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This will be a short question. Let $x$, $y$, $z$ be three elements from any set. Is the following:

$$x \ne y \ne z \tag{1}$$

Equivalent to:

$$x \ne y, ~ ~ y \ne z, ~ ~ z \ne x \tag{2}$$

Or simply:

$$x \ne y, ~ ~ y \ne z \tag{3}$$

Is it even well-defined? I know what $x = y = z$ implies but what about negation? I know I could use $x \ne y \ne z \ne x$ to "make sure" but I was interested in knowing what $(1)$ really means.

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This is more of a personal convention than an actual answer regarding what it commonly means (although I'm pretty sure my convention is other people's too).

Given a binary relation $\triangle$ on some set $X$ and $x_1,\ldots, x_n\in X$ for some $n\in \Bbb N$, I define $$x_1\triangle x_2\triangle \cdots\triangle x_n$$ as $$(x_1\triangle x_2)\wedge\cdots \wedge (x_{n-1}\triangle x_n)$$

Under this convention, $x\neq y\neq z$ should be read $x\neq y\, \wedge y\neq z$.

Note that $x=y=z$ is equivalent to $x=y\wedge y=z \wedge x=z$ only because the equality is transitive.

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the first two are not the same

take $x,y,z$ to be 1,2,1 respectively.

The first and third are equivalent.

If $x\neq y\neq z$ is used to mean (2), it is bad notation. it would be better just to say 'x,y,z distinct' to mean (2)

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We do write often e.g. $x \in y \in z$ as a familiar and acceptable shorthand for $x \in y \land y \in z$. That's evidence that we do allow this kind of notational collapse for conjoined relational propositions even when the relation isn't transitive. (We don't read $x \in y \in z$ as implying $x \in z$.)

Can we similarly write $x \ne y \ne z$ as shorthand for $x \ne y \land y \ne z$, again with no implication that $x \ne z$? You might predict, by parity, that we would: yet I certainly don't think this is common practice. Why not? What makes the difference? Dunno!

I think I have, however, seen the likes of $x \ne y \ne z \ne w $ used as local shorthand for the corresponding symmetric $x \ne y \land x \ne z \land x \ne w \land y \ne z \land y \ne w \land z \ne w$. But I wouldn't adopt such a shorthand without saying what you are doing!

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    $\begingroup$ A somewhat more common (in my experience) situation where are non-transitive relation is used in chains without implying the relation between non-adjacent terms is the normal subgroup relation $\triangleleft$, as used in the subnormal series $1 = A_0\triangleleft A_1\triangleleft \cdots \triangleleft A_n = G$. $\endgroup$ – Marc van Leeuwen Apr 14 '13 at 11:34

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