Straight RAA does not require the "and elimination". It's an additional step when there are multiple assumptions. — Banno
A→ABSURD
∴ "A cannot be affirmed"
...
Introducing ABSURD in the way I did above destroys the LEM of classical logic. — Leontiskos
While you are there, what does "FALSE" mean? — Banno
how do you prove that you may derive ~ρ from ρ→(φ^~φ)? — Lionino
So, what is a "direct proof"? I gather you think using MT is direct, but RAA isn't? WHat's the distinction here? — Banno
But couldn't we just assume B here and get ~A just the same? "If B then ~A," seems to work fine here because the conjunct is still going to come up false. — Count Timothy von Icarus
And then we can do the same thing assuming ~B. That covers all our options assuming LEM. — Count Timothy von Icarus
I guess I'm not seeing the trouble here. I can see the trouble with proofs by contradiction in mathematics that prove things for which no constructive proof exists. That makes sense because, on some philosophies of mathematics, an entity doesn't exist until the constructive proof does (and perhaps it can't exist). — Count Timothy von Icarus
And yet your claim that Reductio is invalid is just wrong. — Banno
The problem is that this A, A→¬B∧B ⊢ ¬A was given as reductio ad absurdum. But this entails anything. — Lionino
What if we reject (1) instead? Then A is made true, but it does not imply (B∧¬B). — Leontiskos
The grain of truth in Leontiskos' position is that reductio arguments need to be used with care. — Banno
1. a → (b ∧ ~b)
2. If b is true (b ∧ ~b) is false. If b is false (b ∧ ~b) is false, so (b ∧ ~b) is false.
3.~a → ~(b ∧ ~b) - contraposition (1)
4. ~a - modus ponens (2,3) — Count Timothy von Icarus
Note that the (analogical) equivocity of 'false' flows into the inferential structure, and we could connote this with scare quotes. (B∧¬B) is "false" and therefore the conclusion is "implied." The argument is "valid." — Leontiskos
It seems we can infer both A and ~A from the same thing. But that's because the two assumptions, A and A→¬B∧B, are inconsistent. — Banno
If ¬(B∧¬B) is true, as it must be, then this is not a valid use of modus tollens. — Banno
Note that we could also do other things, such as treat the second premise as truth incarnate, but this is harder to see:
A→(B∧¬B)
¬(B∧¬B), but now conceived as "true"
∴ ¬A does not follow
...that is, if we conceive of the consequent as a proposition and the second premise as truth incarnate, then ¬A does not follow from the second premise (or from the consequent, absent a premise that negates the consequent qua proposition). — Leontiskos
Why is the second line a quote? — Banno
But A, A→¬B∧B ⊢ A is also valid — Lionino
What if we reject (1) instead? Then A is made true, but it does not imply (B∧¬B). Your proof for ¬A depends on an arbitrary preference for rejecting (2) rather than (1). — Leontiskos
So the question is: how do we choose between either? Isn't it by modus tollens? — Lionino
And what I am after is a straight forward explanation of what "FALSE" is. — Banno
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
What is at stake is meaning, not notation. To draw the modus tollens without ¬(B∧¬B) requires us to mean FALSE. You say that you are not using a modus tollens in the first argument. Fair enough: then you don't necessarily mean FALSE. — Leontiskos
Thus when Banno says that a contradiction (b∧¬b) is false, does he mean that it is false or that it is FALSE?... [argument continues on] — Leontiskos
Rubbish. — Banno
It might as well be a Reductio, although even there it is incomplete. It should be something like:
1. A→(B∧¬B) assumption
2. A assumption
3. B∧¬B 1,2, conditional proof
4. ~A 2, 3 reductio — Banno
Leo, what is "FALSE"? — Banno
It doesn't. Explained last time you made this claim... — Banno
what is "FALSE"? — Banno
If P can only be False, yes; otherwise, no. — Lionino
The matter with modus tollens is that q could be otherwise, while in reductio it is not the case by definition. Then again, I don't think it is meaningful or interesting. — Lionino
I don't see a meaningful difference. — Lionino
a→(b∧¬b)) is True, (b∧¬b)) is False, therefore a is False (from «1»). — Lionino
How is it that both (B∧¬B) and ¬(B∧¬B) can have the exact same effect on the antecedent, allowing us to draw ¬A? How is it that something and its negation can both be false? This is key to understanding my claim that two different senses of falsity are at play in these cases. — Leontiskos
So I ask again: How is it that something and its negation can both [function as the second premise of a modus tollens]? — Leontiskos
We can apply Aristotelian syllogistic to diagnose the way that the modus tollens is being applied in the enthymeme:
((A→(B∧¬B))
∴ ¬A
Viz.:
Any consequent which is false proves the antecedent
(B∧¬B) is a consequent which is false
∴ (B∧¬B) proves the antecedent
In this case the middle term is not univocal. It is analogical (i.e. it posses analogical equivocity). Therefore a metabasis is occurring. — Leontiskos
Ok, Presenting a statement that someone has not made is not presenting an interpretation. — Banno
Thus when Banno says that a contradiction (b∧¬b) is false, does he mean that it is false or that it is FALSE? — Leontiskos
Quite so. So what? It remains that RAA is a valid inference in classical propositional logic. — Banno
A reductio without choosing between them is not yet a reductio. — Leontiskos
Yes, it is truth preserving. That
(A∧(B∧¬B))↔(B∧¬B) is valid
But (A∧C)↔C is invalid
does not make me think rules of logic are conflicting, because the equivalence or not with the second term of an «and-operator» is not a rule of logic. — Lionino
As I replied to sime, interpreting (B∧¬B) as P is not a good move, for P can be True or False, (B∧¬B) cannot be True ever. — Lionino
You would object why I rejected 1 instead of 2? I guess I see your point that it is not valid in a tight sense. After all, from A, A→¬B&B, everything follows, not just ¬A. — Lionino
Presenting a statement that someone has not made is not presenting a translation. — Banno
Either inference, ρ→~μ or μ→~ρ, is valid. — Banno
and see that the choice is not in the reductio but in choosing between the conjuncts. — Banno
If this is right then (b∧¬b) introduces instances of formal equivalence that are not provable. — Leontiskos
Such as? — Lionino
I think that is a valid way to frame it. The thing about (B∧¬B) is that, differently from other formulas, it is always False. — Lionino
I don't think that is logically rigorous. As you say, it is not a term in classical logic, and for good reason.
If you want to say A always implies False, A→(B∧¬B) is good for that. While A→¬(B∧¬B) is "always implies True". — Lionino
If A implies a contradiction, not-A can be stated from LNC.
Dogs are fish. Fish, among other things, is defined as not-mammals. Dog is defined, among other things, as mammal. So we end up with "A mammal is not a mammal". Thus, "dogs are fish" has to be false, so "dogs are not fish" has to be true from LNC.
"... cannot be affirmed" does not stand to me as useful, as the LNC + LEM don't accept a third value. — Lionino
I'd like to explore this idea next — Lionino
-Any consequent which is false proves the antecedent
-(B∧¬B) is a consequent which is false
∴ (B∧¬B) proves the antecedent — Leontiskos
Example:
(A∧(B∧¬B))↔(B∧¬B) is valid
But (A∧C)↔C is invalid. — Lionino
(A→B)↔¬A, ¬B does entail however (A→B)↔¬A, even though (A→B)↔¬A is not True for any B, only when B is False. — Lionino
Difference when the consequent is a contradiction:
(¬A→B)↔¬(A→B) is not valid.
(¬A→(B∧¬B))↔¬(A→(B∧¬B)) is valid.
So when the consequent is a contradiction, the ¬ may be pushed in. But when the consequent is a normal statement, you can't. — Lionino
When I said things like:
Thus when Banno says that a contradiction (b∧¬b) is false, does he mean that it is false or that it is FALSE?
— Leontiskos
...or when Lionino distinguished proposition-qua-variable from proposition-qua-truth-value, we were both pointing to this same valence where a material symbol (b∧¬b) has two legitimately different mental conceptions associated with it. 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
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
There are particular apples and we can generalize about them. There is no apple that is not a particular apple. But we do say things like "If x is an apple, then x has a core". That is not claiming that there is an apple that is not a particular apple, but rather we can make generalizations about apples. — TonesInDeepFreeze
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
Let 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
You lied — TonesInDeepFreeze
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
Unless you think you can use the word "particular" without having any idea what it would mean for something to be non-particular? — Leontiskos
One is a statement in the meta-language and the other in the object language. They are different levels of statement. — TonesInDeepFreeze
1. "If A implies B & ~B, then A implies a contradiction"
2. (a→(b∧¬b))→¬a
(My claim here is that (1) represents a reductio whereas (2) does not, even though ↪Banno thinks his truth table has shown that (2) translates a reductio.) — Leontiskos
You lie again. — TonesInDeepFreeze
...you don't know what you mean by a "particular contradiction." Unless you think you can use the word "particular" without having any idea what it would mean for something to be non-particular? — 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
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
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
'non-particular' is your word. It's up to you to say what you mean by it. — TonesInDeepFreeze
(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
↪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
(3) A ... toward a contradiction — TonesInDeepFreeze