# $T \models \phi(c_1,\ldots,c_n)$ implies $T \models \forall x_1,\ldots,x_n\ \phi(x_1,\ldots,x_n)$?

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?

• 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 $c_i$. – 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 $$M$$.