We have the following indirect implication of form equivalence classes:

345 \(\Rightarrow\) 119
given by the following sequence of implications, with a reference to its direct proof:

Implication Reference
345 \(\Rightarrow\) 43 Topologie, Analyse Nonstandard et Axiome du Choix, Morillon, M. 1988, Universit\'e Blaise-Pascal
43 \(\Rightarrow\) 8 clear
8 \(\Rightarrow\) 27 clear
27 \(\Rightarrow\) 31 clear
31 \(\Rightarrow\) 32 L’axiome de M. Zermelo et son rˆole dans la th´eorie des ensembles et l’analyse, Sierpi'nski, W. 1918, Bull. Int. Acad. Sci. Cracovie Cl. Math. Nat.
32 \(\Rightarrow\) 119 clear

Here are the links and statements of the form equivalence classes referenced above:

Howard-Rubin Number Statement
345:

Rasiowa-Sikorski Axiom:  If \((B,\land,\lor)\) is a Boolean algebra, \(a\) is a non-zero element of \(B\), and \(\{X_n: n\in\omega\}\) is a denumerable set of subsets of \(B\) then there is a maximal filter \(F\) of \(B\) such that \(a\in F\) and for each \(n\in\omega\), if \(X_n\subseteq F\) and \(\bigwedge X_n\) exists then \(\bigwedge X_n \in F\).

43:

\(DC(\omega)\) (DC), Principle of Dependent Choices: If \(S\)  is  a relation on a non-empty set \(A\) and \((\forall x\in A) (\exists y\in A)(x S y)\)  then there is a sequence \(a(0), a(1), a(2), \ldots\) of elements of \(A\) such that \((\forall n\in\omega)(a(n)\mathrel S a(n+1))\).  See Tarski [1948], p 96, Levy [1964], p. 136.

8:

\(C(\aleph_{0},\infty)\):

27:

\((\forall \alpha)( UT(\aleph_{0},\aleph_{\alpha}, \aleph_{\alpha}))\): The  union of denumerably many sets each of power \(\aleph_{\alpha }\) has power \(\aleph_{\alpha}\). Moore, G. [1982], p 36.

31:

\(UT(\aleph_{0},\aleph_{0},\aleph_{0})\): The countable union theorem:  The union of a denumerable set of denumerable sets is denumerable.

32:

\(C(\aleph_0,\le\aleph_0)\): Every denumerable set of non-empty countable sets  has a choice function.

119:

van Douwen's choice principle: \(C(\aleph_{0}\),uniformly orderable with order type of the integers): Suppose \(\{ A_{i}: i\in\omega\}\) is a set and there is a function \(f\) such that for each \(i\in\omega,\ f(i)\) is an ordering of \(A_{i}\) of type \(\omega^{*}+\omega\) (the usual ordering of the integers), then \(\{A_{i}: i\in\omega\}\) has a choice function.

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