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I am aware that the theory of equivalence relations with infinitely many classes, all of which infinite, has quantifier elimination, as can be seen from the answer to this question. However, does the theory which consists only of the axioms stating that $E$ is reflexive, transitive, and symmetric, have quantifier elimination (in a first-order logic with equality)? I'm inclined to say no, but I don't really know how even to begin the proof either way. Any hint would be appreciated.

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Nagase
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1 Answers1

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Let $X = \{a,b,c\}$ and let $E$ be the equivalence relation on $X$ with classes $\{a\}$ and $\{b,c\}$. Then $a$ and $b$ satisfy the same quantifier-free formulas in one free variable (the induced substructures on $\{a\}$ and $\{b\}$ are isomorphic). But $b$ satisfies the formula $\exists x\, [xEy\wedge x\neq y]$, while $a$ does not. So this formula is not equivalent to any quantifier-free formula.

Primo Petri
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Alex Kruckman
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  • Thanks, that's what I thought; what threw me off was that Chang & Keisler ask in an exercise to give "a decision procedure for this theory by the method of elimination of quantifiers" (exercise 1.5.9 of the 3rd edition). I wonder what they could have in mind? – Nagase May 07 '16 at 02:27
  • Doesn't every element of every model satisfy $\exists x ; xEy $? By reflexivity, that is. – hardmath May 07 '16 at 02:29
  • @hardmath what if we include in the conjunct $x \not = y$? – Nagase May 07 '16 at 02:33
  • @Nagase: I thought the Question was about a language (and theory) that only has the predicate $E$, not equality as well as $E$. – hardmath May 07 '16 at 02:38
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    @hardmath - well, equality is generally considered as a logical constant, which is why I didn't bother to mention it explicitly. Sorry if that was confusing. I edited the question to mention it explicitly. – Nagase May 07 '16 at 02:41
  • @hardmath Oh, oops, of course I mean $\exists x, (xEy) \land (x \neq y)$. As Nagese says, first-order logic contains $=$ by default. If you want an example without equality, use $\exists x, \lnot (xEy)$, which is true of any singleton in a structure with more than one class, but not true of any singleton in a structure with just one class. – Alex Kruckman May 07 '16 at 04:26