as false. Note carefully that "implies" here no longer means material implication.)

- We are very far beyond Wikipedia at this point. At this point one can no longer simply appeal to authorities and logic machines. They have to set out the arguments themselves. One must think about the difference between a reductio ad absurdum and a direct proof, as I believe they should have done when the topic came up in logic class.

So, if you KNOW that A doesn't imply (C and ~C), but you also know that if A was false, A has to imply (C and ~C) by the fact that anything follows from falsehood, then you must know that A must be true.

This makes sense in the universe of classic symbolic logic, where everything has explicit truth values and implication means what it means there. — flannel jesus

That is a good explanation. In a way the explanation is sort of a literal translation of the formula, but I was hiccupping at the "then you must know that A must be true" part. As you said:

This makes sense in the universe of classic symbolic logic, where everything has explicit truth values and implication means what it means there. — flannel jesus

Here is something quaint. In modal logic, ¬◇(a→(b∧¬b)) entails □a. But I guess that is simply a consequence of what we are talking about. Not only that, but ¬◇∃x(A(x)→(B(x)∧¬B(x))) entails □∃xA(x). I am confident I am simply misunderstanding what "→(B(x)∧¬B(x)" means, it can't be just "any contradiction", as Tones has pointed.

The main problem for me is, why can we read a→(b∧¬b) as "a implies a contradiction" but not ¬(a→(b∧¬b)) as "a does not imply a contradiction?

I think there is a mystery why we can say it is false in this case. — Leontiskos

Well, if something results in a contradiction, we are able to rule it out, aren't we? At least we do it all the time in physics and mathematics.

We are very far beyond Wikipedia at this point. At this point one can no longer simply appeal to authorities and logic machines. — Leontiskos

No, you asked for the rule of inference from classical logic - it's right there, common knowledge in wikipedia. I don't see any good reason why my answer should be considered unacceptable, other than you just don't want to accept it. You asked for the rule, that's it.

The main problem for me is, why can we read a→(b∧¬b) as "a implies a contradiction" but not ¬(a→(b∧¬b)) as "a does not imply a contradiction? — Lionino

@Banno

Another way to think about it is, "The only way you can be CERTAIN that A doesn't apply a contradiction is if you know A is true."

I'm interested in a system of symbolic logic that doens't deviate that drastically from what we normally mean by those expressions - a system of logic where you can say "I don't think A implies (C and ~C)" without simultaneously saying "A is true". — flannel jesus

Check out the truth tables in Many-Valued Logic section https://plato.stanford.edu/entries/logic-paraconsistent/#ManyValuLogi

In this thread some users also use a third value for variables https://thephilosophyforum.com/discussion/15333/ambiguous-teller-riddle/p1 and, as we discussed, that is basically what my use of "(A or notA)" does as well.

It is also a way that people go around liar paradoxes https://plato.stanford.edu/entries/dialetheism/#SimpCaseStudLiar

Another way to think about it is, "The only way you can be CERTAIN that A doesn't apply a contradiction is if you know A is true." — flannel jesus

Well, there we are going into epistemic territory. But it seems a bit related to what I say in

But I think it might be we are putting the horse before the cart. It is not that ¬(a→(b∧¬b)) being True makes A True, but that, due to the definition of material implication, ¬(a→(b∧¬b)) can only be True if A is true. — Lionino

Well, if something results in a contradiction, we are able to rule it out, aren't we? — Lionino

Sure, so long as we are recognizing that "to rule it out" is a special move, unique and irreducible to any other move in classical logic. Specifically, the implication that we deny is in this case is no longer material implication.

Perhaps my idea is that if someone engages in these sorts of inferences then there should be added an asterisk to their conclusion on account of the fact that this form of metabasis is highly questionable. I mostly want attention to be paid to what we are doing, and to be aware of when we are doing strange things. — Leontiskos

This all goes hand-in-hand with the fact that Banno's reductio has no power, strictly speaking, to draw the conclusion ¬A. A reductio is not a proof in the strict sense, and this is precisely what a metabasis is: a form of non-strict inference. Anyone can deny that reductio without being the worse for wear, logically speaking.

No, you asked for the rule of inference from classical logic - it's right there, common knowledge in wikipedia. — flannel jesus

I'm sorry, but as someone who thinks that material implication is an example of the principle of explosion you are out of your depth here. Material implication is entirely different from the principle of explosion, and an argument from the authority of Wikipedia is not a sufficient answer to the question I am asking.

material implication is an example of the principle of explosion — Leontiskos

I don't think I claimed that. But as you're eager to reject basic reason, I'm not going to be one to stop you.