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Computability and Logic by Boolos and Burgess says that formula $\Gamma_d$ in example 12.12

∀x∀y(∃u(u ≠ x ∧ u ≡ x) ∧ ∃v(v ≠ y ∧ v ≡ y)) → x ≡ y) 

supports models, whose domain is partitioned into equivalence classes of any size, e.g.

enter image description here

is possible. Yet, I do not understand how this is possible. Suppose my x is the first dot. Then, I can find u, which is the second dot. Similarly, let y be the first dot in the second equivalence class and another dot v is in its companion in the eqivalence class. The formula says that x and y must be in the same equivalence class. But they are obviously not!

Are they trying to cheat me?

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1 Answer 1

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They are not trying to cheat you. You need to read the text of Example 12.12, following the formula. It says:

A denumerable model of $\Gamma_d$ will consist of a denumerable set $X$ with an equivalence relation in which any two elements $a$ and $b$ that are not isolated, that is, that are such that each is equivalent to something other than itself, are equivalent to each other.

Thus $x$ and $y$ in your example must be equivalent in the model, and the picture is not an accurate representation of what $\Gamma_d$ captures.

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  • $\begingroup$ The the illustration is indeed inappropriate. My bad. Sorry for the blunder. $\endgroup$
    – Val
    May 12, 2014 at 19:27

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