Suppose that the domain of the propositional function $P(x)$ consists of $−5, −3, −1, 1, 3,$ and $5$. Express the statement $∃(￢()) ∧ ∀(( < 0) → ())$ without using quantifiers, instead using only negations, disjunctions, and conjunctions. since the expression $∀(( < 0) → ())$ can be re-written as $$¬(x<0)∨P(x)\\ (x≥0)∨P(x)$$ by simplification $P(x)$
My solution below, however doesn't seem to be right: $$(¬(−) ∨ ¬(−) ∨ ¬(−) ∨ ¬() ∨ ¬() ∨ ¬()) ∧ ((−) ∧ (−) ∧ (−) ∧ (1) ∧ (3) ∧ (5) $$
On
I hear a few potential misunderstandings based on the title of this question compared to the full question.
First is that when we say that the domain of $P(x)$ is the odd numbers between $-6$ and $6$, we are trying to determine for which of those values $P(x)$ is true, and which ones are false. Phrased in a more intuitive human way, it's asking: "In the eyes of the property $P$ which values of $x$ feel right and true, and which ones are wrong and false"
This is often phrased in math circles as "for what values of $x$ does the property $P$ hold", which is short hand for "for what values of $x$ does $P$ hold true"
So in this case when we write $P(-5)$ we are saying (in the more human phrasing) that $P$ has determined that $-5$ is true. In math terms we would say that "$P$ holds for $-5$." And when we write $\neg P(-5)$ we are saying (again first in the more human way) that $P$ has determined that $-5$ is false. In math terms we would say "$P$ does not hold for $-5$.
We cannot have a situation where $P$ both holds and does not hold for a single value of $x$. $P$ can not determine that $-5$ is both true and false at the same time. $P(x) \land\neg P(x)$ is a contradiction.
In the answer you provided if we are to translate it into English words you are saying: The property $P$ holds for the entire domain (this is the second half of your answer). But there is also a number in the domain that doesn't hold (this is the first half). This implies that for at least one of the numbers: $-5, -3, -1, 1, 3, 5$ it is the case that $P$ thinks it's both true and false, which we just said can't happen.
Let's translate the statement in the question into English. First we have $\exists x(\neg P(x))$. This is saying that there exists some $x$ in your domain where the property $P$ does not hold.
The next part which is connected by a conjunction is $\forall x((x < 0) \to P(x))$ which is saying that "for all elements in your domain, it is the case that if $x$ is less than zero, then the property $P$ holds." In other words: "$P$ holds for all of the elements in your domain less than $0$"
As you figured out, you need to give an answer that only has $P(x)$, $\neg P(x)$, $\lor$, $\land$, and parentheses, for each specific number in the domain. However the problem at hand is writing something that is consistent with the statements "There exists an $x$ such that $P$ does not hold" and "$P$ holds for all $x$ less than $0$".
I hope that helps!
Your conjunctions are wrong. You are saying $P(x)$ is true for all $x$ in the domain, when the statement only asserts that it is so for all the negative values.
The declaration is that there exists an $x$ where $P(x)$ is false but for every $x$ less than $0$ we have $P(x)$ true. That is to say: $P(x)$ is certainly true for each $x\in\{-5,-3,-1\}$ and false for at least one of $x\in \{1,3,5\}$.
Say that.