Oft remarked that two wrongs don't make a right, but never seen anytime establish in general that if you've done n wrongs then n+1 won't fix it. Keep going, dig deeper, there's still hope.
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I think Christian doctrine has something to do with limit stages tbf. "I am the Alpha and the Omega" implies that if you are at, say, Gamma, you just got to keep on going and eventually you come to a Bound.
If wrongs are associative I think you could argue
Wrong+wrong+wrong==(wrong +wrong)+wrong==wrong+(wrong+wrong)
Which must collapse to
Wrong+wrong+wrong=wrong+wrong, allowing inductive proof over the aleph null. Though wrongs are probably not associative, since we don't know what two wrongs make.
Indeed, the order of the wrongs can make all the difference.
The maths would probably be more complicated than just tensors, because daoism tells us that what we might believe is a right or wrong, might turn out to be the opposite.
So, just one wrong can be a right, that might, later, be wrong.
suppose that n+1 wrongs does not ever fix n wrongs. consider any set of n+1 wrongs W and any additional n+2th wrong w*. Order W by wrongness and pick max w\in W. It is well known there is a threshold of wrongness at which a wrong graduates from a "little" to a "big" wrong. So consider two cases:
1. w is a little wrong: there are at least 2 little wrongs in W (since w was maximally wrong). Combining them into one macro-wrong m, let W' = W - {w, w'} \cup {w*}. |W'|=n so by i.h., m does not fix it. But as W' and m = W and w* then w* could not fix W either, so n+2 wrongs don't fix n+1 wrongs.
2. w is a big wrong: consider a set W' = W-{w} \cup {w*}. By inductive hypothesis, it could not be that W', which is of size n+1, fixed any wrongs. But since big wrongs are always wrong, committing w could not fix W' and again n+2 wrongs do not make n+1 rights.
I think you haven't properly shown the inductive step. Two wrongs are two wrongs, not one wrong. So we still need to show the n to n+1 part of the induction.
I assumed that doing only half the induction proof was ops point, sort of advanced nerd sniping so people go 'wait is this induction?' (I think you are correct on the inductive step).
2^aleph_0 wrongs make a right but it’s an open question if aleph_1 wrongs make a right; obviously the continuum hypothesis implies it so it’s consistent and independent of the theory of real numbers
Aleph_1 wrongs make a right in L, but for each set of wrongs there is a wrong-unrighting forcing poset such that the forcing extension satisfies “these wrongs do not make a right.”
This is independent of Peano arithmetic—ZFC proves that in the “real natural numbers” there is no n such that n wrongs make a right, but there are nonstandard models of arithmetic in which there are pseudofinite sets of wrongs that make a right
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Wrong+wrong+wrong==(wrong +wrong)+wrong==wrong+(wrong+wrong)
Which must collapse to
Wrong+wrong+wrong=wrong+wrong, allowing inductive proof over the aleph null. Though wrongs are probably not associative, since we don't know what two wrongs make.
wrong(cheating with a friend's wife) >>> wrong(wearing unmatched socks)
Einstein solved this last equation by wearing no socks, and cheating his own wife.
The maths would probably be more complicated than just tensors, because daoism tells us that what we might believe is a right or wrong, might turn out to be the opposite.
So, just one wrong can be a right, that might, later, be wrong.
https://ncase.me/trust/
Copykitten is the way to go.
Qed
Wrongdoing is an idempotent operator.
Removal of two halves restores the whole
Stephen G. Lomber and Bertram R. Payne
Visual Neuroscience, 1996
lastpositivist: hold my beer while I ruin your life.