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I have a question about existential instantiation.

So, I am familiar with statements like "Let $x$ be an integer" or "Let $x$ $\in$ $\mathbb{Z}$". These statements define $x$ to represent a particular, but arbitrarily chosen integer.

My question is, are we doing a similar thing when we do existential instantion? For example, lets start with the existential statement $P$ = "There exists an integer $x$ such that $x^2 = 4$". When we do existential instantiation on $P$, are we saying "Let $a$ be an element of the set of integers that make $P$ true?"

Pranav Jain
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    "Let $x$ be an integer" and "let $x\in\mathbb{Z}$" are often examples of universal instantiation, not existential. Suppose you're trying to prove a universal statement about integers: "for all integers $x, A(x)$"; here, ""Let $x$ be an integer" is a bit like "Pick an integer, any integer, and I will show you that $A(x)$". However, as you note, sometimes these phrases are examples of existential instantiation. With your statement $P(x) \equiv x^2 = 4$, knowing that there exists an x such that $P(x)$, you can then say, "let $x_0$ be an such an $x$" — that is, such that $P(x_0)$. – BrianO Mar 08 '21 at 16:50
  • So is existential instantiation for the existential statement "there exists an integer x such that P(x)", where "P(x) = x^2 = 4" is essentially "let x ∈ Z such that P(x)" ? – Pranav Jain Mar 08 '21 at 16:57
  • How is this any different than you previous question? https://math.stackexchange.com/questions/4053156/how-to-interpret-let-in-mathematics – fleablood Mar 08 '21 at 17:01
  • @fleablood I'm asking about existential instantiation in this case. So, let's say we know that "there exists an integer x such that P(x)", where "P(x) = x^2 = 4" is true. Then, when we do existential instantion, aren't we just saying "let x ∈ Z such that P(x)"? If we didn't know the truth value of the existential claim, we wouldn't be able to say "let x ∈ Z such that P(x)". – Pranav Jain Mar 08 '21 at 17:03
  • I was just scouring the web, and I found this paper helpful: https://faculty.washington.edu/smcohen/120/Chapter12.pdf – Pranav Jain Mar 08 '21 at 17:27
  • I already told you earlier, you should read this post. If you don't understand actual logical reasoning, you simply cannot understand how wrong those answers you have gotten earlier are. – user21820 Apr 09 '23 at 04:43

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Yes. If you already have a theorem $\exists x.P(x)$ then you can apply existential instantiation to derive $P(a)$, where $a$ is a new symbol. If you don't have the existence theorem, though, then introducing $P(a)$ constitutes making a new assumption that you need to keep track of.

In normal natural-language math exposition, we don't always clearly distinguish between these two situations at the point where $P(a)$ is introduced - the formal proof structure is something the reader reconstructs based on their understanding of what's being explained.

Karl
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