@user170039 Until you stop censoring me from calling out cranks, I will not answer any of your questions pertaining specifically to logic, as I promised.
Suffice to say that nobody can prove mathematical induction non-circularly. I will not elaborate until you stop your censorship, because I do not support cranks especially those who incite violence.
@user21820 I agree that both the comments are extremely offensive (at least it sounds that way to me). So, in this case, it is better to let the site moderators (in this case this) to be aware of its problematic content.
@TungNguyen As LeakyNun showed, they are equivalent if your logic allows for an empty universe. This is because any sentence of the form "forall x exists y ( ... )" is true for an empty structure but "forall x ( ... ) and exists y ( ... )" is false for an empty structure.
In general, you are correct that the order of quantifiers matters. "forall x exists y ( x=y )" is true in any structure, but "exists y forall x ( x=y )" is false in any structure that does not have exactly one element.
That's because you were a party to the conversation :) Yes I suppose I would prefer it be unstarred
user131753
3:21 AM
@user21820 You don't need to elaborate because you have already did so in a previous discussion between me, Mikhail Katz and you. However, there the term "prove induction non-circularly" has been employed there to roughly mean "philosophical justification of mathematical induction non-circularly" and you argued that there is no such justification in your sense.
@user170039 That is the same exact sense I was using in the message you quoted, since that question was about mathematical philosophy, so George Chen's answer is totally wrong. And please stop asking me any question pertaining to logic or philosophy.
@LeakyNun I thought I had answered your question, but apparently not. If your sequence is represented by a function-symbol, and you count quantifiers over nat/rat, then yes indeed whether f is Cauchy is Π3.
The high complexity should not be surprising. After all, if you restrict to integer sequences then the first quantifier is redundant, and then whether they converge is equivalent to whether they are eventually stable.
Given any program P, consider the program Q that on input n will run P on input k for each k from 0 upwards, but for a total of at most n steps, and then output the highest value of k reached. Then Q represents a sequence that is eventually stable iff P is not total, and every item in the sequence is computable (and hence represented by a Σ0-sentence about the input). Recall that non-totality is Σ2, and that's exactly the same as the complexity of the Cauchyness of integer sequences.
