I do not understand why the $\exists$-elimination rule is valid

Question

I have been wondering for a while why the $\exists$-elimination rule $\def\f{\phi} \def\y{\psi} \def\tI{\text{I}}$

$$\frac{\begin{matrix} & [\f[t/x]]\\ \vdots & \vdots \\ \exists x \f & \y \end{matrix}}{\y} \ \ \ \ \ \begin{matrix} \text{provided the term $t$ does not appear in $\exists x\f$,}\\ \text{or in $\y$, or in any other undischarged assumption}\\ \text{other than $\f[t/x]$ in the proof of $\y$} \end{matrix} $$

is valid. I do know that the conditions on $t$ exist to prevent fallacious arguments such as these. It seems to me though that these explanations show that the conditions on $t$ are necessary to prevent invalid arguments, and that it remains to be shown that the conditions are sufficient.

So, why are the conditions sufficient to prevent fallacious arguments?

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Edit: a few days ago I myself asked why these conditions were placed on the term $t$ on a previous post, and the answers (well, comments) explained that it was to prevent fallacious arguments.

This much I understand. Yet, again, it's an explanation as to why the conditions are necessary. The current question asks why they are sufficient. In that sense I think this question is different from the two linked posts.

Answer 1

Those side conditions in the inference rule $\exists_E$ are "sufficient" to prevent fallacious arguments in the following sense: using the rule $\exists_E$ with those side conditions and the other rules of natural deduction for first-order logic, whatever formula you can derive is valid, that is, it is true on all structures. This important result is known as soundness theorem.

Said more precisely and in apparently stronger form, soundness of natural deduction for first-order logic says that if a formula $A$ can be derived in natural deduction from a set of hypotheses $\Gamma$ (this is often noted $\Gamma \vdash A$), then $A$ is a logical consequence of $\Gamma$, that is, every model of $\Gamma$ is also a model of $A$ (this is often noted $\Gamma \models A$). In particular, when $\Gamma$ is the empty set, then the soundness of natural deduction for first-order logic means that if a formula $A$ can be derived in natural deduction without any hypotheses, then $A$ is valid, that is true in every structure.

This means that the rules of natural deduction for first-order logic (in particular, $\exists_E$ with its side conditions) do not allow you to derive any invalid formula, or if you want, to build fallacious arguments. Of course, soundness is not a property of the inference rule $\exists_E$ alone but of its the whole set of inference rules of natural deduction for first-order logic. The fact that the rule $\exists_E$ with the side conditions does not allow us to derive fallacious arguments holds, provided that we do not change the other inference rules of natural deduction.

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For a reference to the proof of the soundness theorem that is freely accessible online (maybe illegally?), see: Dirk van Dalen: Logic and Structure, fourth edition, Springer, 2008. There the theorem "$\Gamma \vdash A$ if and only if $\Gamma \models A$" is called completeness theorem because the terminology is not uniform in the literature. What is often called "completeness" is the direction "$\Gamma \models A$ implies $\Gamma \vdash A$", while the direction "$\Gamma \vdash A$ implies $\Gamma \models A$" is often called "soundness". Since the completeness direction is often considered the most important and challenging to prove, sometimes (this is what van Dalen does, among others) call the whole equivalence "completeness".

Answer 2

Following on from Taroccoesbrocco's answer: when we prove the soundness theorem, we prove by induction on the length of proofs that for any proof $\Gamma\vdash\varphi$ (where $\Gamma$ is the set of hypotheses and $\varphi$ is the conclusion), for any structure $A$, if $A$ satisfies $\Gamma$, then $A$ satisfies $\varphi$. The argument has one inductive step for each proof rule. Let's look at the one for $\exists$-elimination.

Suppose $\Gamma\vdash \psi$, and the proof ends with an application of $\exists$-elimination. Then we have shorter proofs $\Gamma\vdash \exists x\,\varphi$ and $\Gamma\cup\{\varphi[t/x]\}\vdash \psi$.

Let $A$ be a structure, and assume $A\models \Gamma$. We want to show $A\models \psi$.

By induction, since
$\Gamma\vdash \exists x\,\varphi$, $A\models \exists x\,\varphi$. Then there is some $b\in A$ such that if we change $A$ to $A'$ by interpreting $t$ as $b$, $A'\models \varphi[t/x]$. Since $t$ is not free in $\Gamma$, we still have $A'\models\Gamma$.

By induction, since $\Gamma\cup\{\varphi[t/x]\}\vdash \psi$, $A'\models \psi$. Since $t$ is not free in $\psi$, $A\models \psi$ too.