Proving that a propositional theory of any cardinality has an independent set of axioms

This is exercise 1.2.19 from Chang & Keisler's Model Theory, which has been giving me a headache for some time now.

Let $\mathscr{S}$ be a given propositional language of any cardinality (i.e. the set of sentence symbols is allowed to have any cardinality). Call a theory simply any set $\Gamma$ of sentences from $\mathscr{S}$. A set $\Delta$ of sentences is said to be a set of axioms for $\Gamma$ iff $\Gamma$ and $\Delta$ have the same consequences (i.e. they are equivalent). Finally, a set $\Sigma$ of sentences is said to be independent iff for each $\sigma \in \Sigma$, $\sigma$ is not a consequence of $\Sigma - \{\sigma\}$. Exercise 1.2.19 asks us to prove that a theory $\Gamma$ always has an independent set of axioms.

Now, if $|\Gamma| \leq \omega$, the proof is more or less simple; for the finite case, cf. here, whereas, if $\Gamma = \{\gamma_0, \gamma_1, \dots\}$, one first constructs a set $\Gamma' = \{\psi_0, \psi_1, \dots\}$ such that $\psi_0 = \sigma_0$ and $\psi_n = \bigwedge_{i \leq n} \sigma_n$ (i.e. the conjunction of all $\sigma_i$ such that $i \leq n$). Then one constructs another set $\Delta = \{\delta_0, \delta_1, \delta_2, \dots\}$ such that $\delta_0 = \psi_0$ and $\delta_{n+1} = \psi_n \rightarrow \psi_{n+1}$. It's not difficult to show that $\Gamma'$ is a set of axioms for $\Gamma$ and that $\Delta$ is an independent set of axioms for $\Gamma'$.

But I'm completely lost about how to proceed in the transfinite case. It's clear to me that the above construction will not work directly; any ideas about how to adapt that strategy? Or should I look for an entirely different strategy?

EDIT: Notice that this question is different from this one; the other question (and answer) is only asking about the countable case, the solution of which I have already provided a rough sketch in my own question. The general case, however, for theories of any cardinality, is, as C&K remark, much harder.

Solution 1:

In his paper “Tout ensemble de formules de la logique classique est equivalent ´a un ensemble independant”, Iegor Reznikof showed the conclusion of your question, but unfortunately the original paper is in French. However you can find a translation in English here.