Find a model for the given WFF

Question

Find a model for the given WFF:

$\exists xp(x) \rightarrow \forall xp(x)$

I'm interpreting this as saying "There exists an x in the function p(x) which implies For all X in p(x)?

So my solution says that all I have to do is just let Let the Domain D be the set of Natural Numbers, and let $p(x)$ be x equals x + 1...

But I don;t understand how that is a model for the given wff? What is that even implying/saying?

Answer 1

Let $\mathbb{N}$ be the set of natural numbers, and let $p(x)$ be the formula $x=x+1$. For any natural number $a$, $p(a)$ is false. So $\exists x p(x)$ is false in the natural numbers, and therefore for any sentence $\varphi$, the sentence $\exists xp(x)\to\varphi$ is true in the natural numbers. In particular, let $\varphi$ be the sentence $\forall xp(x)$.

Alternately, we could let $p(x)$ be the formula $x=x$. Then $\exists xp(x)\to \forall x p(x)$ is true in the natural numbers. For $\forall xp(x)$ is true, and therefore whatever sentence $\varphi$ we choose, $\varphi\to\forall xp(x)$ will be true.

Answer 2

The statement, $\;\exists x\,P(x) \; \to \; \forall x \,P(x)\;$ is "if there is something that satisfies $P$, then all things satisfy $P$".   (This is not a tautology; its truth depends on what $P$ is.)

An equivalent statement is: $\;\forall x\,\neg P(x) \;\vee\; \forall x\,P(x)\;$ "Either no thing satisfies $P$ or all things satisfy $P$".

So you must find an model where predicate $P$, in a domain of discussion, is such that it is either true for all things or false for all things.

---

All below use the set of natural/integer/rational/real/complex numbers as the domain (pick any one).

$$\exists x\, (x=x+1)\;\to\;\forall x\, (x=x+1)$$

Now the predicate $x=x+1$ is false for all $x$.   Because the antecedent is false, therefore the implication is considered true.   This is a viable model.

$$\exists x\, (x=x+1)\;\to\;\forall x\, (x=x+1)$$

Here the predicate $x=x$ is true for all things. &\nbsp; Because both the antecedent and consequent are true, therefore the implication is validated.   This too is a viable model.

$$\require{cancel} \color{red}{ \cancel{ \color{black}{ \exists x\, (x=5)\;\to\;\forall x\, (x=5) } } }$$

However, the predicate $x=5$ is true for one thing, but not for all things.   Because the antecedent is true but the consequent is false, therefore the implication does not hold.   So this is not a viable model.