We don't say: If Gu{P} |- Q and Gu{P} |- ~Q, then ~P |- G.
— TonesInDeepFreeze
I made a mistake. I meant to say:
"The bulk of the debate here between Banno and Leontiskos (and me interjecting sometimes) is why say G |- ~P instead of P |- ~G." — Lionino
Banno — Lionino
It has happened before in the history of science where we had to reject G when finding out that Gu{P} is contradictory, because P was so evidently true. — Lionino
You got it wrong. I know what I meant with my posts. "We" there refers to me, I was not talking about anyone else. The specific post you quoted did not help sort out the issue, specifically the nitpick on "translation", which is why I had to make a whole new thread for that topic specifically. — Lionino
In fact in my thread you corrected yourself about something midway through the discussion — Lionino
if we reach a contradiction we know there is a false assumption — Banno
I lightly jibed, aiming at myself as much as anyone — TonesInDeepFreeze
But since your revisionist attack on me is ill-premised I respond. — TonesInDeepFreeze
I couldn't tell since you often jump into a conversation that happened several pages before the last post of the respective thread. — Lionino
alright however you wanna fly it — Lionino
So, the point you are making could be stated ('\' stands for set difference): — TonesInDeepFreeze
You can use the notation however you wish, but in my formulations, G is a set of formulas not a formula, and on the left side of the turnstile is a set of formulas while on the right side of the turnstile is a formula. — TonesInDeepFreeze
Yep. Good point.We know that the assumptions are inconsistent — TonesInDeepFreeze
More like the point Leontiskos is making. — Lionino
Leontiskos was further saying that the RAA is not strictly logical because it does not tell you which side of the conjunct to rule out. I disagreed in the last post of page 21. — Lionino
(2) Get the conversation completely wrong by falsely saying that I had not explained why your translation was incorrect — TonesInDeepFreeze
If I recall correctly, I was thinking of specifically my post with the several examples compared line by line which was what finally established that that translation was not possible. — 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
Yes, granted. I used the word "translation" wrong in basically all of my posts. I meant "is a true statement about..." instead. — Lionino
Now, the conclusion that I arrived at is that "A does not imply a contradiction" is not an accurate statement about ¬(A→(B and ¬B)), it would be a true statement about (A→¬(B and ¬B)) instead. — Lionino
When it comes to ¬(A→(B and ¬B)), as it is the same as (¬A→(B and ¬B)), "not-A implies a contradiction" is a true statement about it. — Lionino
Amazing! — Lionino
and the tech agent would say "Amazing!" in a kind of lilting modern way when I executed each step in the walkthrough so that I got to the right element each time. As if each click to the next element was a cause for a moment of minor celebration. I found that charming. — TonesInDeepFreeze
I guess it is a thing. — TonesInDeepFreeze
Ok. That is much clearer than your other posts. I suppose I agree with what is conveyed. However, the RAA has formally either rho or mu as a premise, so no choice between the conjunct is needed within the RAA, the RAA is logical. — Lionino
From the supposition we learn (P→¬Q), at which point P must be further asserted beyond supposition if we are to actually arrive at ¬Q: — Leontiskos
B1. ρ
B2. Suppose: μ
B3. Contradiction, therefore ¬μ — Leontiskos
That's obviously not the reductio. — Banno
It's not even a valid argument.B1. ρ
B2. Suppose: μ
B3. Contradiction, therefore ¬μ — Leontiskos
If P, along with possibly other lines, shows a formula Q and a formula ~Q, then infer ~P and charge it with all the lines used to show Q and to show ~Q except the line for P.
If ~P, along with possibly other lines, shows a formula Q and a formula ~Q, then infer P and charge it with all lines used to show Q and used to show ~Q, except the line for ~P. [not intuitionistic]
Those rules are equivalent with (formulated equivalently this time with conjunction):
If Gu{P} |- Q & ~Q , then G |- ~P
If Gu{~P} |- Q & ~Q, then G |- P [not intuitionistic]
There is no mention of 'premise', 'assumption' or 'supposition' nor, for that matter, 'contradiction'. — TonesInDeepFreeze
1. A→(B∧¬B) assumption
2. A assumption
3. B∧¬B 1,2, conditional proof
4. ~A 2, 3 reductio — Banno
1. A→(B∧¬B) assumption
2. A assumption
3. B∧¬B 1,2, conditional proof
4. ~A 2, 3 reductio — Banno
Without the word 'assumption':
1. A -> (B & ~B) {1}
2. A {2}
3. B & ~B {1, 2}
4. ~A {1} — TonesInDeepFreeze
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
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