Propositional Logic: How to negate one of the ... statement

Question

I am trying to negate a statement expressed as something like one of the members satisfy ... property.

My understanding is that one of statement is equivalent to the existential logic, so its negation is supposed to be the universal statement like all of them does not satisfy ... property.

Is my understanding correct? I feel it's right but not 100% so sure.

Thx.

Answer 1

Nice intuition! Here is a formal representation of what you're saying...

Let $P(x)$ be a the following statement: $x$ has such and such property.

$\exists x P(x)$ is the formal way of saying "there exists at least one $x$ such that $x$ has such and such property."

$\forall x P(x)$ is the formal way of saying "for every $x$, $x$ has such and such property."

Now,

$\neg \exists x P(x) \Leftrightarrow \forall x \neg P(x)$

Is the formal way expressing that these two statements are logically equivalent:

1. It is not the case that there exists at least one $x$ such that $x$ has such and such property.

2. For every $x$, it is not the case that $x$ has such and such property.

Answer 2

To add to the RyRy the Fly Guy answer, below is a proof of the tautology $\neg\exists x P(x) \leftrightarrow \forall x\neg P(x)$:

\begin{align} & \text{First we prove that $\neg \exists x P(x) \rightarrow \forall x \neg P(x)$:}\\ &(1)\quad \neg\exists x P(x) & \text{assumption}\\ &(2)\quad P(x) & \text{assumption}\\ &(3)\quad \exists P(x) & \text{E.G.}\\ &(4)\quad \exists P(x)\land\neg\exists P(x) & \text{conj. intro.}\\ &(5)\quad \bot & \text{$\bot$ intro. (4)}\\ &(6)\quad \neg P(x) & \text{R.A.A (2, 5)}\\ &(7)\quad \forall x\neg P(x) & \text{U.S.(1, 7)}\\ &(8)\quad \neg \exists x P(x) \rightarrow \forall x \neg P(x)&\text{C.P.}\\ \\ & \text{Now prove that $\forall x \neg P(x)\rightarrow\neg \exists x P(x)$:}\\ &(9)\quad \forall x \neg P(x) & \text{assumption}\\ &(10)\quad \exists x P(x) & \text{assumption}\\ &(11)\quad P(\alpha) & \text{E.S.}\\ &(12)\quad \neg P(\alpha) & \text{U.S.}\\ &(13)\quad \neg P(\alpha)\land P(\alpha) & \text{conj. intro}\\ &(14)\quad \bot & \text{$\bot$ intro. (13)}\\ &(15)\quad \neg\exists x P(x) & \text{R.A.A (10, 14)}\\ &(16)\quad \forall x \neg P(x)\rightarrow\neg \exists x P(x)&\text{C.P. (9, 15)}\\ \\ & \text{Conclude that $\neg\exists x P(x) \leftrightarrow \forall x\neg P(x)$}\\ \end{align}

E.G. stands for Existential Generalization, E.S. - Existential Specification, U.S. - Universal Specification, C.P. - Conditional Proof, R.A.A. - Reductio Ad Absurdum.