Banno's reductio — Leontiskos

Where

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.

These are just paradoxes of material implication, no?

The negation of a contradiction is always true, and being true it is implied by anything, true or false.

As I noted earlier in response to Tones' reductio, a reductio is an indirect proof which is not valid in the same way that direct proofs are. You can see this by examining your conclusion. In your conclusion you rejected assumption (2) instead of assumption (1). Why did you do that?

Same here, it seems to me to be a paradox of material implication that is the source of confusion.

In a normal conversation, we might ask "but what if A really only implies B and not B and not-B?" Or conversely: "what if A only implies not-B but does not actually imply B?" But the way implication works here it is not an additional premise we can reject, we don't assign a truth value to it except in virtue of the truth values of A and B themselves.

If A is true then B always implies it, whether B is true or false.

In a relevance logic or per Aristotle's comments, we might turn around and question if A truly implies either B or not-B regardless of the truth value of either A or B alone.


For example, we could set up something like:

Elvis is a man - A
Elvis is a man implies that Elvis is both mortal and not-mortal. - A → (B and ~B)
Therefore Elvis is not a man.

Obviously, the common sense thing to do would be to reject "Elvis is a man implies Elvis is non-mortal," while affirming that "Elvis is a man does imply Elvis is mortal."

A truth table won't lay those possibilities out, it will only tell us how things work in terms of A or B being true.





However, there is a quite good reason not to do this in symbolic logic. Once you start getting into "what 'really' entails what," you get into judgement calls and a simple mechanical process won't be able to handle these.

But of course, you still need judgement to make sure your statements aren't nonsense, so you just kick that problem back a level. A proof from contradiction is only going to be convincing if we believe that A really does imply both B and not-B. I know plenty of skepticism has been raised against proofs from contradiction in general, outside of this issue, but for many uses they seem pretty unobjectionable to me.

The negation of a contradiction is always true, and being true it is implied by anything, true or false. — Count Timothy von Icarus

I think that would be (A→¬(B∧¬B)), which is True for any value of A and B. While we are talking about ¬(A→(B∧¬B)), True only when A is True.

Elvis is a man - A
Elvis is a man implies that Elvis is both mortal and not-mortal. - A → (B and ~B)
Therefore Elvis is not a man. — Count Timothy von Icarus

What about.
¬(A→(B and ¬B))?

When A is true ~A implies anything, including contradictions.

Where — Lionino

Here:

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

A reductio is not truth-functional. If we want to stick to strict truth functionality then we cannot accept reductios. In that case we can only think of them as indicating the inconsistency of a system, not as grounds for denying one thing or another. Technically a reductio is a form of special pleading (i.e. “Please let me reject this premise rather than that one, for no particular reason”). I think I can only be wrong about this if there is some principled difference between a supposition and a premise (or an assumption). A reductio is a kind of bridge between formal logic and the real world, and this is because of the background conditions it presupposes. In a purely formal sense no proposition is inherently more or less plausible than another, and therefore there is no reason to reject one premise rather than another in the event of a contradiction.

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

In general, the consistency of an axiomatic system isn't provable in an absolute sense due to Godel's second incompleteness theorem; the upshot being that consistency is a structural property of the entire system that isn't represented as a theorem by the system if it is sufficiently powerful.

Suppose that the logic concerned is weaker than Peano arithmetic, 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.

But if the axiomatic system contains Peano arithmetic such that the second incompleteness theorem holds, then a proof of ¬¬a does not necessarily imply the absence of a proof of ¬a, since Peano arithmetic cannot prove its own consistency.

Let's see.

Elvis is a man – A
Elvis is a man does not imply that Elvis is both mortal and immortal – ¬(A → (B and ¬B))
Therefore Elvis is a man. – A
A, ¬(A → (B∧ ¬B)) entails A. That makes sense.

Let's say now.

Elvis is not a man – ¬A
Elvis is a man does not imply that Elvis is both mortal and immortal – ¬(A → (B and ¬B))
Therefore Elvis is not a man – ¬A
¬A,¬(A→(B∧¬B)) entails ¬A. That makes sense.

Elvis is not a man – ¬A
Elvis is a man does not imply that Elvis is both mortal and immortal – ¬(A → (B and ¬B))
Therefore Elvis is a man – A
¬A, ¬(A → (B∧ ¬B)) entails A. That doesn't make sense to me.
But I guess it does make sense when we consider that ¬(A→(B∧¬B))↔(¬A→(B∧¬B)) is valid. So, ¬A implies (B∧¬B), from where we can say A.
But if ¬A→(B∧¬B), it is a bit strange that we can derive ¬A above in the second schema. Perhaps because from a contradiction everything follows?

However, A,¬(A→(B∧¬B)) does not entail ¬A. So from {Elvis is a man} and {Elvis is a man does not imply that Elvis is both mortal and immortal}, you can't derive Elvis is not a man, because there is no contradiction being states here from where we can affirm anything.

Therefore, if ¬(A→(B∧¬B))↔(¬A→(B∧¬B)) is true, and ¬A→(B∧¬B) can be read as not-A implies a contradiction, it must be that ¬(A→(B∧¬B) cannot be read as A does not imply a contradiction.

The negation of a contradiction is always true, and being true it is implied by anything, true or false. — Count Timothy von Icarus

Yes, good. :up: Kreeft's point comes back.

In a normal conversation, we might ask "but what if A really only implies B and not B and not-B?" Or conversely: "what if A only implies not-B but does not actually imply B?" But the way implication works here it is not an additional premise we can reject, we don't assign a truth value to it except in virtue of the truth values of A and B themselves. — Count Timothy von Icarus

Right. As I have been saying it, "falsity incarnate" and "truth incarnate" are reifications. <FALSE> is a new idea, more or less foreign to classical logic.

However, there is a quite good reason not to do this in symbolic logic. Once you start getting into "what 'really' entails what," you get into judgement calls and a simple mechanical process won't be able to handle these. — Count Timothy von Icarus

Yep.

But of course, you still need judgement to make sure your statements aren't nonsense, so you just kick that problem back a level. A proof from contradiction is only going to be convincing if we believe that A really does imply both B and not-B. I know plenty of skepticism has been raised against proofs from contradiction in general, outside of this issue, but for many uses they seem pretty unobjectionable to me. — Count Timothy von Icarus

I think metabasis is useful, but I don't close my eyes to the fact that it is metabasis:

Every time we make an inference on the basis of a contradiction a metabasis eis allo genos occurs (i.e. the sphere of discourse shifts in such a way that the demonstrative validity of the inference is precluded). Usually inferences made on the basis of a contradiction are not made on the basis of a contradiction “contained within the interior logical flow” of an argument. 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

Reductio ad absurdum is useful and important, but it is not formally valid in the same way that direct proofs are (and because of this it is a (useful) metabasis). We need to recognize that we are doing something special with a reductio, and that a reductio-inference to ¬A is quite different from a direct inference to ¬A. So if someone wants to say that ¬A is implied they must put an asterisk next to implied*.

Suppose that the logic concerned is weaker than Peano arithmetic, 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.

But if the axiomatic system contains Peano arithmetic such that the second incompleteness theorem holds, then a proof of ¬¬a does not necessarily imply the absence of a proof of ¬a, since Peano arithmetic cannot prove its own consistency. — sime

Thanks. This is what I was trying to remember but could not find online (i.e. the complexities surrounding proofs of ¬¬a).

So I guess that, in order to say "A does not imply a contradiction", we would have to say instead (A→¬(B∧¬B)). From there things start to make more sense.

Since ¬(A→(B∧¬B)) does not translate to "A does not imply B and not-B". I have to fix my post above.

Elvis is a man – A
Elvis is a man does not imply that Elvis is both mortal and immortal – ¬(A → (B and ¬B))
Therefore Elvis is a man. – A
A, ¬(A → (B∧ ¬B)) entails A.
[...]
Elvis is not a man – ¬A
Elvis is a man does not imply that Elvis is both mortal and immortal – ¬(A → (B and ¬B))
Therefore Elvis is not a man – ¬A
¬A,¬(A→(B∧¬B)) entails ¬A.
[...]
Elvis is not a man – ¬A
Elvis is a man does not imply that Elvis is both mortal and immortal – ¬(A → (B and ¬B))
Therefore Elvis is a man – A
¬A, ¬(A → (B∧ ¬B)) entails A. — Lionino

to:

Elvis is a man – A
Elvis is not a man implies that Elvis is both mortal and immortal – ¬(A → (B and ¬B))
Therefore Elvis is a man. – A
A, ¬(A → (B∧ ¬B)) entails A. A entails A.

Reminder that ¬(A→(B∧¬B)) is the same as (¬A→(B∧¬B))

Elvis is not a man – ¬A
Elvis is not a man implies that Elvis is both mortal and immortal – ¬(A → (B and ¬B))
Therefore Elvis is not a man – ¬A
¬A,¬(A→(B∧¬B)) entails ¬A, from contradiction everything follows.

Elvis is not a man – ¬A
Elvis is not a man implies that Elvis is both mortal and immortal – ¬(A → (B and ¬B))
Therefore Elvis is a man – A
¬A, ¬(A → (B∧ ¬B)) entails A, from contradiction everything follows.

Elvis is a man – A
Elvis is not a man implies that Elvis is both mortal and immortal – ¬(A → (B and ¬B))
These two premises do not entail that Elvis is not a man, because there is no contradiction. A has to entail A.

I think, keeping explosion in mind, this makes much more sense in natural language.

So let's look at the cases with (A→¬(B∧¬B)), which is finally translated properly as "A does not imply a contradiction".

Elvis is a man – A
Elvis is a man does not imply that Elvis is both mortal and immortal – (A → ¬(B and ¬B))
Therefore Elvis is a man – A
A, (A → ¬(B∧ ¬B)) |= A

Elvis is not a man – ¬A
Elvis is a man does not imply that Elvis is both mortal and immortal – (A → ¬(B and ¬B))
Therefore Elvis is not a man – ¬A
¬A, (A → ¬(B∧ ¬B)) |= ¬A

Elvis is a man – A
Elvis is a man does not imply that Elvis is both mortal and immortal – (A → ¬(B and ¬B))
These two do not entail that Elvis is not a man – ¬A.

Elvis is not a man – ¬A
Elvis is a man does not imply that Elvis is both mortal and immortal – (A → ¬(B and ¬B))
These two do not entail that Elvis is a man.

I think I finally solved my own problem. When translating it to natural language, I was misplacing the associativity of the → operator in this case.
So ¬(A → (B∧ ¬B)) is the same as (¬A) → (B∧ ¬B), which may be read as "Not-A implies a contradiction", it can't read as "A does not imply a contradiction". We would have to say something like A ¬→ (B∧ ¬B), which most checkers will reject as improper formatting, so we just say A → ¬(B∧ ¬B), which can be read as "A implies not-a-contradiction", more naturally as "A does not imply a contradiction".

@flannel jesus :cool: gigabrain has done it once again

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

Are you interpreting "a does not imply a contradiction" as the basis of a reductio (i.e. "Suppose a; a implies a contradiction; reject a")? If so, then I again think it is because a reductio is not reducible to a truth-functional move. A reductio requires more than negation and falsity.

Edit:

I think I finally solved my own problem. When translating it to natural language, I was misplacing the associativity of the → operator.
So ¬(A → (B∧ ¬B)) is the same as (¬A) → (B∧ ¬B) — Lionino

Does this support my claim that what is at stake is something other than a material conditional? The negation does not distribute to a material conditional in the way you are now distributing it.

I was misplacing the associativity of the → operator.
So ¬(A → (B∧ ¬B)) is the same as (¬A) → (B∧ ¬B) — Lionino

Do you believe for, for all statements (A -> B), you can do ¬(A -> B) and transform that into ¬A -> B?

I solved my main problem just right above. See if that works.
In the meanwhile, I can finally go cook with peace of mind.

(i.e. "Suppose a; a implies a contradiction; reject a") — Leontiskos

But for the record I do accept this as a valid rhetorical move. However when it comes to propositional logic, from
P1: A
P2: A→contradict
The conclusion can be whatever we want, from explosion
https://www.umsu.de/trees/#A,(A~5(B~1~3B))|=A
https://www.umsu.de/trees/#A,(A~5(B~1~3B))|=A
https://www.umsu.de/trees/#A,(A~5(B~1~3B))|=C
https://www.umsu.de/trees/#A,(A~5(B~1~3B))|=D
https://www.umsu.de/trees/#A,(A~5(B~1~3B))|=P
https://www.umsu.de/trees/#A,(A~5(B~1~3B))|=Z
https://www.umsu.de/trees/#A,(A~5(B~1~3B))|=Z(P(G(F(x))))

(A -> B), you can do ¬(A -> B) — flannel jesus

Edited:

These two are different statements. By the way that the syntax of these operators is made, (¬(A → (B∧¬B))) is the same thing as (¬A→(B∧¬B)). It is like 2(x*y)=2x*y, but 2(x*y)≠x*2y

¬(A → B) is the same thing as ¬A→B — Lionino

That's what I was asking, thank you.

I don't believe that's correct.

I don't believe that's correct. — flannel jesus

Yeah I messed that up.

Those two are not the same thing.

But ¬(A → (B∧¬B))) and (¬A→(B∧¬B)) are the same thing.

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.

understand that you'd think that B∧¬B should be able to be replaced by any proposition P — Lionino

Me? You understand that I think that?

But what just happened is that you did that, and I told you it's incorrect...

You asked me

for all statements (A -> B), you can do ¬(A -> B) and transform that into ¬A -> B? — flannel jesus

The answer is not for all statements. I never replied positively to the question, I copy pasted incorrectly in a post that is now edited.

I never replied positively to the question — Lionino

Well you gave what certainly looked like an affirmation. If I ask you "is lemonade your favourite flavour", and you say "lemonade is the same as my favourite flavour", most people are gonna think that's pretty much a "yes" to the question.

It is like 2(x*y)=2x*y, but 2(x*y)≠x*2y — Lionino

Is * multiplication here? I don't think this is right either.

I don't see where I did that.

* is any operation. I know multiplication doesn't work like that obviously

I don't see where I did that. — Lionino

You wrote
¬(A → B) is the same thing as ¬A→B

The very proposition of "there both a) is a self and b) is no self" has (a) and (b) addressing the exact same thing - irrespective of how the term "self" might be defined or understood as a concept, the exact same identity is addressed — javra


The point is that if there is no determinate entity that 'the self' refers to, if there is only the concept, and if there is no actual entity, then saying that we are speaking about the same thing is incoherent. On the other hand, if you stipulate that the self is, for example, the body, then what would A be in the proposition (A implies B) where B is 'there is a self' ? Let's say that A is 'the perception of the body': this would be 'the perception of the body implies that there is a self". 'The perception of the body implies that there is no self' would then be a contradiction to that. — Janus

I’ll offer that the commonsense notion of “self” necessarily pivots around what we westerners commonly term “consciousness” - "the self" here always entailing the subject of one’s own experience of phenomena (or, in for example the more philosophical jargon of Kant, the “empirical ego” (via which empirical knowledge is possible) - this as contrasted to what he specifies as the underlying “transcendental ego”). And, in this commonsense understanding of "the self", the body of which one is aware is then not the “self” which is in question - "the self" instead holding as referent that which is so aware. Nor does the notion you present in any way cohere with the descriptions of self as they are addressed in the Buddhist doctrine of anatta: this being the very metaphysical understanding of reality from which we obtain propositions along the lines of “neither is there a self nor is there not a self”. Which, on the surface, do at least at first appear to be contradictory (though, as I've previously argued, do not need to so necessarily be).

But, since, we each hold our own - sometimes more divergent than at other times - understandings of what terms signify, I’ll here say that were the term “self” to be devoid of any referent outside of the occurrence of empirically observable physical bodies (maybe needless to add, that are living and so normally endowed with a subject of awareness, this rather than being dead and decaying), then I might find agreement with your general reasoning here.

I’m however not one to find the term “self” - and, hence, terms such as “I”, “you”, “us” and “them” - devoid of referents, unempirical (imperceivable) though I take these referents as "subjects of experience" to be. All the same, this thread is not the place to engage in debates regarding what “the subject of one’s own experiences” might specify or else be - although I do agree that it is not "an entity" in the sense of being a thing.

It is like 2(x*y)=2x*y, but 2(x*y)≠x*2y — Lionino

This doesn't make sense if * is "any operator" either. Replace * with + and 2(x*y)=2x*y is not true

You wrote
¬(A → B) is the same thing as ¬A→B — flannel jesus

I didn't, you are referring to this , which I already said was a copypaste mistake, it has been edited. I don't see what the issue is.

"Any operator" is not any mathematical operator you want. In that case, the → operator is working like an operator ⊙ where z(x⊙y)=zx⊙y, but z(x⊙y)≠x⊙zy, it is a syntactic rule and nothing more.

didn't, you are referring to this ↪Lionino, which I already said was a copypaste mistake, it has been edited. I don't see what the issue is. — Lionino

The issue is you said you never wrote it, but you did write it. I understand it's a mistake. Therefore it's not correct to say you never wrote it, it's correct to say you wrote it by mistake.

"Any operator" is not any mathematical operator you want. — Lionino

I don't know the rules of that game, my bad

And
since (a→b) is the same as (¬a∨b),
(a→(b∧¬b)) is the same as (¬a∨(b∧¬b)),
and ¬(a→(b∧¬b)) the same as ¬(¬a∨(b∧¬b)), which is the same as (a∨(b∧¬b)).
So ¬(a→(b∧¬b)) is the same as (a∨(b∧¬b)), so I think now it is a bit more clear why ¬(a→(b∧¬b)) is True only when A is True, the second member is always False and the or-operator returns True when at least one variable is True.