Existential quantifier looks fallacious

Question

I was reading about existential quantifier on Wikipedia whereby I came across this explanation for it's negation.

Generally, then, the negation of a propositional function's existential quantification is a universal quantification of that propositional function's negation; symbolically,

¬∃x∈X P(x) = ∀x∈X ¬P(x)

A common error is stating "all persons are not married" (i.e., "there exists no person who is married"), when "not all persons are married" (i.e., "there exists a person who is not married") is intended.

¬∃x∈X P(x) = ∀x∈X ¬P(x) ≠ ¬∀x∈X P(x) = ∃x∈X ¬P(x)

Now, let X = {all persons}, p = x is married and e = ∃x∈X, p is true

The above example "not all persons are married" seems logically equivalent to "some persons are married", i.e., "e". Here, the negation of e seems to imply e itself. But that doesn't make any sense. How can e be logically equivalent to the negation of e?

Maybe I'm missing something or maybe I am not able to think of quantifiers and logical expression in a formal or rigorous manner. What is the correct way to think logically (no pun intended)?

Answer 1

You've incorrectly negated the existential statement: "not all people are married" is equivalent to "some people are not married," not "some people are married." I suspect your confusion comes from two assumptions you're making which are unjustified, one about the term "some" in general and one about the specific nature of marriage. Specifically:

• That "some" means "some but not all."

• That there are some married people in the first place.

The first reflects a possible discrepancy between the natural-language meaning of "some" and its meaning as a translation of the precise quantifier "$\exists$." This is just something you have to move past (or avoid using the word "some" in this context - but recognize that others will use it). The second reflects the danger of implicitly allowing "outside knowledge" to creep into the logical analysis of a statement; this is something we always have to avoid doing.

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Actually there's a second issue re: "some" as a translation for "$\exists$," which doesn't play a role in your specific question but is still worth noting: namely, that the latter doesn't specify whether only one or more than one example exists. So both "some $X$ is $Y$" and "some $X$ are $Y$" may feel a bit off. Again, this is just something we have to learn to work with: that insofar as we use "some" as a translation for "$\exists$," there will be discrepancies between its usage in the context of logic and its natural-language connotations.

Answer 2

“Not all persons are married” is logically equivalent (in standard logic) to “there exists a person who is not married”. On the other hand, “Some persons are married” is logically equivalent to “there exists a person who is married”. So these two statements are not negations of one another. Indeed, as long as there are two people in the universe, one of whom is married and one of whom isn’t, then both statements are simultaneously true.

Answer 3

In ordinary language, some usually implies some not, otherwise it would be pointless to mention the fact. Nobody would say "some giants are tall".

That restriction does not exist in logics. Some can be all (but not none).

Answer 4

Rather than using the word "some ... are ..." for $\exists$, think about it with the language "there exists ... such that ...".

You see that:

$$\neg \exists x \in X: P(x) \iff \forall x \in X: \neg P (x)$$

can be expressed as:

"It is not the case that there exists an $x$ in $X$ that has $P$" means the same thing as "Every $x$ in $X$ does not have $P$."

That is:

"No $x$ have $P$" means the same as "All $x$ do not have $P$".

Now:

$$\neg \forall x \in X: P(x) \iff \exists x \in X: \neg P(x)$$

can be expressed as:

"It is not the case that all $x$ in $X$ have $P$" means the same as "There exists (at least one) $x$ in $X$ which does not have $P$"

That is:

"Not all $x$ have $P$" means the same as "At least one $x$ does not have $P$."

But if you say "It is not the case that all $x$ have $P$" is completely consistent with the statement "No $x$ have $P$"

If I were to say: "All people are over $4$ metres tall", you could instantly disprove it by pointing to yourself, and say, "No they are not all over $3$ metres tall, look at me, I'm under $3$ metres tall." (Probably).

But in fact "all people are not over $4$ metres tall."