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I don't really understand when we're allowed to use $\exists$-elimination when making first-order natural deduction proofs. I understand that the criteria are that the variable must be free in the conclusion, and in the "context" of the implication. When reading this, I feel like I understand, but when looking at examples, I get confused. For example,

$\exists x P(x)$ and $P(x) \implies \forall y P(y)$, we cannot use the elimination rule; why is this?

Another example would be $\exists x P(x)$ and $P(x) \implies P(x)$. Here I understand that, since x is free in $P(x)$, we cannot use the elimination. However, would it be possible if we instead had $\exists x P(x)$ and $P(x) \implies \forall xP(x)$?

If so, why? Wouldn't x still be free in the context?

Any help is appreciated. Thanks!

Adding picture:

$\begin{array}{lllll} 1&1&\text{Assume }\exists x\,P(x)\land(P(x)\to\forall y\,Q(y))&\qquad\\ 1&2&\exists x~P(x) &&\land E_1,1\\ 1 &3& P(x)\to\forall y~Q(y)&&\land E_2,1\\ \color{red}1&4&\forall y~Q(y)&&\exists E~2, 3~\color{red}{\text{ERROR}} \end{array}$

Graham Kemp
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iaskdumbstuff
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  • @MauroALLEGRANZA Ah, I see. So, for example, if our first line simply had $\exists x P(x)$, then x would not be free in the undischarged assumption, and therefore the elimination would be valid? Thanks very much for your help. – iaskdumbstuff May 17 '19 at 20:49
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    The formula you assume in the image is equivalent to $(\exists z.P(z))\land(P(x)\Rightarrow \forall y.Q(y))$. The fact that there is some value of which $P$ holds does not mean that $x$ is that value. – Derek Elkins left SE May 17 '19 at 21:32
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    I'm surprised that your initial assumption was accepted as syntactically correct with x being used as both a bound and a free variable. This may be the source of your confusion. You might consider changing the bound variable x's to anything but x. When you eliminate the $\exists$, you should name the new free variable anything but x or y. (I like to reserve the letters a-j for bound variables, the remaining letters for free variables.) – Dan Christensen May 18 '19 at 14:57

2 Answers2

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The $\exists$-elimination rule :

$$ \frac{\begin{array}{ccc} &&\phi[x_0/x]\\ &&\vdots\\ \exists x\phi&&\psi \end{array}}{\psi}\exists\text{-elim} $$

holds with the following proviso :

$x_0$ does not occur in $\psi$, $\phi$ or any undischarged assumption of the subderivation of $\psi$, except $\phi[x_0/x]$.

Here is an example showing the need for the proviso in the $\exists$-elim rule :

1) $\exists x(x=0)$ --- assumed

2) $(x=0)$ --- assumed for $\exists$-elim

3) $x=0$ --- from 2) and 1) by $\exists$-elim, discharging 2) : illegal ! $x=0$ is "the $\psi$" and we have the variable $x$ used in the $\exists$-elim application free in it

4) $\forall x(x=0)$ --- $\forall$-intro : correct ; there are no free occurrence of $x$ in the only assumption left, i.e. into 1)

5) $\exists x(x=0) \to \forall x(x=0)$ --- $\to$-intro.

The conclusion is clearly false in $\mathbb N$.

The intuition about the rule is this : we know that there is something that is $P$. Call it $x_0$ (a new term; this is required by the proviso asking for a term that does not occur in any undischarged assumption ).

Form $P(x_0)$ we derive a conclusion $\psi$ that does not "depend" on $x_0$ (and this is for sure satisfied if $x_0$ does not occur in $\psi$).

Thus, we can discard the temporary assumption $P(x_0)$ and the conclusion follows from the premise $\exists x P(x)$.

In your example, $\forall y Q(y)$ is the conclusion $\psi$ and $x$ is free into the hypothesis 1) that is an undischarged assumption of the subderivation of $\psi$.

Thus, the proviso has been violated.

Mauro ALLEGRANZA
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  • Thanks very much for your response. There is still something I'm confused about. When you say "does not depend on $x_0$, what does this mean exactly? For example, I've added a picture to the post; could you explain to me why this does not work? Thanks again. – iaskdumbstuff May 17 '19 at 19:45
  • I have trouble understanding how, in $\forall y Q(y)$ depends on x. – iaskdumbstuff May 17 '19 at 19:48
  • @iaskdumbstuff I found this notion of dependency between variables to be the most difficult concept of predicate logic. If you follow some natural guidelines, you can forget all about it. (1) Do not use Universal Specification to introduce new free variables. (2) Before you discharge a premise (assumption), make sure that there will be no free variables in your conclusion that were introduced after that premise. – Dan Christensen May 18 '19 at 22:11
  • (3) Immediately after discharging that premise, use Universal Generalization (∀ Intro) to generalize on any free variables in the conclusion that were introduced in that premise. I incorporate these guidelines in the Conclusion Rule in my DC Proof program. – Dan Christensen May 18 '19 at 22:12
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Using my DC Proof 2.0

enter image description here

DC Proof will not allow the use of $x$ as both a bound and a free variable within a statement, saving much confusion. As you can see, you won't be able to apply Detachment (Modus Ponens) here.

Suggestion: Try to use the lowercase letters a-j for bound variables, the remaining lowercase letters for free variables. This also avoids some confusion.

Dan Christensen
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