In
``Steel, John R. On Vaught's conjecture. Cabal Seminar 76–77, pp. 193–208''
the following is proved:


**Theorem.** Let $\phi\in L\_{\omega\_1,\omega}.$ If every model of $\phi$ is a tree, then $\phi$ has either $\leq \aleph\_0$ models or perfectly many countable models.


In page 206 of the paper the following is stated:


It would be natural to attempt to extend the above theorem from tree to arbitrary partial orders. However Arnold Miller has shown Vaught's conjecture for partial orders is equivalent to the full conjecture.



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> **Question 1.** How Miller's result can be proved?
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**Added remarks and new questions:**


According to Enayat's answer, even the Vaught's conjecture for lattices, implies the full Vaught's conjecture.
So it is natural to ask how much we can weaken our assumptions, for example 



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> **Question 2.** Does Vaught's conjecture for Boolean algebras imply the full Vaugh's conjecture? 
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In general:



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> **Question 3.** For which structures it is known that the full Vaught's conjecture is implied by the Vaught's conjecture for those structures?
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Vaught's conjecture for partial orders

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> This answer is an elaboration of the comment by Emil Jeřábek.
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The reduction of Vaught's conjecture to the special case of partial orders is an immediate consequence of the fact that every structure in a finite language $L$ that has at least two elements is bi-interpretable with a special type of partial order, namely a *lattice* (see [here](http://en.wikipedia.org/wiki/Lattice_%28order%29#Lattices_as_partially_ordered_sets) for a definition of a lattice).


The above bi-intepretability result can be found as Theorem 5.5.2 in Hodges' text *Model Theory* (p.230), and it is attributed Taitslin (on p.261), who published the result in 1962.


The reduction of Vaught's conjecture to lattices continues to hold even if $L$ is a *countable* language, as explained in the last paragraph of p.231 of Hodges' text.


