# interpret first-order formula

To the following formula. As I can give structure to be true?

R(x) $\longleftrightarrow$ $\forall{x}$ ¬R(x)

I tried to break it down but still can not understand how I can interpret this formula to be true.

for example if I take the predicate R (x) as true when x is prime, equivalence tells me that for all x, R (x) is false. Someone can help me.

(R(x) $\rightarrow{}$ $\forall{x}$ ¬R(x)) $\wedge$ ( $\forall{x}$ ¬R(x) $\rightarrow{}$ R(x))

I think that the only structure that can satisfy this formula is the empty one.

Consider the interpretation $I$ defined on the domain $D=\emptyset$.

We have to interpret the predicate $R$ with a subset $R^I$ of $D$; clearly : $R^I=\emptyset$.

For the RHS, we have that it is equivalent to :

$\lnot \exists x R(x)$.

The formula is true in $I$ because $R^I$ is empty; thus it is true that there are no $a \in R^I$.

Consider now the LHS; it is an open formula.

A "reasonable" semantic condition is that $R(x)$ is true in $I$ if for all $a, a \in R^I$ [i.e. $R$ holds for all objects in the domain].

But trivially, there are no $a$ such that $a \notin R^I$.

Thus we can conclude that also $R(x)$ is true in $I$.

In an empty domain both $\forall x (x=x)$ and $\forall x (x \ne x)$ holds [see this post].

Thus, if we agree on the semantics for open formulae, i.e. that $R(x)$ is true iff $\forall x R(x)$ is, we can consider a simlar formula, with $=$ in place of $R$ :

$x = x \leftrightarrow \forall x (x \ne x)$.