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In How To Prove It?, Velleman explains that, in order to prove a statement with a goal of the form $\exists x P(x)$, we must declare an arbitrary variable $y$ and assign a value $a$ to it and use the assignment to prove the substitution instance $P(y/x)$ of the goal. Hence, assuming $\Gamma$ is a well-founded formula, we have:

  1. $\Gamma \rightarrow \exists x P(x)$
  2. $(\Gamma \land (y=a)) \rightarrow P(y/x)$

My question concerns the formal structure of such proof technique and the relation between statements 1 and 2.

My first attempt was this one: since $P(a)$ is logically equivalent to $\forall y[(y=a) \rightarrow P(y)]$, then $\Gamma \rightarrow P(a)$ is equivalent to $\Gamma \rightarrow \forall y[(y=a) \rightarrow P(y)]$. If $y$ does not occur in $\Gamma$, then the last statement is equivalent to $\forall y[\Gamma \rightarrow ((y=a) \rightarrow P(y))]$, which means $\forall y[(\Gamma \land (y=a)) \rightarrow P(y)]$.

Now, since $\Gamma \rightarrow P(a)$ implies $\Gamma \rightarrow \exists x P(x)$, it is fair to believe that the relation between statements 1 and 2 is $\forall y[(\Gamma \land (y=a)) \rightarrow P(y)] \rightarrow [\Gamma \rightarrow \exists x P(x)]$.

However, some members on this site informed me that this implication is a weaker version of the more general assertion $\forall y[[(\Gamma \land (y=a)) \rightarrow P(y)] \rightarrow [\Gamma \rightarrow \exists x P(x)]]$. In this case, $y$ appears as a free variable, whose quantification ranges upon the entire proof. However, I can't tell how it is possible to derive such stronger implication, especially because the formula $(\Gamma \land (y=a)) \rightarrow P(y)$ seems meaningless to me.

TylerD007
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2 Answers2

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IMO, your analysis of Existential Introduction is unnecessarily complicated...

The intuitive reasoning is the following:

if we know that Socrates is a Philosopher, then we are licensed to assert that a Philosopher exists.

In symbols:

$\dfrac { P(s) } { \exists x P(x)}$.

Thus, there is no need to add a "declaration" about the term (i.e. a "name") $s$.


Things are different with Existential Elimination, where we have $\exists x P(x)$ as premise.

In this case, we introduce a new term $a$ to "temporary" refer to the unknown individual that we know to exists.

The "new" condition is formalized through the proviso that $a$ must not be used in the "context" $\Gamma$, nor in the conclusion of the proof.



With ref to Velleman's book, page 112, I do not see big differences...

The author says:

"Try to find a value of $x$ for which you think $P(x)$ will be true."

This amounts to saying: "let $x= \text {Socrates}$".

"... and proceed to prove $P(x)$ for this value of $x$."

This amounts to saying: "prove $\text {Philosopher}(\text {Socrates})$".

Then apply EI to conclude that $\exists x (\text {Philosopher}(x))$.

  • I totally agree with you, Mr Allegranza, but that's what Velleman does, he uses value assignments to prove goals of the form $\exists x P(x)$. All I want is to establish the relation between statements in which such assingments are assumed and the original proofs. EDIT: I suppose my original problem does not relate with existential elimination, at least not at this point of the proof. – TylerD007 Aug 25 '20 at 11:29
  • About your second edit: I think that's not exactly what Velleman proposes. He says, if we have a proof like $\Gamma \rightarrow \exists y (y + 2x =0) $, then we assume a value of a free variable y that makes the formula $(y + 2x =0)$ true and that implies $\exists y (y + 2x =0)$. Like I said, a member of this site told me that the first implication as a whole implies the existentially quantified goal. Hence, $[(y = -2x) \rightarrow (y + 2x =0)] \rightarrow \exists y (y+2x=0)$. However, that reasoning makes little sense to me, even after much consideration about it. – TylerD007 Aug 25 '20 at 11:45
  • Your reasoning makes much more sense to me. The two problems that I formalize in my post are: 1) from where does Velleman introduce the value assignment into the proof? 2) does he really use EI to derive $\exists x P(x)$ or does he use this kind of "stronger" implication? – TylerD007 Aug 25 '20 at 11:51
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    @TylerD007 - Frankly speaking, I'm not sure that that is the right reading of Velleman's text... He has a predicate $P(y) =: (y+2x=0)$ and he wants to prove $\exists y P(y)$. In order to do so, he assume $y=a$ for a suitable $a$; in this case the suitable value is $a=-2x$. Then he check (no special logic here: only algebra) that $P(a)$ holds, i.e. that $a+2x=0$. Thus, having proved $P(a)$, he apply EI to conclude with $\exists y P(y)$. – Mauro ALLEGRANZA Aug 25 '20 at 15:41
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Suppose you are trying to prove $\exists y(y+2x = 0)$. Here are two ways you might write the proof:

  1. Prove $(-2x)+2x=0$.
  2. Start with the sentence "Let $y = -2x$" and then prove $y+2x=0$.

The logic of these two approaches is exactly the same; the difference is merely stylistic. In the second approach, the letter $y$ is just being introduced as a notational shorthand for $-2x$.

Which style is better? It may depend on the example, and it may be a matter of taste. But if the conclusion of the theorem being proven has been stated with an explicit existential quantifier, then I think many mathematicians would prefer the second style.

For example, suppose you are giving an $\epsilon$-$\delta$ proof of a limit statement--say, $\lim_{x \to 2} 3x = 6$. The statement to be proven is $\forall \epsilon>0 \exists \delta>0 \forall x(0 < |x-2| < \delta \to |3x-6|<\epsilon)$. I think most mathematicians would start the proof like this: "Let $\epsilon$ be an arbitrary positive number. Let $\delta = \epsilon/3$. Then $\ldots$." There is really no need for the sentence "Let $\delta = \epsilon/3$." One could simply prove that $\epsilon/3$ has the property required for $\delta$, and then conclude that the limit statement is true. But most readers would find the sentence "Let $\delta = \epsilon/3$" helpful, because it tells the reader that $\epsilon/3$ is the number that is being proposed as the value of $\delta$ in the $\epsilon$-$\delta$ definition.

Note that How To Prove It is not a book on formal logic, although there is some logic in the book. It is a book about how to write proofs in English. So although the proof techniques that are discussed often correspond to rules of formal logic, the explanations focus on how to use those techniques to write proofs in English, not how to write proofs in formal logic.

Dan Velleman
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