[Math] Written as disjuctions, conjunctions and negations

discrete mathematicspropositional-calculus

With a domain from -2 to 2 I'm trying to write the following using disjunctions conjunctions and nagations. I'm not sure how correct I am and wanted to know if I did them correct? Could someone help with the last one I just cant figure out how to start that expression?

$∀ x\ P(x) = P(-2) ∧ P(-1) ∧ P(0) ∧ P(1) ∧ P(2)$

$\neg ∀x\ P(x) = (P(−2)∨P(−1)∨P(0)∨P(1)∨P(2))∧¬(P(−2)∧P(−1)∧P(0)∧P(1)∧P(2))
In English doesn't that mean: At least one element is true and one element is false?

$∃x\ \neg P(x) = $

Edit: The last two with the correct answer of "At least one is not true".

∃x¬P(x) = ¬P(-2) ∨¬P(-1) ∨¬P(0) ∨¬P(1) ∨¬P(2)

¬∀xP(x) = ¬(P(-2) ∧ P(-1) ∧ P(0) ∧ P(1) ∧ P(2))

Best Answer

The last two should have the same answer, since the second one, $\lnot\forall x.P(x)$ says that it is not true that $P(x)$ holds for every $x$, while the third one, $\exists c.\lnot P(x)$ says that there exists an $x$ for which $P(x)$ does not hold. These mean exactly the same thing. "Not every crow is black" is the same as "There is a crow that is not black."

But your answer for the second one is not correct.

Mouse over for hint:

$\lnot\forall x.P(x)$ says that it is not true that every element of the domain satisfies $P$. But $P(-2)\lor P(-1)\lor\ldots\lor P(2)$ says that $P(x)$ is true for -2, or for -1, or…

EDIT: Now your second one is correct, if you and I have the same idea for the part you indicated by "…", but it could be much simpler.

EDIT: Now your second one is incorrect again. $\lnot\forall x.P(x)$ says that $P(x)$ is not true for every $x$. It does $not$ say that $P(x)$ is true for any $x$; it might be false for all $x$.

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