How to interpret predicate logic if no context is given about possible values of variables?

Question

Is the following true?

$$\forall x \forall y(Rxy) \models \forall x (Rxx)$$

I'm wondering, because I' not sure if $x$ and $y$ come from the same 'set'.

If $$x \in X \space \text{and} \space y \in Y \space \text{and} \space X \cap Y = \emptyset $$ than $Rxx$ may not be true.

How should I interpret statements like this if no context is given about whether $x$ may be equal to $y$?

Answer 1

How to interpret predicate logic if no context is given about possible values of variables?

Of course, in order to interpret a formula we need a "context", but we are not "forced" to interpret it into set theory.

The "standard" semantics for first-order logic needs a domain where individual variables "range over" (e.g.the set $\mathbb N$ of natural numbers) and an interpretation for each predicate letter and each individual constant (and function letter, if any).

Thus, with your example :

$∀x∀y(Rxy) \vDash ∀x(Rxx)$

we have that the individual variables $x,y$ "range over" elements of the domain $\mathbb N$, i.e. natural numbers.

We have to assign an interpretation to the binary predicate letter $R$. We can use "$=$" ("equality") and see what happens, or $<$ ("less than") ...

We can try with the above intepretations, to check :

$∀x∀y(x = y)$ and $∀x(x = x)$

and :

$∀x∀y(x < y)$ and $∀x(x < x)$.

Of course, "checking" the truth of a sentence in an interpretation do not "establish" (in case of positive answer) its validity: only satisfiability, while a single counterexample it's enough to conclude in the negative.

For the validity we need a proof: at least an "informal" compelling argument; better a formal proof with some first-order logic proof systems.

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The way we can "informally" argument for the validity of this law is this.

Consider a domain $D$ whatever; because we have $\forall x \forall y R(x,y)$, if we pick an object $a \in D$ whatever, we have that $\forall y R(a,y)$ holds.

But this holds for all objects whatever in the domain $D$; thus, also for $a$, i.e. $R(a,a)$ holds.

But if $R(a,a)$ holds for an object $a \in D$ whatever, we may conclude that $\forall x R(x,x)$ holds.

This argument holds for a domain $D$ whatever. Thus it applies to every (non empty) domain.

But, having shown that, for every domain, if $\forall x \forall y R(x,y)$ holds, then $\forall x R(x,x)$ holds, we are licensed to conclude with :

$\forall x \forall y R(x,y) \vDash \forall x R(x,x)$.