The discrete math question (pg 47; #20;c) from Rosen's text is:
"Suppose that the domain of the propositional function P(x) consists of $-5, -3, -1, 1, 3, 5$. Express these statements without using quantifiers, instead only using negations, disjunctions, and conjunctions."
For instance:
c) $\forall x \left(\left(x \neq 1 \right)\right) \rightarrow P\left(x\right)$
So, in my initial thoughts on the answer, I imagined it to be:
$P(-5) \wedge P(-3) \wedge P(-1) \wedge \neg P(1) \wedge P(3) \wedge P(5)$
But now from doing some research the popular answer to this problem is omitting the restricted values. Perhaps "restricted values" isn't the proper way to phrase it. So now we have:
$P(-5) \wedge P(-3) \wedge P(-1) \wedge P(3) \wedge P(5)$
So insofar as I can see, this is the popular answer and it makes sense, too. The rationale to my understanding is $P\left(1\right)$ is no longer part of the set within $P\left(x\right)$, thus it no longer exists. Therefore, it does not make sense to negate it. As such it is left out of the answer. This correct?
I’m assuming that the expression is supposed to read $\forall x\big((x\ne 1)\to P(x)\big)$.
Remember that $(x\ne 1)\to P(x)$ is logically equivalent to $\neg(x\ne 1)\lor P(x)$, i.e., to $(x=1)\lor P(x)$. This is automatically true when $x=1$, whether $P(1)$ holds or not. For any other $x$ in the domain of $P$, however, the statement is true only if $P(x)$ is true. Thus, the original statement is equivalent to the conjunction
$$P(-5)\land P(-3)\land P(-1)\land P(3)\land P(5)\;;$$
there is no requirement that $P(1)$ be true or that it be false.