And just a few posts ago, I discussed the no one is disallowed from choosing what to refute with RAA, as long as the refutation is valid. — TonesInDeepFreeze
1. A -> (B & ~B) {1}
2. A {2}
3. B & ~B {1, 2}
4. ~A {1} — TonesInDeepFreeze
The problem is that this proof of yours was invalid:
1. A -> (B & ~B) {1}
2. A {2}
3. B & ~B {1, 2}
4. ~A {1}
— TonesInDeepFreeze — Leontiskos
[fixed quote in edit]is no rule of inference that allows us to draw (4) from (1) and (2). — Leontiskos
It is valid. — TonesInDeepFreeze
Every interpretation in which "A -> (B & ~B)" is true is an interpretation in which ~A is true. — TonesInDeepFreeze
The argument doesn't draw (4) from (1) and (2). The argument draws (4) from (1) as (2) is discharged. — TonesInDeepFreeze
There is no rule of inference that allows us to draw (4) from (1) and (2). — TonesInDeepFreeze
What to conclude except that Leo does not understand validity.But it's not. — Leontiskos
That he asks this is quite odd, since our purpose here was A -> (B & ~B) ⊢ ~A.Heh. Why is (2) "discharged" and not (1)? — Leontiskos
There is no rule of inference that allows us to draw (4) from (1) and (2).
— TonesInDeepFreeze
Yep. The example Leo gave is not an example of RAA. — Banno
That was obvious. A mere typo.Tones was quoting me — Leontiskos
I didn't.If he had you would not have inadvertently agreed with me. — Leontiskos
This is inane. (4) cannot be invalid on its own. The argument is valid in classical prop logic.Note that (4) is originally your conclusion, and we now both agree that it is invalid. — Leontiskos
This is inane. (4) cannot be invalid on its own. — Banno
The argument is valid in classical prop logic. — Banno
What rule of inference do you think you used to draw (4)? — Leontiskos
I didn't. — Banno
And so, A -> (B & ~B) ⊢ ~A. This is a valid argument.1. A -> (B & ~B) {1}
2. A {2}
3. B & ~B {1, 2}
4. ~A {1} — TonesInDeepFreeze
B1. ρ
B2. Suppose: μ
B3. Contradiction, therefore ¬μ — Leontiskos
It was this with which I was agreeing.There is no rule of inference that allows us to draw (4) from (1) and (2). — TonesInDeepFreeze
1. A -> (B & ~B) {1}
2. A {2}
3. B & ~B {1, 2}
4. ~A {1} — TonesInDeepFreeze
If one looks at previous posts by me, one would see that I also directly, explicitly and formally addressed the matter that RAA also provides:
1. A -> (B & ~B) {1}
2. A {2}
3. B & ~B {1, 2}
4. ~(A -> (B & ~B)) {2} — TonesInDeepFreeze
You made the claim that this was RAA: — Banno
Which, as Tones pointed out, leaves out 3: — Banno
There is no rule of inference that allows us to draw (4) from (1) and (2). — Leontiskos
And you would be wrong.I would simply say that both of these proofs are invalid. — Leontiskos
The conversation I am having with Tones revolves around <your argument>, which is an instance of the form of reductio that I gave. — Leontiskos
It is valid.
— TonesInDeepFreeze
But it's not. — Leontiskos
All you are saying is, "ρ→¬μ," but this does not make the proof valid. — Leontiskos
What rule of inference do you think you used to draw (4)? (4) adjudicates the and-elimination. — Leontiskos
The modus tollens and the reductio are two different things:
A1. μ→¬ρ
A2. ρ
A3. Therefore, ¬μ
B1. ρ
B2. Suppose: μ
B3. Contradiction, therefore ¬μ
You can say that "the RAA is logical," but the fact remains that B3 is not as secure as A3... — Leontiskos
The poster continues to indicate that he does not know what validity is in this context and that he is unwilling to read the posts to which responds. He skips that I just stated exactly why the argument is valid. If he won't look at a truth table as suggested, then there's little hope he'll understand anything here. — TonesInDeepFreeze
The poster seems to not know what validity is and that he is unwilling to read the post to which he responded. He skips that I stated exactly why the argument is valid. If he won't look at a truth table as suggested, then there's little hope he'll understand anything here. — TonesInDeepFreeze
/quoteWhat rule of inference do you think you used to draw (4)? (4) adjudicates the and-elimination.quote — TonesInDeepFreeze
1. A -> (B & ~B) {1}
2. A {2}
3. B & ~B {1, 2}
4. ~A {1} — TonesInDeepFreeze
The argument doesn't draw (4) from (1) and (2). The argument draws (4) from (1) as (2) is discharged.
— TonesInDeepFreeze
Heh. Why is (2) "discharged" and not (1)? — Leontiskos
He skips that I stated exactly why the argument is valid. If he won't look at a truth table as suggested, then there's little hope he'll understand anything here. — TonesInDeepFreeze
Has everyone agreed by this point that ↪Banno's truth table does not fully capture what a reductio is? (See bottom of post for truth table)
((a→(b∧¬b)) ↔ ¬a) is truth-functionally valid, but the implication in the first half of the biconditional is not the same implication that is used in a reductio ad absurdum. — Leontiskos
And in bringing clarity to what classical logic actually is, one needs to explain. — TonesInDeepFreeze
A proposition can be invalid qua conclusion — Leontiskos
Again and again the simple questions go unanswered:
What rule of inference do you think you used to draw (4)?
— Leontiskos — Leontiskos
Another inane misattribution. Nowhere have I said that, and certainly not in the post linked....much less ↪Banno's half-baked reductio:
ρ
μ
Contradiction, therefore ¬μ — Leontiskos
From a different angle, Tones says:
1. A -> (B & ~B) {1}
2. A {2}
3. B & ~B {1, 2}
4. ~A {1}
— TonesInDeepFreeze
If one looks at previous posts by me, one would see that I also directly, explicitly and formally addressed the matter that RAA also provides:
1. A -> (B & ~B) {1}
2. A {2}
3. B & ~B {1, 2}
4. ~(A -> (B & ~B)) {2}
— TonesInDeepFreeze
Tones thinks that ¬(1) and ¬(2) both follow from (1, 2, 3). — Leontiskos
From a different angle, Tones says:
1. A -> (B & ~B) {1}
2. A {2}
3. B & ~B {1, 2}
4. ~A {1}
— TonesInDeepFreeze
If one looks at previous posts by me, one would see that I also directly, explicitly and formally addressed the matter that RAA also provides:
1. A -> (B & ~B) {1}
2. A {2}
3. B & ~B {1, 2}
4. ~(A -> (B & ~B)) {2}
— TonesInDeepFreeze — Leontiskos
It goes without saying that there is no rule of inference that forces one rather than the other. — Leontiskos
I would simply say that both of these proofs are invalid. — Leontiskos
There is no rule of inference to justify (4) on either count. — Leontiskos
This all goes to the misunderstandings of reductio ad absurdum in this thread — Leontiskos
and in particular to Tones' recent claim that there is no need to advert to a difference between an assumption/premise and a supposition. — Leontiskos
Not so. What has been clearly demonstrated is that you do not have a grasp of propositional logic.So many of your claims have already been debunked in this thread. — Leontiskos
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