1. A→(B∧¬B) assumption
2. A assumption
3. B∧¬B 1,2, conditional proof
4. ~A 2, 3 reductio
— Banno
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. — Leontiskos
You can see this by examining your conclusion. In your conclusion you rejected assumption (2) instead of assumption (1). Why did you do that? In fact it was mere whim on your part, and that is the weakness of a reductio. — Leontiskos
I am attributing the modus tollens to you because you are the one arguing for ¬A. If you are not using modus tollens to draw ¬A then how are you doing it? By reductio? — Leontiskos
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
If A is false, then A implies anything. — flannel jesus
You can check the truth-table on implication: A -> B is always true if A is false. — flannel jesus
His ready-made approach doesn't answer the questions that are being asked — Leontiskos
They will be required to examine the logic machine itself instead of just assuming that it is working. — Leontiskos
A reductio requires special background conditions. In this case it would require the background condition that (1) is more plausible than (2). — Leontiskos
in your reductio you do not treat the contradiction as false — Leontiskos
One must think about the difference between a reductio ad absurdum and a direct proof — Leontiskos
I am simply misunderstanding what "→(B(x)∧¬B(x)" means, it can't be just "any contradiction", as Tones has pointed. — Lionino
why can we read a→(b∧¬b) as "a implies a contradiction" — Lionino
A reductio is not truth-functional. — Leontiskos
In general, the consistency of an axiomatic system isn't provable in an absolute sense due to Godel's second incompleteness theorem — sime
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. — sime
"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
(3) A ... toward a contradiction — TonesInDeepFreeze
↪Leontiskos I solved my main problem just right above. — 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. — 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 — Lionino
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
...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. — Lionino
(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)) — Lionino
The one who performs the reductio sees an opportunity to produce a contradiction and then decides to pursue it in order to achieve the inference desired (which inference is, again, a metabasis). — Leontiskos
(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. — TonesInDeepFreeze
Thus when Banno says that a contradiction (b∧¬b) is false, does he mean that it is false or that it is FALSE? — Leontiskos
(Tones called his move a supposition whereas Banno called the same move an assumption). — Leontiskos
what is a non-particular contradiction? — Leontiskos
'non-particular' is your word. It's up to you to say what you mean by it. — TonesInDeepFreeze
Perhaps now you are beginning to see the point? — Leontiskos
does he mean that it is false or that it is FALSE? — Leontiskos
In your language we would say that it can be conceived as a particular contradiction or a non-particular contradiction (non-particular being, in my terms, "falsity incarnate," or FALSE, or ABSURD, and in Lionino's earlier phrasing, contradiction-proposition-qua-truth-value, which truth value is necessarily false as opposed to contingently false). — Leontiskos
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? In fact it was mere whim on your part, and that is the weakness of a reductio.*
* A reductio requires special background conditions. In this case it would require the background condition that (1) is more plausible than (2). — Leontiskos
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". — Lionino
"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
Or if you think it is only truth-functional if it fits in a truth-table: — Banno
Banno may speak for himself, but I don't know what difference in reference you mean by spelling 'false' without caps and with all caps. — TonesInDeepFreeze
...and such is in line with ↪Banno's claim that "(p ^ ~p) is false in classical propositional logic," as if we could formally translate a contradiction as "false" (whatever that is supposed to mean). — Leontiskos
If "A implies a contradiction" were a translation of the sentences — Leontiskos
It's not a translation of the sentences discussed. — 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
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