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user131753
8:55 AM
In this question I don't think what Mauro's answer aims at is what we want to show (assuming that $\vdash \alpha \leftrightarrow ∀x \alpha$ implies that for some underlying structure $S$ there is a proof of the formula $\alpha \leftrightarrow ∀x \alpha$). Am I wrong @user21820?
 
user21820
user21820
11:37 AM
@user170039: I see that your last comment states that you understand his answer. Is that the case?
Also, you can see that in general mathematicians, even logicians, do not precisely specify what they mean. The use of ellipsis ("...") is never precise, but making the desired notion precise is probably too troublesome and not worth the effort. In this case however, it caused a little confusion because it suggested that n >= 1.
It is indeed true that the notation seems to specify a sequence from "x[0]" to "x[n-1]", which would be meaningless if n = 0, but the intended meaning is that when n = 0 the sequence is empty.
Wait a minute, the definition as stated must be wrong.
It requires x[n-1] to be free in A. So the definition does not apply to closed formulae; it does not specify the closure of a closed formula A because there is no n such that x[n-1] is free in A. Mauro's answer clearly missed the point as well.
 
 
2 hours later…
user131753
1:25 PM
I forgot to delete the last comment. I understand his answer in the sense that what he seems to write is perfectly clear to me. But I don't understand his answer in the sense that I don't see how his answer exactly proves that the "closure of a closed formula" (from the definition given I can't even make sense what the formal representation of it will be) is the formula itself. I think that we need to define it, rather than trying to prove it. What do you think @user21820?
 
user131753
By the way, I told junior what you told me to tell him. Fortunately for me, he didn't ask anymore questions like I did.
 
user21820
user21820
2:47 PM
@user170039: Ah good to hear about your junior. As I said, the definition as stated in your question is in fact invalid, since it is self-contradictory when applied to closed formulae.
Of course, Mauro probably didn't think though the question carefully enough to its logical end.
(Since from my memory his answers are almost always correct.)
 
user131753
3:32 PM
So, what will be the answer of the question @user21820 (you may also write an answer instead of commenting here, but that's upto you)?
 
user21820
user21820
@user170039: I have written an answer. See if it answers your question. =)
Do comment on my answer to clarify anything, so that it will benefit other readers. I've to go off now though, so I'll respond when I'm back on Math SE.
 

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