The conclusion still always follows with p=1 and must in a way be "contained" in what one starts with. — Count Timothy von Icarus
Everything contained in the conclusion must be contained in the premise; we learn nothing from deduction. — Count Timothy von Icarus
Hence, deduction is informative because it involves communication. — Count Timothy von Icarus
Perhaps mathematics is better thought of in terms of signs and relations instead of identity. — Count Timothy von Icarus
It perhaps depends on how much you embrace pancomputationalism in physics and information theoretic explanations of the other special sciences. If we think number only appears in nature in terms of computation and process then that seems suggestive at the very least. — Count Timothy von Icarus
His argument is that intuitionist assumptions — Count Timothy von Icarus
you do not know what you mean by 'particular.' — Leontiskos
No one in this thread has been able to understand what that concept means — Leontiskos
to talk about a particular contradiction without a sense of a non-particular contradiction does not make sense. — Leontiskos
A contradiction is an assertion of Propositional Logic that is false in all situations; that is, it is false for all possible values of its variables. — Tautologies and Contradictions
The conclusion of a reductio is like, "This is an apple." — Leontiskos
G be a set of premises and a sentence P is not a member of G. And we want to show that G proves ~P. Then we may use any of the members of G in our argument. But, along with members of G, we also may suppose P to derive a contradiction, thus to show that G proves ~P.
— TonesInDeepFreeze
The fiction in the reductio for the formalist is that there is some formal difference between an assumption or premise and a supposition. I say that there is not. — Leontiskos
if from ¬(A → (B ∧ ¬B)) we infer that A implies no contradiction — Lionino
(A¬→(B∧¬B)), if such a thing were proper writing, — Lionino
The truth-functionalist is likely to object to me, “But your claims are not verifiable within classical logic!” — Leontiskos
My hunch here is that the classical logic system treats everything in this purely relational and context-dependent way and assumes that every non-simple proposition can be cast as a simple 'p' while preserving all of the validity relations. — Leontiskos
P→Q
∴¬P
The revised textbook would need to add an asterisk: "This is a fallacy*."
" *Except in that case where Q = (b∧¬b)" — Leontiskos
This is the path that Banno and @TonesInDeepFreeze have chosen:
(a→(b∧¬b)) → ¬a — Leontiskos
One is a statement in the meta-language and the other in the object language. They are different levels of statement.
— TonesInDeepFreeze
A direct proof requires no recourse to the meta-language. When the reductio identifies a contradiction it is dipping into the meta-language. That exchange earlier with Tones was about whether the reductio is truth-functional. It turns out that you cannot represent a reductio in the object language. — Leontiskos
↪TonesInDeepFreeze pointed out, a reductio is meant to show the inconsistency of some assumption given a set of premises (i.e. more than 1!). — Leontiskos
We have established that "A does not imply a contradiction" is not a good reading of ¬(a→b∧¬b). — Lionino
TonesInDeepFreeze [has] chosen:
(a→(b∧¬b)) → ¬a — Leontiskos
What I have consistently said is that reductio is not valid in the same way that a direct proof is. — Leontiskos
how do you prove that you may derive ~ρ from ρ→(φ^~φ)? — Lionino
how do you prove that you may derive ~ρ from ρ→(φ^~φ)?
— Lionino
I consider it an open question as to whether this question is answerable. — Leontiskos
When we do a reductio
A, A→¬B∧B ⊢ ¬A is valid
But A, A→¬B∧B ⊢ A is also valid
So the question is: how do we choose between either? — Lionino
This is the RAA, innit?
(S∧¬P)→(B∧¬B)
S
∴ P — Lionino
lying is not the same as saying something that is false. — Leontiskos
I have been ignoring your posts, and have only read a handful of them. — Leontiskos
As Philosophim said:
Not exactly the model of a sage and wise poster. You came on here with a chip on your shoulder to everyone. I gave you a chance to have a good conversation, but I didn't see a change in your attitude.
— Philosophim — Leontiskos
Not exactly the model of a sage and wise poster.
— Philosophim
You leave out that I went on to give a proof in two versions. And it is appropriate to ask whether a poster is really serious asking for something that is, as far logic is concerned, as simple as showing that 4 is an even number. If in a thread about number theory someone happened to write "4 is even", and then another said "Prove it", you think that would not be remarkable enough to reply "Are you serious? You don't know how how to prove it?", let alone to then go on to prove it anyway.
You came on here with a chip on your shoulder to everyone.
— Philosophim
Where is here? This thread? I came with no shoulder chip, not to anyone, let alone "everyone". If I permitted myself to do as you do - to posit a false claim about interior states - I would say that you do so from your own umbrage at having been corrected.
And my point stands that I did not insult you, whereupon you insulted me.
I gave you a chance to have a good conversation
— Philosophim
By saying "don't be a troll".
You can converse as you please. I'm not stopping you. And I have read your subsequent posts, even after your insulting "don't be a troll" and have given you even more information and explanation. I have not shut down any conversation. — TonesInDeepFreeze
I don't have time for silly spats and allegations. — Leontiskos
If you can't answer simple questions without telling me that I am lying a dozen times then I will just put you back on ignore. — Leontiskos
I answer sincere and coherent questions as best I can. — TonesInDeepFreeze
Then why do I have 13 new replies from you today, 11 of which are in a single thread? You're a spammer and I don't have time for this stupid shit. Get someone else to teach you how a reductio works. Maybe they can also teach you how to interact without spamming. Adios! — Leontiskos
You're welcome for that. (Not too very bumptious of me.) — TonesInDeepFreeze
1. (S & ~P) -> (B & ~B) {1}
2. S {2}
3. ~P {3}
4. B & ~B {1,2,3}
5. ~~P {1,2} — TonesInDeepFreeze
With the non-intuitionistic form we can have the sentence on the last line be P. — TonesInDeepFreeze
(1) If Gu{P} |- Q and Gu{P} |- ~Q, then G |- ~P — TonesInDeepFreeze
Do (A implies B) and (A implies notB) contradict each other? — flannel jesus
1. (S & ~P) -> (B & ~B) {1}
2. S {2}
3. ~P {3}
4. B & ~B {1,2,3}
5. ~~P {1,2}
— TonesInDeepFreeze
Ok, that is the derivation. The source I quoted at least is correct when abriding it. The RAA however is not how it was being presented in this thread by others before, which is what I was trying to confirm. — Lionino
This is the RAA, innit?
(S∧¬P)→(B∧¬B)
S
∴ P — Lionino
With the non-intuitionistic form we can have the sentence on the last line be P.
— TonesInDeepFreeze
We are all speaking non-intuitionistically here, which is standard at least in amateur circles. — Lionino
(1) If Gu{P} |- Q and Gu{P} |- ~Q, then G |- ~P
— TonesInDeepFreeze
The bulk of the debate here between Banno and Leontiskos (and me interjecting sometimes) is why say G |- ~P instead ~P |- G. — Lionino
You're welcome for that. (Not too very bumptious of me.)
— TonesInDeepFreeze
The post you quoted there was before you joined these threads. So there is no connection to you. "We" there simply means "I" — not bumptious of me, the greatly humble person I am. — Lionino
We have established that "A does not imply a contradiction" is not a good reading of ¬(a→b∧¬b). — Lionino
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.
— Lionino
That's not what I said.
If I recall correctly, you said that "A -> (B & ~B)"* may be translated as "A implies a contradiction". (*Or it might have been a related formula; not crucial since my point pertains to all such examples.)
That is not the case as follows:
(1) The sentence has a sub-sentence that is a contradiction, but the sentence itself does not mention the notion of 'contradiction'.
(2) To say "a contradiction" is to implicitly quantify: "There exists a contradiction such that A implies it". And that quantifies over sentences. If we unpack, we get: "There exists a sentence Q such that Q is a contradiction and A implies it".
A translation of "A -> (B & ~B)" is:
If A, then both B and it is not the case that B.
and not
"A implies a contradiction".
(3) "B & ~B" is a particular contradiction, not just "a contradiction". Even though all contradictions are equivalent, a translation should not throw away the particular sentences that happened to be mentioned.
(4) If we have that A implies B & ~B, then of course, we correctly say "A implies a contradiction". But that is a statement about A, not part of a translation.
"If A implies B & ~B, then A implies a contradiction" is true, but it is a statement about the sentences, not a translation of them. — TonesInDeepFreeze
"If A implies B & ~B, then A implies a contradiction" is true, but it is a statement about the sentences, not a translation of them.
— TonesInDeepFreeze
Yes, granted. I used the word "translation" wrong in basically all of my posts. I meant "is a true statement about..." instead. — Lionino
Thank you for recognizing my point. — TonesInDeepFreeze
We have established that "A does not imply a contradiction" is not a good reading of ¬(a→b∧¬b). — Lionino
Ugh. — TonesInDeepFreeze
I'm dumping this: — TonesInDeepFreeze
We don't say: If Gu{P} |- Q and Gu{P} |- ~Q, then ~P |- G. — TonesInDeepFreeze
Indeed, the "problem" is not with reduction, but with and-elimination. And-elimination has this form
ρ^μ ⊢ρ, or ρ^μ ⊢μ. We can choose which inference to use, but both are quite valid.
We can write RAA as inferring an and-sentence, a conjunct:
ρ,μ ⊢φ^~φ⊢ (ρ→~μ) ^ (μ→~ρ)
(ρ^μ) →φ^~φ⊢ (ρ→~μ) ^ (μ→~ρ)
(fixed error)
...and see that the choice is not in the reductio but in choosing between the conjuncts. — Banno
In a formal sense a reductio in no way determines that we must take one side of the fork rather than the other. Yet the reason a reductio is not usually controversial is because there usually is a set of axioms that both interlocutors are committed to, and a quality reductio will place those axioms in one side of the scale, thus persuading such interlocutors to choose the side of the fork that favors the axioms. I maintain—as I said at the outset—that a reductio involves a <background of plausibility> in order to adjudicate between the two sides of the fork. If there is no background rationale for adjudicating in favor of one side of the fork rather than the other, then the reductio will have no rational force <as a choice between (ρ→~μ) ∨ (μ→~ρ) >. — Leontiskos
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