Let $L$ be a language and $T$ be an $L$-theory.

Let $L'=L \cup C$ where $C$ is a set of new constant symbols.

Suppose $\phi(x_1,\ldots,x_n)$ is an $L$-formula and $(c_1,\ldots,c_n) \in C^n$.

(1) Prove that if $T \models \phi(c_1,\ldots,c_n)$ in $L'$ then $T \models \forall x_1,\ldots,x_n\ \phi(x_1,\ldots,x_n)$ in $L$.

How can this statement be true? Shouldn't it be $T \models \phi(c_1,\ldots,c_n)$ in $L'$ implies $T \models \exists x_1,\ldots,x_n\ \phi(x_1,\ldots,x_n)$ in $L$?

Moreover, prove that $T$ admits quantifier elimination as an $L'$-theory if and only if $T$ admits quantifier elimination as an $L$-theory.

I proved the $(\Rightarrow)$ part by assuming (1) is true. So is (1) really true? And how can I prove the converse?

  • $\begingroup$ A theory is a set of sentences; thus, $T \vDash \phi$ means that sentence $\phi$ is a logical cons of $T$. If $T$ ia an $L$-theory, a model $\mathcal M$ for $T$ does not specifiy how to interpret the new constants $c_i$. Thus, $T\vDash \phi$ means that every model $\mathcal M$ of $T$ satisfies also $\phi$, whatever the values of the domain $M$ of $\mathcal M$ are assigned to the new constants $c_i$. $\endgroup$ – Mauro ALLEGRANZA Nov 28 '18 at 12:38

The trick is that you didn't add any axioms about the constants in $C$.

If $T$ is inconsistent, it proves anything, so might as well assume that $T$ is consistent. Take any model of $T$, and any $m_1,\ldots,m_n$ in that model, now interpret the constants $c_i$ as $m_i$ and any other constant symbol as $m_1$.

Since $T$ proved that $\phi(c_1,\ldots,c_n)$, it means that in $M$ (as an $L'$-structure), $\phi(m_1,\ldots,m_n)$ holds. But this is true for any such $n$-tuple in $M$. In particular, $M\models\forall x_1\ldots\forall x_n\phi(x_1,\ldots,x_n)$. And now this is true for all $M$.


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