Proof that adding witnesses to $S$ keeps $S$ consistent

Question

For context: this may help me resolve this previous post.

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Definition: for any set of sentences $S$ in the language $L$, define$\def\f{\phi}$ $$w(S) := S\cup\bigg\{\phi[c/x] : (\exists x\phi)\in S\bigg\}$$ in the language $w(L)$ consisting of $L$ with enough constants $c$ added (one such constant $c$ per $\exists x\phi$ statement in $S$ is sufficient).

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Is the following result and proof correct?:

Result: if $S$ is consistent, then so is $w(S)$.

Proof: suppose $w(S)$ is inconsistent, so that $$S,\psi_1[t_1\setminus x_1],\ldots,\psi_m[t_m\setminus x_m]\vdash \bot$$ for sentences $\psi_j[t_j\setminus x_j]$ for which $(\exists x_j \psi_j(x_j))\in S$. Adding an extra step to the proof of $\bot$ if necessary (recall that everything follows from $\bot$) we assume that $\bot$ is a sentence in $L$. The goal is to construct a proof of $\bot$ in $S$. We use the following rule of natural deduction:

$$\frac{ \begin{matrix} & & \left[\phi[t/x]\right]\\ & & \vdots\\ \exists x\phi(x) & & \psi \end{matrix}} {\psi}$$ for a newly introduced constant $c$ into the language (note this implies that $c$ is not in $\exists x\phi(x)$, nor in $\psi$, nor in any axiom, as these are all sentences of the language, which does not contain $c$).

Substituting into the rule, we get

$$\frac{ \begin{matrix} & & \left[\psi_1[t_1/x_1]\right]\\ & & \vdots\\ \exists x_1\psi_1(x_1) & & \bot \end{matrix}} {\bot}$$

providing us with $S,\psi_2[t_2\setminus x_2],\ldots,\psi_m[t_m\setminus x_m]\vdash \bot$. Applying the argument $m$ times gives $S\vdash \bot$, which is a contradiction.

Answer 1

Yes, your proof is correct, and $w(S)$ is consistent per your proof. It's worth noting, however, that your result follows almost immediately from the model existence theorem, and is roughly equivalent to it. Since $S$ is consistent then it has a model, any model for $S$ can be trivially converted into a model for $w(S)$, and $w(S)$ having a model implies it's consistent. Conversely, if we know $w(S)$ is consistent, then a model for $S$ can be directly constructed from the set of all constant symbols included by $w(S)$.

This all becomes very trivial once you familiarize yourself with how the model existence theorem is conventionally proven. Simply extend $S$ to a maximally consistent set of sentences $S'$, then let the model $M$ be the set of all true existence statements in $S'$, modulo provable equality. All relations and functions are defined on $M$ in accordance with whatever is proven by $S'$, and the maximality of $S'$ guarantees that everything works out. So, each formula $\phi$ obeying $S'\vdash \exists x, \phi(x)$ admits a corresponding equivalence class $t_\phi$ as an object in $M$, which necessarily obeys $M\models\phi(t_\phi)$. In other words, the model $M$ is literally the set of all witnesses to the existence statements of $S'$.

Even if the model is not constructed in exactly that way, however, you can still construct $w(S)$ straightforwardly from any model of $S$. Assume $M$ models $S$, and without loss of generality we also assume $M$ is disjoint from the language $L$. For each provable existence statement $S \vdash \exists x, \phi(x)$, it must hold that $M\models \exists x,\phi(x)$ since $M$ models $S$. Consequently, there exists $t\in M$ such that $M\models \phi(t)$. Since $M$ is disjoint from $L$, we can simply include each $t\in M$ as a constant symbol in the extended language $w(L)$, and include the axiom $\phi(t)$ to the extended theory $w(S)$ whenever it holds that $M\models \phi(t)$. Minor technicalities notwithstanding, $M$ is now a model for $w(S)$, and so $w(S)$ is consistent.