2

I recently learned that "$\forall x \in S, \phi(x)$" is shorthand for "$\forall x \big(x \in S \rightarrow \phi(x)\big)$". Does the same idea apply for the shorthand "$\exists x \in S, \phi(x)$"?

i.e. is "$\exists x \in S, \phi(x)$" logically equivalent to "$\exists x \big(x \in S \rightarrow \phi(x)\big)$"?

I am somewhat tempted to think that it may actually denote "$\exists x\big( x \in S \land \phi(x)\big )$" but am not positive! Any insight is greatly appreciated. Cheers~

S.C.
  • 4,984

2 Answers2

5

I am somewhat tempted to think that it may actually denote "$\exists x\big( x \in S \land \phi(x)\big )$"

Yes, that's what it means.

i.e. is "$\exists x \in S, \phi(x)$" logically equivalent to "$\exists x \big(x \in S \rightarrow \phi(x)\big)$"?

That would not be very useful, because $x \in S \rightarrow \phi(x)$ is true for every $x \notin S$, no matter what $\phi$ is, and so $\exists x \big(x \in S \rightarrow \phi(x)\big)$ is true for any $S$ that is not the whole universe.

Nate Eldredge
  • 97,710
  • That makes perfect sense. One question, however (more of a "please confirm the following"). It is NOT the case that $\exists x \big(x \in S \rightarrow \phi(x)\big)$ is a "wrong" statement. It does not, for example, break any first order logic rules. All that it IS...as you stated...is "not [...] very useful". Which I take to mean, "in most circumstances, this won't help you with the construction of an argument". Is that correct? – S.C. Mar 20 '20 at 00:06
  • 1
    Yes, it is a well-formed sentence. It just isn't one you'd be likely to ever care about, and so it would be silly to take it as the definition. In particular, it certainly doesn't fit the intuitive meaning of "there exists an $x$ in $S$ such that $\phi(x)$." – Nate Eldredge Mar 20 '20 at 00:07
  • And that intuitive meaning is: "There exists an $x$ that is in $S$ and satisfies $\phi(x)$". – Graham Kemp Mar 20 '20 at 00:24
2

$\exists x{\in}S~.\phi(x)$ is intended to be read as "There is something, call it $x$, which is in $S$ and satisfies $\phi(x)$," and so is indeed synonymous with $\exists x~.(x\in S\wedge\phi(x))$

Moreover, we want quantifier duality to hold in restricted domains, so will need $\neg\forall x{\in}S~.\neg\phi(x)$ and $\exists x{\in}S~.\phi(x)$ to be logical equivalences.

Now because $\neg \forall x{\in}S~.\neg\phi(x)$ is synonymous with $\neg\forall x~.(x\in S\to \neg\phi(x))$, which by quantifier duality is equivalent to $\exists x~.\neg(x\in S\to\neg\phi(x))$, and thus with $\exists x~.(x\in S\wedge\phi(x))$, therefore we shall want this to be synonymous with $\exists x{\in}S~.\phi(x)$ .

Graham Kemp
  • 129,094