# Step 2 of Completeness proof in Endertons' "Mathematical Introduction to Logic"

Step 2 of the completeness proof in Enderton's "Mathematical Introduction to Logic" (p. 136) reads:

"For each wwf $$\phi$$ (in the new language) and each variable $$x$$ we want to add to $$\Gamma$$ the wff $$\neg\forall x\phi \to \phi[c/x]$$ where $$c$$ is one of the new constants symbols. We can do this in such a way that $$\Gamma$$ together with the set $$\theta$$ of all the added wff is still a consistent set."

He continues by supplying the following detials:

Adopt a fixed enumeration of the pairs $$\langle \phi, x\rangle$$ where $$\phi$$ is a wff of the expanded language and $$x$$ is a variable: $$\langle \phi_1,x_1\rangle, \langle \phi_2,x_2\rangle, ...$$ This is possible since the language is countable. Let $$\theta_1$$ be $$\neg\forall x_1\phi_1 \to \phi_1[c_1/x_1]$$ where $$c_1$$ is the first of the new constant symbols not occurring in $$\phi_1$$. In general $$\theta_n = \forall x_n \phi_n \to \phi_n[c_n/x_n]$$ where $$c_n$$ is the first of the new constant symbols not occurring in $$\phi_n$$ or $$\theta_k$$ where $$k < n$$.

My question: How does this construction ensures that every formula $$\phi$$ in the expanded language in which only one variable occurs freely actually gets a witness?

More specifically, I am thinking that $$\phi_1$$ is simply the first formula by some enumeration. This means that $$\phi_1$$ could have $$x_1$$ free or not. It could be that $$\phi_1$$ has $$x_2$$ free. Say for instance $$\phi_1 = P_1^1(x_2)$$. Then, $$\theta_1 = \forall x_1 \phi_1 \to \phi[c_1/x_1]$$ would simply be $$\theta_1 = \forall x_1 P_1^1(x_2) \to P_1^1(x_2)$$. Since $$\theta_2$$ and all other $$\theta_n$$ are talking about different $$\phi_i$$ it seems to me that $$\phi_1$$ is left without witness in the overall construction. Why am I wrong? what am I missing?