What is the calculation law in proposition logical?

Question

How is $¬(∃x¬P(x))$ calculated to $∀xP(x)$ ?

In my understanding,
$¬(∀xP(x))$ can be calculated to $∃x¬P(x)$, and
$¬(∃xP(x))$ can be calculated to $∀xP(x)$ ?

Is my calculated result right? What's the exact way to do the logical calculations?

Answer 1

Check it intuitively : when we say $[\forall x P(x)]$ , we mean that $P(x)$ is true for all elements.
When we negate it , we are saying that there is some element for which $P(x)$ is not true , hence $[\lnot \forall x P(x)] \equiv [\exists x \lnot P(x)]$.

Equivalently , when we say $[\exists x \lnot P(X)]$ we are saying that there is some element for which $P(x)$ is not true , we are saying that it is not true that $P(x)$ is always true , hence $[\exists x \lnot P(X)] \equiv [\lnot \forall x P(x)]$.

(1) ... $\lnot (\forall x P(x))$ can be calculated to $\exists x \lnot P(x)$ : CORRECT
(2) ... $\lnot (\exists x P(x))$ can be calculated to $\forall x P(x)$ : NOT CORRECT , it should be $\forall x \lnot P(x)$

Check out references :

https://www.csm.ornl.gov/~sheldon/ds/sec1.6.html
The negation of a universal statement ("all are") is logically equivalent to an existential statement ("Some are not").
The negation of an existential statement ("some are") is logically equivalent to a universal statement ("all are not").

http://www.sci.brooklyn.cuny.edu/~mate/misc/logic.pdf
Page 5 : "¬(∀x) ≡ (∃x)¬" & "¬(∃x) ≡ (∀x)¬"

https://sites.math.washington.edu/~aloveles/Math300Winter2011/m300Quantifiers.pdf
"Negation Rules: When we negate a quantified statement, we negate all the quantifiers first, from left to right (keeping the same order), then we negative the statement."

More general references :

https://en.wikipedia.org/wiki/Universal_quantification
https://en.wikipedia.org/wiki/Existential_quantification
https://math.libretexts.org/Courses/SUNY_Schenectady_County_Community_College/Discrete_Structures/02%3A_Logical_Reasoning/2.04%3A_Quantifiers_and_Negations

Answer 2

Let's prove that $\sim (\exists x)P(x)$ equivalent to $(\forall x)(\sim P(x))$.

Suppose that there exist $P(x)$ such that $|\sim (\exists x)P(x)| = 0$ (1) and $|(\forall x)\sim P(x)| = 1$ (2). From (1): $|(\exists x)P(x)| = 1$ which means that for some $a$: $P(a) = 1$ or, equivalently, $\sim P(a) = 0$. But then $|(\forall x)\sim P(x)| = 0$ which contradicts (2).

The same thinking for other case: Suppose that there exist $P(x)$ such that $|\sim (\exists x)P(x)| = 1$ (3) and $|(\forall x)\sim P(x)| = 0$ (4). From (4): for some $b$ $\sim P(b) = 0$ or, equivalently, $P(b) = 1$. Then $|(\exists x) P(x)| = 1$ and it contradicts to (3).