Consider binary relation operators
d b
q p
(with a direct correspondence by generalization of:
< >
≮ ≯
these are a subset of the above, thus they have mandatory rules as transitivity, which are not required is the generalized operators, whose FULL set of rules is described/directly deducible from the general statement below)
where $\forall x,\forall y,\ \ x\ \text d\ y \Leftrightarrow y\ \text b\ x\Leftrightarrow\lnot(x\ \text q\ y) \Leftrightarrow \lnot(y\ \text p\ x)$
There is a set of objects (which is the set of real numbers, for example, and real relation operators above whose generalization is made) such that transitivity applies; that is $x\ \text d\ y \land y\ \text d\ z\Leftrightarrow x\ \text d\ z\Leftrightarrow\lnot(x\ \text q\ z)$. For the same example, a sort of "inantitransitivity" also applies, that is, $x\ \text d\ y \land y\ \text d\ z\Leftrightarrow x\ \text p\ z\Leftrightarrow\lnot(x\ \text b\ z)$. Note: check the correspondence between the general operators and the actual operators that apply to this example
Is there a set of objects such that intransitivity applies; that is $x\ \text d\ y \land y\ \text d\ z\Leftrightarrow x\ \text q\ z\Leftrightarrow\lnot(x\ \text d\ z)$?
Is there a set of objects such that antitransitivity applies; that is $x\ \text d\ y \land y\ \text d\ z\Leftrightarrow x\ \text b\ z\Leftrightarrow\lnot(x\ \text p\ z)$?
Thanks for reading. Please do comment if you see something that doesn't make sense (or some bad nomenclature, that's the thing that is wrong the most amount of times when one learns on the internet; I'm not sure those words exist).
The use of equivalence instead of the more common implication in the definition of transitivity allows the following chain of reasoning: $a\operatorname{d} b \Rightarrow (\forall y)a\operatorname{d} y \Rightarrow (\forall y)(\forall x)x\operatorname{d} y$ (using the "back" arrow of the equivalence in both cases). In other words, if any two elements are comparable, all pairs compare... both ways; making the $d$-relation trivial. Did you really really want to use the equivalence?
In case you did, consider a small modification of the same argument applied to the "antitransitivity" case: $a\operatorname{d} b \Rightarrow (\forall x)b\operatorname{d} x\Rightarrow (\forall x)(\forall y)x\operatorname{d} y$ (again, the "back" arrow was used in both cases). Once again, any two elements being comparable imply all elements being comparable both ways (collapsing the relation to a trivial one).
Finally, the "intransitivity" turns out to be self-contradictory. Consider any element $a$ and the condition imposed by intransitivity for $(x,y,z)=(a,a,a)$: $(a\operatorname{d} a) \wedge (a \operatorname{d} a) \Leftrightarrow \neg(a \operatorname{d} a)$; a contradiction.