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I've encountered this question, and I'm not sure if my interpretation is right because if it is, seems like there would be very trivial models (and there would no problem at all).

Ex 1: $\{\forall x \neg Rxx\}$. Here wouldn't be possible to have as the domain the empty set? (or because its empty, the condition $Rxx$ is true for every element in the set?). If not, seems like the domain can't be only one element, then an structure with the less cardinal possible could be $\{\{1,2\};c_1,c_2;\{<c_1,c_2>\}\}$?; and since the domain here is $\{1,2\}$, then $\{\{1,2\};c_1,c_2;\{<c_1,c_2>,<c_2,c_1>\}\}$ works as well?.

Ex 2: $\{\exists y \exists z Ryz, \forall x \forall y \forall z ((Rxy\wedge Ryz)\implies Rxz)\}.$

Here I have the same question as before, if it needs to be true, can the empty set as domain satisfy the conditions?, or it would be a valid solution if it weren't because of the $\exists y \exists z$ because of which I need at least two elements in the domain?, and because of $Rxy\wedge Ryz$ needs to be another one?. Then a possible structure could be $\{\{1,2,3\};c_1,c_2,c_3;\{<c_1,c_2>,<c_2,c_3>,<c_1,c_3>\}\}$ with three elements in the domain?.

Ex 3: $\{\forall x Rxx\}$

Here with an structure with one element in the domain would suffice, but again, is this the smallest?, am I allowed to have the empty set as domain?.

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The "standard" semantics for first-order logic does not allow for interpetations with empty domains.

But there other approachs : see Free Logic

Regarding Ex 1 : $∀x¬Rxx$,

if we stay with the usual approach (no empty domain), we still have an interpretation with a sinle element.

Consider the interpretation $\mathfrak A$ with domain $A = \{ 1 \}$ and $R^{\mathfrak A}$ is $ < $ ("less than").

We have $\lnot (1 < 1)$; thus : $\mathfrak A \vDash ∀x¬Rxx$.

The same for Ex 3 : $∀xRxx$.

In this case, we may interpret $R$ as $=$ (or $\le$), and again, from $1=1$ we may conclude with : $\mathfrak A \vDash ∀xRxx$.


Note

Regarding the above two examples, being them universally quantified formulas, they are true in the empty domain.


Regarding Ex 2 : $∃y∃z,∀x∀y∀z((Rxy \land Ryz) \rightarrow Rxz)$, we have "too many" quantifiers.

Rewriting it as :

$∀x∀y∀z((Rxy \land Ryz) \rightarrow Rxz)$

it still holds in a one-element domain with $R$ interpreted as $=$ (or $\le$).

If instead we rewrite it as :

$∃y∃z∀x((Rxy \land Ryz) \rightarrow Rxz)$

it is still true in the above interpretation (because there is no condition like $y \ne z$, imposing us that $y,z$ must be different).

In this case, being existentially quantified, it is nor more true in the empty domain.

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