I have been given a language L in which there are only finitely many relations. I am told that for every primitive formula $\phi(y_1\dots y_n) = \exists x(\bigwedge_i \psi_i(y_1\dots y_n))$ where $\psi_i$ is an atomic formula or a negation of an atomic formula, where every atomic formula of L (or its negation) is in the conjunction of $\phi$, $\phi$ is T-equivalent to a quantifier free formula.
I need to show that this implies that every primitive formula is equivalent to a quantifier free formula, but I can't for the life of me figure out how. It seems like it should be obvious.
My intuition screams that I should just select a primitive formula $\alpha$ and find a $\phi$ (defined above) that is an extension of $\alpha$ and simply delete all atomic formulas in $\phi$ that aren't in $\alpha$, then conclude that $\alpha$ is equivalent to a quantifier free formula. I doubt that this is the right way to go about it, but is it at least on the right track?
Just to clarify terminology: A formula is primitive if it has the form $\exists \overline{x} \bigwedge_{i = 1}^n \psi_i$, where each $\psi_i$ is atomic or negated atomic in the variable context $\overline{x}\overline{y}$. We'll say a primitive formula $\varphi$ extends a primitive formula $\theta$ if they have the same variable context $\overline{y}$, their existential quantifiers are over the same variables $\overline{x}$, and every atomic and negated atomic formula in the conjunction in $\theta$ is also in the conjunction in $\varphi$. Let's say a primitive formula (in variable context $\overline{y}$, with existential quantifiers over $\overline{x}$) is complete if for every atomic formula $\psi$ in variable context $\overline{x}\overline{y}$, either $\psi$ or $\lnot \psi$ appears in the conjunction.
This breaks down at the "simply delete" part. Let $\alpha$ be a primitive formula, and find a complete primitive formula $\varphi$ which extends $\alpha$. Then $T\models \forall \overline{y} (\varphi\rightarrow \alpha)$, and by assumption there is some quantifier-free formula $\theta$ such that $T\models \forall \overline{y} (\theta\leftrightarrow \varphi)$. But it's not at all clear how to modify $\theta$ to find a quantifier-free formula which is $T$-equivalent to $\alpha$.
Instead, the key observation is that $\alpha$ is $T$-equivalent to $\bigvee_{\varphi\in \Phi}\varphi$, where $\Phi$ is the set of all complete primitive formulas extending $\alpha$. Once you have proven this, it's easy to see that $\alpha$ is $T$-equivalent to $\bigvee_{\varphi\in \Phi}\theta_\varphi$, where $\theta_\varphi$ is a quantifier-free formula $T$-equivalent to $\varphi$ for each $\varphi\in \Phi$.