Logic - 2016-08-28

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user21820

02:08

@amWhy: Just curious, do you specialize in logic?

2 hours later…

user131753

03:50

I was going through again our last discussion. You wrote, "Classical first-order logic is meaningful." and "So any formal system with quantifier switch is useless." But from a strictly formal point of view, no formal system is more "meaningful" than other. Similarly "usefulness" of a formal system is immaterial for a formlaist (in principle, at least). Isn't it @user21820?

user21820

05:09

@user170039: It depends on what you mean by "strictly formal".

Remember I talked about the meta-system MS. From the point of view of MS, there are certainly formal systems that are meaningful (have a model) and others that are meaningless.

If you don't work in MS, you'll have to get down to defining what you mean by formal..

There are some mathematicians who treat mathematics as purely a game that has absolutely no relation to the real world. In that sense yes indeed there are no useful formal systems!

But most mathematicians cannot reasonably uphold such a viewpoint in light of the fact that the whole internet is very much a product of at least elementary number theory. The reason you can have HTTPS connections is because Fermat's little theorem works for numbers of the size in use.

user131753

Yes. That's true.

user21820

@user170039: So yea that's why I believe all mathematics becomes circular when you get to the natural numbers.

Because even the pure formalist that treats mathematics as a game of symbol pushing needs to know how to push symbols.

user131753

Exactly.

user21820

Yea.

@user170039: By the way, if you're just starting to learn logic, you may benefit from doing the same exercises that I gave toto.

user131753

05:26

By the way, recently I was reading the definition of closure of a formula from Angelo Margaris's book. There it is written, "A statement or a closed formula is a formula with no free variables. If $P$ is a formula and $v_1,v_2,\ldots, v_n$ are the distinct variables that are free in $P$ in that order (from left to right) in which they first occur free in $P$ then $\forall v_1\forall v_2\ldots\forall v_n P$ is the closure of $P$."

user131753

My question is: don't we need to prove the uniqueness of closure of $P$?

user21820

Hmm the closure is defined as a specific formula, so why is there a need to prove uniqueness?

I mean, the definition you gave already gives a precise specification of one sentence as the closure of the formula. So it's automatically unique, even though we never need that fact in practice.

user131753

Otherwise how can we say that it is "the" closure and not just "a" closure (much like in Real Analysis we prove that it is the supremum and not a supremum)?

user21820

Yes i see you're pedantic. Very good. Note that the author is justified to do so without proof because the procedure described is deterministic and produces a single sentence as the closure.

Just like you can say "Let f(x) = x^2+1, for each x in R."

Okay I see the issue; the author is not very precise, and should have said:
> For each formula P with free variables v[1..n] occurring in that order in P, let the closure of P be "forall v[1] ( forall v[2] ( ... forall v[n] ( P ) ) )".

In other words the author is defining "closure" of a formula, not stating something about it.

user131753

Yes that would be more precise. Or we could say that, "For all formula $P$ we define the closure of $P$ as $∀v_1∀v_2…∀v_nP$ where $v_1,v_2,…,v_n$ are the distinct variables that are free in $P$."

user131753

05:36

Probably this is one of the pitfalls of $\sf{NL}$ - the ambiguity of sense of words.

user21820

Yes.

I have a particular dislike for using "if" in a definition when what is intended is "iff".

But it's so standard that it's unlikely to change anytime soon.

user131753

Anyway, thanks for helping. See you soon.

user21820

Sure you're welcome and see you!

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