Are $\forall x\forall y(p(x,y)\leftrightarrow p(y,x))$ and $\forall x\forall y p(x,y) \leftrightarrow \forall y\forall x p(x,y)$ tautologies?

Question

$$\forall x\forall y( p(x,y) \leftrightarrow p(y,x) )$$

I don't know if this formula is a tautology or not.

I think the order inside the predicates is not important and therefore, it's a tautology.

Thanks for your help!

What about this one ?

$$\forall x\forall y p(x,y) \leftrightarrow \forall y\forall x p(x,y).$$

The order in the all-quantifier isn't important too, so i think this one is a tautology too, right ?

Answer 1

$$\forall x\forall y( p(x,y) \leftrightarrow p(y,x) )$$

Consider the interpretation $\mathcal{I}(P) = \{\langle x, y \rangle: x \text{ loves }y\}$. Sadly enough, just because you love someone doesn't always mean they loves you back: We can easily think of a structure where John loves Peter (i.e. $\langle John, Peter \rangle \in \mathcal{I}(P)$) but Peter doesn't love John ($\langle Peter, John \rangle \not \in \mathcal{I}(P)$). Thus $P(x,y)$ does not imply $P(y,x)$. Since we found is at least one interpretation that makes the formula false, it is not tautological.

The interpretation of a predicate is a set of tuples. Tuples are ordered: $\langle a, b \rangle \neq \langle b, a\rangle$. Since the order of the objects in the interpretation of the predicate matters, so does the order in which we write the terms down in the expression.

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$$\forall x\forall y p(x,y) \leftrightarrow \forall y\forall x p(x,y).$$

The order in the all-quantifier isn't important

A bit sloppy, but yes: In general, $\forall x \forall y \phi$ is logically equvivalent to $\forall y \forall x \phi$; that is, we can commute quantifiers of the same type. So yes, this formula indeed is tautological.

Answer 2

Consider the domain $\{0,1\}$ with $p=\{(0,1)\}$.

For the second, it is valid, if that's what you mean; see this:

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