= Solution
The requested examples are <binary relations> on <sets>, rather than <partially ordered sets>: a <partially ordered set> necessarily has all three properties. This corrects the wording of the source.<Calculus, linear algebra, discrete mathematics, logic and proof, and models of computation examination paper by the University of Oxford (2005)>[Printed p. 5, Q7(b), requesting a “partially ordered set” without the third axiom]{ref}
For an <antisymmetric relation> and <transitive relation> that is not a <reflexive relation>, take $S=\{0,1\}$ and $R=\{(0,1)\}$. No distinct pair relates both ways, so the <antisymmetric relation> condition holds. There are no composable pairs, so the <transitive relation> condition holds vacuously. Neither $(0,0)$ nor $(1,1)$ belongs to $R$, so the <reflexive relation> condition fails.
For a <reflexive relation> and <transitive relation> that is not an <antisymmetric relation>, take $S=\{0,1\}$ and $R=S\times S$. All diagonal pairs and all composite pairs belong to $R$, but $R(0,1)$ and $R(1,0)$ both hold with $0\ne1$.
For a <reflexive relation> and <antisymmetric relation> that is not a <transitive relation>, take
$$
S=\{0,1,2\},\qquad R=\{(0,0),(1,1),(2,2),(0,1),(1,2)\}.
$$
All diagonal pairs are present, and no off-diagonal pair has its reverse present. However $R(0,1)$ and $R(1,2)$ hold while $R(0,2)$ fails. Thus each pair of axioms can hold without the third.
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