Why isn't $\forall x \forall y ~ P(x,y) \rightarrow \forall x \forall y~P(y,x)$ a tautology?

Question

From an Indian entrance exam:

Which one of the following well-formed formulae is a tautology?

$$\forall x \, \exists y \, R(x,y) \, \leftrightarrow \, \exists y \, \forall x \, R(x, y)\tag A$$ $$( \forall x \, [\exists y \, R(x,y) \, \rightarrow \, S(x, y)]) \, \rightarrow \, \forall x \, \exists y \, S(x, y)\tag B$$ $$[ \forall x \, \exists y \, \left( P(x,y) \, \rightarrow \, R(x, y) \right)] \, \leftrightarrow [ \forall x \, \exists y \left(\neg P(x, y) \, \lor R(x, y) \right)]\tag C$$ $$\forall x \, \forall y \, P(x,y) \, \rightarrow \, \forall x \, \forall y \, P(y, x)\tag D$$

The given answer was option C, which I understand.

This means that option D is not a tautology. But the bound variables $x,y$ cover the same universe, don't they? If so, why would the order of the terms in the function/predicate matter?

Answer 1

The given answer was $$∀x∃y( P(x,y)→R(x,y))↔ ∀x∃y (¬P(x,y)∨R(x,y))\tag C$$

Yes, option (C) is a tautology.

$$\color{green}{\forall x \forall y ~ P(x,y)} \rightarrow \color{red}{\forall x \forall y~P(y,x)} \tag D$$ why would the order of the terms in the function/predicate matter?

Yes, option (D) is a validity:$$\color{green}{\forall x \forall y ~P(x,y)} \equiv \forall y \forall x~P(x,y) \equiv \color{red}{\forall x \forall y~P(y,x)}.$$

Why isn't option (D) a tautology?

In propositional logic, ‘logical truth’ and ‘tautology’ are synonyms.

In predicate logic, ‘logical truth’ and ‘validity’ are synonyms.

Authors who never need to discuss and don't care about tautologies in the propositional-logic sense use the three terms interchangeably; in this sense, the formulae $∀x\;(x=x)$ and option (D) are “tautologies”. But since their truth-functional forms are $A$ and $(A\to B),$ respectively, a stricter usage does not consider them (propositional-logic) tautologies, merely validities.