I want to prove that $Th(\mathbb{Q},+,0)$ has quantifier elimination. Please point out where i go wrong or how to finish my reasoning:
We can bring the formula in disjunctive normal form and distribute the existential quantifier. So we are left with a conjunction of literals. Now the basic formula's can be brought to 'additive normal form' (I don't know if this is usual terminology), so $m_0 x_0 + ... + m_{n-1} x_{n-1} + m x = 0$ or an inequality, where $x$ is the variable from the existential quantifier.
But then?
via Recent Questions - Mathematics - Stack Exchange http://math.stackexchange.com/questions/290504/theory-of-q-0-has-qe
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