I am looking for a natural deduction proof for above question. I have formalized the argument in the following way:
$$ \forall x \neg Rxx, \ \forall x\forall y \forall z (Rxy\land Ryz \rightarrow Rxz) \vdash \forall x \forall y \forall z \neg(Rxy\land Ryz \rightarrow Rzx) $$ I may have made an error in the formalization, and if this is the case I would be very happy for anybody to point this out. I currently have problems with showing that a relation is not three-cycle when there are no elements a,b,c such that $$ \neg (Rab\land Rbc) $$ I have already managed to show this from the premises, with only $$ Rab \land Rbc \rightarrow Rca $$ as undischarged premiss.
Suppose that $R\subseteq S\times S$ is a relation on $S$ that is irreflexive and transitive.
Now if $\langle x,y\rangle,\langle y,z\rangle,\langle z,x\rangle\in R$ then the transitivity of $R$ will lead to the conclusion $\langle x,x\rangle\in R$ contradicting that $R$ is irreflexive.
Our conclusion is that no elements $x,y,z\in S$ exist such that $\langle x,y\rangle,\langle y,z\rangle,\langle z,x\rangle\in R$.
In words: $R$ does not contain any $3$-cycles if it is irreflexive and transitive.