Suppose that the domain of the propositional function $P(x)$ consists of $-5,-3,-1,1,3,5$. Express $\forall x((x\ne1\rightarrow P(x))$ without using quantifiers, instead using only negations, disjunctions, or conjunctions. Select the choice or choices that correspond to a correct solution.
- $P(-5)\land P(-3)\land P(-1)\land P(3)\land P(5)$
- $P(-5)\lor P(-3)\lor P(3)\lor P(5)$
- $(x=1)\lor P(1)$
- $\neg P(-1)$
- $\neg P(1)$
- $(x=-5)\land(x=-3)\land(x=-1)\land(x=3)\land(x=1)$
I've managed to work out the following:
When all the elements of in the domain can be listed—say, $x_1,x_2,\ldots,x_n$—it follows that the universal quantification $\forall xP(x)$ is the same as the conjunction $P(x_1)\land P(x_2)\land\cdots\land P(x_n)$. Thus,
$\forall x((x\ne1\rightarrow P(x)) \\ \Rightarrow(-5\ne1\rightarrow P(-5))\land(-3\ne1\rightarrow P(-3))\land\cdots\land(5\ne1\rightarrow P(5)) \\ \Rightarrow(\neg(-5\ne1)\lor P(-5))\land(\neg(-3\ne1)\lor P(-3))\land\cdots\land(\neg(5\ne1)\lor P(5)) \\ \Rightarrow((-5=1)\lor P(-5))\land((-3=1)\lor P(-3))\land\cdots\land((5=1)\lor P(5)) \\ \Rightarrow(\mathbf{F}\lor P(-5))\land(\mathbf{F}\lor P(-1))\land(\mathbf{F}\lor P(-1))\land(\mathbf{T}\lor P(1))\land(\mathbf{F}\lor P(3))\land(\mathbf{F}\lor P(5)) \\ \Rightarrow P(-5)\land P(-3)\land P(-1)\land \mathbf{T}\land P(3)\land P(5) \\ \Rightarrow P(-5)\land P(-3)\land P(-1)\land P(3)\land P(5) $
This shows that choice 1 is correct.
Then choice 2 would have to be incorrect because it allows for any three of $P(-5),P(-3),P(3),P(5)$ to be false, which contradicts choice 1.
I would think choice 3 is correct because $1$ is in the domain.
I think choice 4 is incorrect because it contradicts choice 1.
Choice 5 I'm not sure about. Does choice 1 mean that $P(1)$ is false, or is it left open that it could be true or false?
An then choice 6 I have a similar uncertainty; does the exclusion of $x=5$ mean that $x=5$ is false, or is that left open?
Thank you.