No existential elimination rule

Question

I am reading Logic of Mathematics by Adamowicz and Zbierski and there is no inference rule called Existential Elimination / Instantiation in this text. Incidentally, one of the exercises is to show that $\vdash F(t/x)\to \exists x F$.

However, the third axiom of logic (in this text) is $$\forall F\to F(t/x)$$

and my text also defines $\exists x F$ as $\neg \forall x \neg F$. In any of my previous courses on basic logic I would be allowed to infer $F(t)$ for some $t$ from $\exists x F$.

But I'm only reading this text independently (not as part of a course) so I usually use axiom three (above) whenever I want to instantiate a variable and do some wrangling.... of course I pick up these extra negation symbols. I would like to have the following inference rule $$\exists x F \Big /\therefore F(t/x)$$ or equivalently the following theorem $$\exists F \to F(t/x)$$

but I'm not sure if this is allowed or why.

Any help is appreciated, thank you. Part of my concern is what language the instantiation must occur if at all... I think the exercise below is related and maybe I can "pull back" results after expanding to $L\{c\}$... I dunno.

I proved the following exercise: Let $L\{c\}$ be an expansion of the language L by a new constant c. Show that for $F, G \in Fm(L)$ if $x$ is not a free variable in $G$ then the condition $T, \exists x F \vdash G$ in $L$ is equivalent with the condition $T, F(c/x) \vdash G$ in $L\{c\}$.

Answer 1

In the comments, you linked to the Wikipedia page for Existential instantiation. Here, the following rule of inference is presented: Given a premise of the form $\exists x\, \varphi(x)$, conclude $\varphi(c)$, where $c$ is a new constant symbol which has not occurred earlier in the proof.

This rule is not sound. For example, suppose we have a constant symbol $d$ in the language. Let's start with the valid sentence $\exists x\, (x = d)$. The rule tells us that we can conclude $c = d$, since $c$ is a constant symbol which has not occurred earlier in the proof. But the sentence $c = d$ is not valid! It's true in some models but not in others (depending on whether $c$ and $d$ are interpreted to be the same element).

To fix this problem, we have to place an additional restriction on our proofs: If $c$ is a constant symbol introduced through an existential instantiation, then $c$ is not allowed to appear on the conclusion of the proof. This is an awkward kind of restriction, and I think it leads to the kinds of confusions you've expressed in the question and the comments. But unfortunately it seems some textbooks do take this approach. I've just made a small edit to Wikipedia to correct the error. Probably the whole page should be rewritten explaining various proof rules for existential quantifiers in various proof systems.

Here are some more elegant ways of handling existential elimination:

• A sequent calculus system might have the bidirectional rule $\exists x\, \varphi \vdash \psi$ if and only if $\varphi \vdash \psi$, where $x$ does not occur free in $\psi$.

• A natural deduction system might have the following rule: Suppose we can prove $\exists x\, \varphi(x)$, and after introducing the hypothesis $\varphi(y)$, where $y$ is a new variable, we can prove $\psi$, where $y$ does not occur free in $\psi$. Then discharge the hypothesis and conclude $\psi$.

• A Hilbert style system might have the axiom $\exists x\, \varphi\rightarrow ( (\forall x\,(\varphi \rightarrow \psi)) \rightarrow \psi)$, where $x$ does not occur free in $\psi$.

Your book (by Adamowicz and Zbierski) uses a Hilbert style system, so let me show you how the last form of the rule is derived from their axioms for $\forall$ and the definition of $\exists$ as $\lnot \forall \lnot$.

Suppose we know $\exists x\, \varphi$. This means (1) $\lnot \forall x\, \lnot \varphi$. Suppose we also know (2) $\forall x\, (\varphi\rightarrow \psi)$, where $x$ does not occur free in $\psi$. We want to prove $\psi$. Toward a proof by contradiction, let's add a third assumption (3) $\lnot \psi$.

By "Propositional Calculus" axiom 3, $\forall x\,((\varphi\rightarrow \psi)\rightarrow (\lnot \psi\rightarrow \lnot \varphi))$, and by the "Distributivity of a Quantifier" axiom, $\forall x\, (\varphi\rightarrow \psi) \rightarrow \forall x\, (\lnot \psi\rightarrow \lnot \varphi)$. By Modus Ponens and our assumption (2), we have $\forall x\, (\lnot \psi\rightarrow \lnot \varphi)$. By "Distributivity of a Quantifier" again, $\forall x\,\lnot \psi \rightarrow \forall x\, \lnot \varphi$.

By the "Adding a Redundant Quantifier" axiom and assumption (3), since $x$ is not free in $\psi$, we have $\forall x\, \lnot \psi$. By Modus Ponens, $\forall x\, \lnot \varphi$. This is a contradiction to assumption (1).

Applying Theorem 9.11 (Reductio ad absurdem), we obtain $\{\exists x\, \varphi, \forall x\, (\varphi\rightarrow \psi)\}\vdash \psi$. Applying Theorem 9.1 (On deduction) twice, we obtain $\vdash \exists x\, \varphi\rightarrow ( (\forall x\,(\varphi \rightarrow \psi)) \rightarrow \psi)$.