separating propositons with commas?

From Kaye's Mathematics Logic, about notation for propositional logic:

Another place where we relax notation is in the notation on the left hand side of a turnstile symbol $\vdash$. Instead of using set theory notation with $\{\ldots\}, \cup, ∅$, etc., it is traditional to list formulas and sets of formulas, separating them with commas, and regard the list as a single set of formulas, so the order of formulas in the list and any repetitions in it is ignored. This applies to both the turnstile $\vdash$ of this chapter and the turnstile that will be introduced in the next. Thus, with all the conventions in place, the previous example would be written as $a ∧ b \vdash ¬(¬ a ∨ ¬ b)$. The empty set is written as an empty list, as in $\vdash (a ∨ ¬ a)$.

1. Does "separating them with commas" mean to represent $a ∧ b \vdash ¬(¬ a ∨ ¬ b)$ as $a, b \vdash ¬(¬ a ∨ ¬ b)$ ? Then what is the difference between $a ∧ b$ and $a, b$ ?

I can't figure out what the paragraph is saying from the two examples.

2. Also a propositional language doesn't have comma as a punctuation symbol, while a first order language does. Is it true?

Thanks.

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Well, I think in this case it turns out that $\{a \land b\}$ and $\{a,b\}$ are equivalent, but I imagine the author meant that you write $a \land b \vdash ...$ instead of $\{ a \land b \} \vdash ...$? – copper.hat Jul 23 '14 at 18:10
(1) First of all, by $\{a∧b\}$ do you mean it is a set? why is the LHS a set? Isn't it a boolean term? (2) WHere is "comma"? – Tim Jul 23 '14 at 18:13
I think this was very badly explained. I agree with copper.at, the author means that instead of writing $\{\varphi _1, \ldots ,\varphi_n\}\vdash \text{Whatever}$, you can write $\varphi _1, \ldots ,\varphi_n\vdash \text{Whatever}$. – Git Gud Jul 23 '14 at 18:13
I mean a set of formulae. If the author's example was $a \lor b$ then you would write $a \lor b \vdash ...$. It just so happens that in this particular example we have $a \land b \vdash ...$ is equivalent to $a,b \vdash ...$. – copper.hat Jul 23 '14 at 18:14
@Git: (1) why is the LHS a set? I think it is a boolean term, not a set. (2)since we have $∧$, why do we need comma? – Tim Jul 23 '14 at 18:15

The "official" notation is :

$\Gamma \vdash \varphi$

where :

$\Gamma = \{ φ_1,\ldots,φ_n \}$.

The "abbreviation" licenses us to write it as :

$φ_1,\ldots,φ_n \vdash φ$.

Also, when $\Gamma = \emptyset$, instead of :

$\emptyset \vdash \varphi$

we will write :

$\vdash \varphi$.

Another abbreviation is the following :

$\Gamma, \alpha \vdash \varphi$

in place of the "official" :

$\Gamma \cup \{ \alpha \} \vdash \varphi$.

Comment

Regarding the example discussed, it must be :

$\{ a∧b \} \vdash a$

abbreviated as :

$a∧b \vdash a$

and not : $a, b \vdash a$.

We have to note an important distinction: that between object language and meta-language.

In the formal system of propositional logic, assuming that the propositional letters are the $p_i$'s, we have that e.g. $p_1 \land p_2$ is a formula.

The symbol :

$\Gamma \vdash \varphi$

is not part of the formal system. It is part of the meta-language, and says that there is a derivation (i.e.a sequence of formulae of the system satisfying certain rules) of the formula $\varphi$ from the assumptions in $\Gamma$.

Thus, when we license ourselves to "abbreviate" :

$\{p \rightarrow q, q \rightarrow r \} \vdash p \rightarrow r$

as :

$p \rightarrow q, q \rightarrow r \vdash p \rightarrow r$

we are introducing an abbreviation in the metalanguage.

$f_i(x_1, \ldots, x_n)$
where $f_i$ is the $i$-th function letter (in the enumeration of the alphabet) and has $n$ argument-places.
If we choose to make explicit the number of arguments, writing : $f_i^n$, we can avoid comma and parentheses, writing terms as :
$f_i^nx_1 \ldots x_n$.