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Take a proposition like $\neg(\forall x \in S, P(x))$. If negation is not distributive, how does one determine which connective to be negated? Here are the options I've seen:

$$\neg(\forall x \in S, P(x)) \iff \neg\forall x \in S, P(x) \iff \exists x \in S, \neg P(x)\tag1$$

$$\neg(\forall x \in S, P(x)) \iff \forall x \not \in S, P(x) \tag2 $$

$$\neg(\forall x \in S, P(x)) \iff \forall x \in S, \neg P(x) \tag3$$

$$\neg(\forall x \in S, P(x)) \iff \neg\forall x \not \in S, \neg P(x) \iff \exists x \in S^c, P(x) \tag4$$

Here, $(4)$ is the version in which negation is distributive. If $(4)$ is not correct, how does one determine what is to be negated in cases like the LHS above?

EDIT:

All even numbers are divisible by three: $\forall n \in \Bbb N, n | 3$

Obviously not true. What is true? Following the Law of Excluded Middle, $P \lor \neg P$. Since $P$ is false, its negation must be true. The following RHS expressions are true:

$\neg(\forall n \in \Bbb N, n | 3) \iff (\neg \forall n \in \Bbb N, n | 3 ) \iff (\exists n \in \Bbb N, \displaystyle \frac n3 \not \in \Bbb N)$

However, it is also true that there exists a number outside of the naturals that is divisible by three, tagged as $(4)$ here. So, if we assume LEM, my above list is shortened:

$$\neg(\forall x \in S, P(x)) \iff \neg\forall x \in S, P(x) \iff \exists x \in S, \neg P(x)\tag1$$

$$\neg(\forall x \in S, P(x)) \iff \neg\forall x \not \in S, \neg P(x) \iff \exists x \in S^c, P(x) \tag4$$

It is not the case that all naturals are disivisble by three. It is the case that there exists naturals that aren't divisble by three. It is also the case there exists non-natural numbers that are divisible by three. Both $(1)$ and $(4)$ satisfy LEM in the given example, and both of them seem like ways of negating the example proposition. Which one is correct? When not assuming LEM, which is correct then?

user110391
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  • Have you tried testing it with some examples? E.g. let $S$ be the even numbers, and let $P$ mean "divisible by 3", or alternatively let $P$ mean "divisible by 4" – Vsotvep Apr 10 '22 at 10:02
  • @Vsotvep See my edit. I was going to use your example, but is mis-remembered even numbers as the natural numbers. I think my confusion was still communicated, but maybe I missed a point you were making by using a different example? – user110391 Apr 10 '22 at 10:20
  • Well, by using the even numbers and divisibility by 4, you can exclude (4) as being a possibility as well. What rests, is proving that (1) is actually valid. Proving that requires knowing which axiomatic system you use, or you can argue semantically. – Vsotvep Apr 10 '22 at 10:23
  • [1/2] @Vsotvep How so? $\forall 2n \in \Bbb N,2n|4$ is false, and both $\exists 2n\in \Bbb N, 2n \not | 4$, and $\exists 2n \not \in \Bbb N,2n|4$ are true. You could say that if a number is divisible by $4$, then it is necessarily in the naturals. However, any natural has a corresponding number in the other sets. Strictly speaking, from a set-theoretic point of view, $n_{\Bbb N} \neq n_{\Bbb R}$. However, if we disregard that, I think I get your point. So basically, in my example, even though $(4)$ has the opposite truth value, and seems like a negation, it isn't, because it generally isn't? – user110391 Apr 10 '22 at 11:59
  • [2/2] If that's the case, are all negations of statements with the antecedent quantifier being $\forall$, always of the form "$\neg \forall x (...) \iff \exists x \neg$ (...)"? Looking at Wuestenfux answer also, it seems so. @Vsotvep – user110391 Apr 10 '22 at 11:59
  • See De Morgan Laws – Mauro ALLEGRANZA Apr 10 '22 at 12:48
  • @user110391 I believe ryang's answer already answer the question, but to comment on $n_{\Bbb N}\neq n_{\Bbb R}$: usually we work in a model where the domain of discourse only contains the relevant objects. If we talk about divisibility of numbers, you only really need natural numbers, since divisibility is not a very useful concept for real numbers in general, hence I assumed that the universe consisted of only natural numbers. When talking from a set theoretic viewpoint, indeed $n_{\Bbb N}\neq n_{\Bbb R}$, but then $S^c$ is not well-defined (it's a proper class, and not a set). – Vsotvep Apr 10 '22 at 13:13

2 Answers2

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$$\neg(\forall x \in S, P(x)) \iff \neg\forall x \in S, P(x) \iff \exists x \in S, \neg P(x)\tag1$$

Only option $(1)$ is correct:

\begin{align}&\neg\,\forall x {\in}S \; P(x) \\\equiv{}&\neg\,\forall x\; \Big(x \in S \to P(x)\Big) \\\equiv{}&\exists x\; \neg\Big(x \in S \to P(x)\Big) \\\equiv{}&\exists x\; \neg\Big(x \notin S \:\lor\: P(x)\Big) \\\equiv{}&\exists x\; \Big(x \in S \:\land\: \neg P(x)\Big) \\\equiv{}&\exists x{\in}S\; \neg P(x). \end{align}

(Please refer to Quantifier notational shorthand.)

And so, $$\neg\,\exists x{\in}S\; Q(x)\equiv\ \forall x {\in}S \; \neg Q(x).$$

EDIT:

All even numbers are divisible by three: $$\forall n \in 2\Bbb N, n|3$$

Obviously not true. What is true?

Some even number is indivisible by three: $$\exists n {\in} 2\Bbb N\quad n\not|\,\:3.$$

This makes sense, since you would negate

  • Every person has a mother  (T)

not with (option 4)

  • Some non-person has a mother  (T)

but with (option 1)

  • Some person has no mother.  (F)

Is negation distributive?

No, negation is not straightforwardly "distributive", but it can be—using De Morgan's laws—if we first convert every conditional $\Big(A(x)\to B(x)\Big)$ to $\Big(\lnot A(x) \lor B(x)\Big).$

ryang
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Hint: Expand the restricted quantifiers as follows:

$\forall x (x\in S \Rightarrow P(x))$ and $\exists x(x\in S\wedge P(x))$.

Moreover, $\exists x$ corresponds to $\neg \forall x (\neg ...)$ and $\forall x$ corresponds to $\neg \exists x (\neg ...)$.

Tankut Beygu
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Wuestenfux
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    Your general forms confuse me a bit. Isn't it rather so that $\exists x$ corresponds to $\neg \forall x \neg(...)$ and likewise for the other quantifier? The way you wrote it, it appears as if the negation is placed on whatever is first in the sequent, yet isn't it rather place on the whole of the sequent? Perhaps your notation does mean the latter, and I'm just misinterpreting it? – user110391 Apr 10 '22 at 11:32
  • In case my formulation was a bit confusing; ryang's answer shows what I meant by my version of the general forms. – user110391 Apr 10 '22 at 12:29