(a→b) ↔ (¬a∨b)
¬(a→b) ↔ ¬(¬a∨b)
However (a∨b) and ¬(¬a∨b) aren't the same
So ¬(a→b) and (a∨b) aren't the same

(a→(b∧¬b)) ↔ (¬a∨(b∧¬b))
¬(a→(b∧¬b)) ↔ ¬(¬a∨(b∧¬b))
(¬a→(b∧¬b)) ↔ ¬(¬a∨(b∧¬b))
Since ¬(¬a∨(b∧¬b)) is the same as (a∨(b∧¬b))
(¬a→(b∧¬b)) ↔ (a∨(b∧¬b))

almost always a sign of latent homosexuality — Mikie

Strange though, is that supposed to be a bad thing?

Obsession with whatever goofy ideas of masculinity they have — almost always a sign of latent homosexuality and subsequent fear — Mikie

"If you point how I am oestrogenated it means you are obsessed with masculinity!"

Awful argument as always, Mikie-chan.

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.

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.

Not ZF but ZF-inf. It requires removing and denying the axiom of infinity. — Tarskian

That was a typing mistake obviously as the very post I quoted said "ZF-inf" instead of just "ZF". Regardless, ZF-inf and PA are not "two completely different ways". One is tied to the other.

Two things that are mutually exclusive cannot "turn out to be simultaneously correct as well". It is absurd. Besides, formalism is not an ontology of mathematics, it is an approach to foundations.

I don't see where I did that.

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

It is not feedback, I am not interested in your self-improvement as you are an incorrigible fool. I am showing once again you have no clue what you are talking about.

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 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.

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.

(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

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))))

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

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".

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.

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

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))?

But then how can there be any final beatific return, apokatastasis, the accomplishment of exitus and reditus in salvation history? Won't people always just turn away again eventually? — Count Timothy von Icarus

It is clear how that is troublesome for Christian eschatology.

The idea here is that a higher good (and for man full conformity to the image of God) requires a sort of self-transcedence and not merely the fulfillment of what is desired by nature. — Count Timothy von Icarus

Though this might work as the basis of a useful fiction to solve the above trouble for Christian eschatology, is self-transcedence permanent? Can't one fall back into imperfection?
At the same time, given apokatastasis is not orthodox, I don't give it much merit.

But let's say it is permanent. That still doesn't go past the point here:

It is not about the arbitrary level of suffering in this world, but about making so that the amount of pain in life isn't so horribly ridiculous — Lionino

Is this much suffering really needed for self-transcendence? Couldn't our tribulations be less gruesome, and more like a journey of the hero? Sometimes it feels like life is in a dark fantasy setting, but without the fantasy.

Banno's reductio — Leontiskos

Where

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

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

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

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.

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?

That doesn't make sense automatically because formalism is a program for foundations, platonism is an ontological claim. And idk what post of MU it is.

What's it derived from? — Moliere

From what mathematical objects are.

What of us who think it is both created and discovered? — jgill

Sounds contradictory to me, unless you are saying the application of it is discovered.

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.

— Donarudo-san. Get up, Donarudo-san.
— ...Shinzo?

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.

If that is true, then Von Neumann ordinals cannot exist. Example 3 = {{},{{}},{{},{{}}}}. It wouldn't work because it would be reduced to the empty set which does not even exist according to Leśniewski, So, that view is very unproductive. — Tarskian

That is one single nominalist out of the several nominalists. Stop misrepresenting the view that you found out about by reading a Redditpedia article two hours ago.
And your criticism is spurious.

ITT: People arguing about whether a French loanword entails another French loanword. Monumental indeed.

I had a divine call from the heavens. No one can stop this bullet now.

PA and ZF-inf are two completely different ways of looking at things. "Everything is a natural number" versus "Everything is a set". — Tarskian

No. PA can be built from ZF but not the converse.

Furthermore, besides formalism, there are several other competing ontologies for mathematics. They all turn out to be simultaneously correct as well. For example, Platonism is not wrong either. It is just another way of looking at things. — Tarskian

Impossible, those are two mutually exclusive views.

Nor are your comparisons — Paine

Comparisons are not a "good thing or not", they are either true or not. And my comparison is true.

and you hit a member of a militia group — Paine

That is not a good thing though.

A implies B and C. C is not B. A implies both, B and not B. No contradiction. — creativesoul

10 pages later...

The issue about you folx/folkettes is that you always stay in the "wanted"

Man perishes; his corpse turns to dust; all his relatives return to the earth. But writings make him remembered in the mouth of the reader. A book is more effective than a well-built house or a tomb-chapel, better than an established villa or a stela in the temple! — Ancient Egypt, 12th century BC

.