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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?

ryang
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user2277550
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    Who said it isn't a tautology? – Thomas Andrews Aug 10 '22 at 09:30
  • @ThomasAndrews Hi, this was asked in an Indian entrance exam. https://gateoverflow.in/8259/gate-cse-2015-set-2-question-55 The given answer was option C, which I understand. That meant that option D, which was the statement in this question, is not a tautology. – user2277550 Aug 10 '22 at 09:34

1 Answers1

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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.

ryang
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  • That last sentence would make sense if we didn't have the $\forall$ but if $P(X,Y)$ is "X likes Y" then $\forall x\forall y P(x,y)$ and $\forall y\forall x P(x,y)$ mean the same thing. – Thomas Andrews Aug 10 '22 at 15:30
  • In any event, I'd be curious why there is this distinction between tautology and validity. Why is this distinction made? It is not at all obvious to me. Looking at a first order statement, I am unlikely to confuse it for a propositional calculus statement. – Thomas Andrews Aug 10 '22 at 15:52
  • @ThomasAndrews 1. Your assertion (with the quantifiers) is correct, and my final sentence (without the quantifiers) is also correct, and there's absolutely no inconsistency between them. $\quad$ 2. When one needs to discuss FOL formulae in a PL setting? "$ψ$ is a validity but not a tautology" is a stronger hint than "$ψ$ is a tautology in both PL and FOL" for guessing that $ψ:=(∀x,x=x).$ I don't know, and I'm not advocating anything. – ryang Aug 10 '22 at 16:26
  • It is a stronger statement, but it doesn't seem a useful statement. I think your last line misses the point - the OP isn't really asking what the difference is between $P(x,y)$ and $P(y,x),$ but the difference in the context of the FOL statements. The actual distinction you give implies that it is important in this case, which it is not. Stick with the focused and correct distinction between validity and tautology. – Thomas Andrews Aug 10 '22 at 16:33
  • @ThomasAndrews Good idea; edited! – ryang Aug 11 '22 at 05:41