Inequality with respect to transitivity

Question

Given a relation R, R is said to be transitive if aRb ∧ bRc, then aRc.

The unequal relation (≠) is not transitive, for instance a≠b ∧ b≠c, then a≠c is an invalid consequent of the antecedent (a≠b ∧ b≠c).

For instance, 1 ≠ 2 and 2 ≠ 1, then it must be the case that 1 ≠ 1; however, this isn't the case because the relation ≠ is not transitive.

Since the ≠ relation is symmetric, (i.e. a≠b → b≠a), is a chain of inequality thus transitive?

a≠b≠c≠d≠..≠z, notice how each predicate variable is not isolated but chained together such that the property of symmetry can now come into play, is this chain transitive?

A somewhat naive application: For some reason I want to say terms, a, b, c, and d are mutually exclusive/disjoint, or unique, I could simply say "Let a, b, c, and d but unique" but I could also say "a≠b ∧ a≠c ∧ a≠d" ... ... "d≠a ∧ d≠b ∧ d≠c" but that is four expressions to write out formally, and 16 different inequalities for the four terms I have to account for.

If I wanted to write an expression without using English to specify uniqueness between the terms could I write a≠b≠c≠d as a chain exploiting ≠'s symmetrical property as shorthand?

NOTE: specifically I am stating the expression (a≠b≠c≠d) should not be considered equal, a substitution instance of, or equivalent in form to (a≠b, b≠c, c≠d)

Answer 1

1. $\neq$ is a binary relation. $a\neq b \neq c$ has no natural meaning apart from a slightly more compact way of writing $a\neq b$ and also $b\neq c$.

2. If you choose to define a 3-ary relation $\neq:(a,b,c)$ to denote $a\neq b$ and $b\neq c$ and $a\neq c$, that is your business. It is well-defined, although I would not recommend writing it as $a\neq b\neq c$, due to confusion with the binary relation $\neq$.

3. The symmetry of the binary relation $\neq$ has nothing to do with the 3-ary relation $\neq:(a,b,c)$ as defined above.

Answer 2

A convenient way of writing this in one fell swoop without defining new ternary or whatnot relations (while avoiding the apparent ambiguity of chained notequal signs) would be to have your variables written as $x_1, x_2, x_3, x_4, \cdots$ instead of $a, b, c, d, \cdots$. In doing so you can say for your set of variables $\{x_i\}$ that

$$x_i \neq x_j ~~~~\forall i\neq j$$

In doing so, this allows for an arbitrary number of variables to be used and is still quite compact.