TITLE: Proving Undecidability of first order logic without first proving it for arithmetic.
QUESTION [6 upvotes]: All text I have read prove the Undecidability of first order logic a bit as an afterthought and  after having proved the incompleteness and Undecidability of (Peano) Arithmetic.
This proof also hardly works in philosophy, I would like a proof that first order logic is undecidable, but without using arithmetic , Godels incompleteness theorems and all of that.
It is just such a tortuous route and get many will lost halfway, it is about first order logic why do we need to add arithmetic to proof undecidability?
My own first ideas of a proof: 
(not sure even if it is correct, I posted this earlier in  little different form at , Gödel's Completeness Theorem and satisfiability of a formula in first-order logic , in the hope of comments)

There are some formula's where the answer about validity depends on the size of the domain.

$\exists x R(x) \to \forall x R(x) $ is true when the domain only contains one element, but false in any larger domain.
Suppose now we have a formula $\phi $ that is false in every finite domain but is true in an infinite domain.  

We cannot construct a model of $\phi $ just because it requires an infinite domain.
Also we cannot proof $\lnot \phi $ because it is just not a true statement

So by soundness  $ \not \vdash \lnot \phi $ and also $ \phi $ has no finite (constructable)  model so $ \phi $ and $\lnot \phi $ are undecidable statements
And for $\phi $ we can just use $\lnot ( \forall x \forall y \forall z (((Rxy \land Ryz) \to Rxz ) \land \lnot Rxx) \land \forall y \exists x  Rxy )$
I agree $\phi $ is a kind of (minimal)  arithmetic in disguise, interprete Rxy  as $ x > y $ , but it is not full arithmetic and the proof is much shorter.
But is this a correct line of reasoning? 
Has a model to be constructive? ($ \phi $ is "obviously" satisfiable it is just not constructable) 
Are there other and better ways to explain it? (references are very welcome)

REPLY [4 votes]: The book Computability and Logic by Boolos and Jeffrey relates the undecidability of first-order logic directly to the halting problem: by providing a first-order logic description of any turing machine and input, the halting problem can be stated as a first-order logic argument. Thus, if first-order logic is decidable, the halting problem is solvable. Since it is not, first-order logic is not decidable.